summary of trigonometric identities
TRANSCRIPT
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Summary of trigonometric identities
You have seen quite a few trigonometric identities in the past few pages. It is convenient to have a summary of them for reference. These identities mostly refer to one angle denoted t, but there are a few of them involving two angles, and for those, the other angle is denoted s..
More important identities
You don't have to know all the identities off the top of your head. But these you should.Defining relations for tangent, cotangent, secant, and cosecant in terms of sine and cosine.
tan t = sin t
cos t cot t =
1
tan t =
cos t
sin t
sec t = 1
cos t csc t =
1
sin tThe Pythagorean formula for sines and cosines.
sin2 t + cos2 t = 1
Identities expressing trig functions in terms of their complements
cos t = sin( /2 – t) sin t = cos( /2 – t)
cot t = tan( /2 – t) tan t = cot( /2 – t)
csc t = sec( /2 – t) sec t = csc( /2 – t)
Periodicity of trig functions. Sine, cosine, secant, and cosecant have period 2 while tangent and cotangent have period .
sin (t + 2 ) = sin t
cos (t + 2 ) = cos t
tan (t + ) = tan t
Identities for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions.
sin –t = –sin t
cos –t = cos t
tan –t = –tan t
Sum formulas for sine and cosine
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sin (s + t) = sin s cos t + cos s sin t
cos (s + t) = cos s cos t – sin s sin t
Double angle formulas for sine and cosine
sin 2t = 2 sin t cos t
cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t
Less important identities
You should know that there are these identities, but they are not as important as those mentioned above. They can all be derived from those above, but sometimes it takes a bit of work to do so.The Pythagorean formula for tangents and secants.
sec2 t = 1 + tan2 t
Identities expressing trig functions in terms of their supplements
sin( – t) = sin t
cos( – t) = –cos t
tan( – t) = –tan t
Difference formulas for sine and cosine
sin (s – t) = sin s cos t – cos s sin t
cos (s – t) = cos s cos t + sin s sin t
Sum, difference, and double angle formulas for tangent
tan (s + t) = tan s + tan t
1 – tan s tan t
tan (s – t) = tan s – tan t
1 + tan s tan t
tan 2t = 2 tan t
1 – tan2 tHalf-angle formulas
sin t/2 = ± ((1 – cos t) / 2)
cos t/2 = ± ((1 + cos t) / 2)
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tan t/2 = sin t
1 + cos t =
1 – cos t
sin t
Truly obscure identities
These are just here for perversity. Yes, of course, they have some applications, but they're usually narrow applications, and they could just as well be forgotten until, if ever, needed.Product-sum identities
sin s + sin t = 2 sin s + t
2 cos
s – t
2
sin s – sin t = 2 cos s + t
2 sin
s – t
2
cos s + cos t = 2 cos s + t
2 cos
s – t
2
cos s – cos t = –2 sin s + t
2 sin
s – t
2Product identities
sin s cos t = sin (s + t) + sin (s – t)
2
cos s cos t =
cos (s + t) + cos (s – t)
2
sin s sin t =
cos (s – t) – cos (s + t)
2
Aside: weirdly enough, these product identities were used before logarithms to perform multiplication. Here's how you could use the second one. If you want to multiply xtimes y, use a table to look up the angle s whose cosine is x and the angle t whose cosine is y. Look up the cosines of the sum s + t, and the difference s – t. Average those two cosines. You get the product xy! Three table look-ups, and computing a sum, a difference, and an average rather than one multiplication. Tycho Brahe (1546-1601), among others, used this algorithm known as prosthaphaeresis.
Triple angle formulas. You can easily reconstruct these from the addition and double angle formulas.
sin 3t = 3 sin t – 4 sin3 t
cos 3t = 4 cos3 t –3 cos t
tan 3t = 3 tan t – tan3t
1 – 3 tan2tMore half-angle formulas. (These are used in calculus for a particular kind of substitution in integrals sometimes called the Weierstrass t-substitution.)
sin t = 2 tan t/2
1 + tan2 t/2
cos t = 1 – tan2 t/2
1 + tan2 t/2
tan t = 2 tan t/2
1 – tan2 t/2
A Table of Exact Trig values
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that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
π24
7.5√6 – √3 + √2 – 2(√2 – 1)(√3 – √2)
82.5
11 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
π10
18
10 + 2
√5
=
5 +
√5
8 8
4cos2(18°) = [0; 1, 9, [2, 8]]
1 –
2√5
5
tan2(18°) = [0; 9, [2,8]]
72 2 π 5
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36 √5 1 = 3 √5
√5 – 2√5
54 3 π 10
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+4 4+
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
tan2(36°) = [0; 1, 1, [8, 2]]
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π4
45
2
4
=
1
√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
1 45 π2
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
3 π 10
54
10 – 2
√5
=
5 –
√5
8 8
4cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72 √5 –
1 = 3 √5
√5 + 2√5
18 π10
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4 4–8 8
cos(72°) = [0; 3, [4]]cos2(72°) = [0; 10, [2, 8] ]
tan2(72°) = [9; [2, 8]]
5 π 12
75
√6 –
√2
4 4
=
1 –
√3
2 4
=
4 –
√12
8 8cos2(75°) = [0; 14, [1, 12]]
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]15 π
12
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d + ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
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( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
1. enter the angle as a number in one box leaving the other empty2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60√3
230
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos( 1
Select a trig function:
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9°)= 2
2 + 2 + Φ
cos(18°)
=
1
2
2 +2 + φ
=
1
22 + Φ
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cos(27°)
=
1
2
2 + 2 – φ
cos(36°)
=
1
2
2 +2 – Φ
=
1
22 + φ
=
Φ
2
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cos(54°)
=
1
2
2 –2 – Φ
=
1
2
2 – φ
cos(63°)
=
1
2
2 – 2 – φ
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cos(72°)
=
1
2
2 –2 + φ
=
1
2
2 – Φ
=
φ
2
cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
Angle
cosinesine
Angle
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90√
2 – √4 = 0
2
0
75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
72√2 – Phi
2
18
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
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Here are two simple triangles which give us the formulae for the trig values of these three angles:-
This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)
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For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)
= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
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If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
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Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
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The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
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By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
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(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product?
3. This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now
available as a Dover book(1988). A Table of Exact Trig values
that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
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π24
7.5√6 – √3 + √2 – 2(√2 – 1)(√3 – √2)
82.5
11 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
π10
18
10 + 2
√5
=
5 +
√5
8 8
4cos2(18°) = [0; 1, 9, [2, 8]]
1 –
2√5
5
tan2(18°) = [0; 9, [2,8]]
72 2 π 5
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36
√5 +
1
4 4
=
3 +
√5
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
√5 – 2√5
tan2(36°) = [0; 1, 1, [8, 2]]54 3 π
10
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π 45 1 1 45 π
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4
2
4
=√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
2
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
3 π 10
54
10 – 2
√5
=
5 –
√5
8 8
4cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72
√5 –
1
4 4
=
3 –
√5
8 8cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
√5 + 2√5
tan2(72°) = [9; [2, 8]]
18 π10
5 π 12
75 √6 –
√2
4 4
= 1
–√3
= 4
–√12
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]
15 π12
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2 4 8 8cos2(75°) = [0; 14, [1, 12]]
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d + ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
Select a trig function:
![Page 23: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/23.jpg)
1. enter the angle as a number in one box leaving the other empty2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60√3
230
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos(9°)
=
1
2
2 + 2 + Φ
![Page 24: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/24.jpg)
cos(18°)
=
1
2
2 +2 + φ
=
1
2
2 + Φ
cos(27°)
=
1
2
2 + 2 – φ
![Page 25: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/25.jpg)
cos(36°)
=
1
2
2 +2 – Φ
=
1
2
2 + φ
=
Φ
2
cos(54°)
=
1
2
2 –2 – Φ
=
1
22 – φ
![Page 26: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/26.jpg)
cos(63°)
=
1
2
2 – 2 – φ
cos(72°)
=
1
2
2 –2 + φ
=
1
22 – Φ
=
φ
2
![Page 27: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/27.jpg)
cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
90√
2 – √4 = 0
2
0
Angle
cosinesine
Angle
72√2 – Phi
2
18
![Page 28: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/28.jpg)
75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
Here are two simple triangles which give us the formulae for the trig values of these three angles:-
![Page 29: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/29.jpg)
This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
![Page 30: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/30.jpg)
a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
![Page 31: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/31.jpg)
If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
![Page 32: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/32.jpg)
Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
![Page 33: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/33.jpg)
The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
![Page 34: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/34.jpg)
By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
![Page 35: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/35.jpg)
(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
A Table of Exact Trig values
that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
π 7.5 √6 – √3 + √2 – 2 82. 1
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24 (√2 – 1)(√3 – √2) 5 1 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
π10
18
10 + 2
√5
=
5 +
√5
8 8
4cos2(18°) = [0; 1, 9, [2, 8]]
1 –
2√5
5
tan2(18°) = [0; 9, [2,8]]
72 2 π 5
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36
√5 +
1
4 4
=
3 +
√5
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
√5 – 2√5
tan2(36°) = [0; 1, 1, [8, 2]]54 3 π
10
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π4
452
=
1 1 45 π2
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4√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
3 π 10
54
10 – 2
√5
=
5 –
√5
8 8
4cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72
√5 –
1
4 4
=
3 –
√5
8 8cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
√5 + 2√5
tan2(72°) = [9; [2, 8]]
18 π10
5 π 12
75√6
–
√2
4 4
=
1 –
√3
2 4
=
4 –
√12
8 8
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]
15 π12
![Page 38: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/38.jpg)
cos2(75°) = [0; 14, [1, 12]]
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d + ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
1. enter the angle as a number in one box leaving the other empty
Select a trig function:
![Page 39: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/39.jpg)
2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60√3
230
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos(9°)
=
1
2
2 + 2 + Φ
![Page 40: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/40.jpg)
cos(18°)
=
1
2
2 +2 + φ
=
1
2
2 + Φ
cos(27°)
=
1
2
2 + 2 – φ
![Page 41: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/41.jpg)
cos(36°)
=
1
2
2 +2 – Φ
=
1
2
2 + φ
=
Φ
2
cos(54°)
=
1
2
2 –2 – Φ
=
1
22 – φ
![Page 42: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/42.jpg)
cos(63°)
=
1
2
2 – 2 – φ
cos(72°)
=
1
2
2 –2 + φ
=
1
22 – Φ
=
φ
2
![Page 43: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/43.jpg)
cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
90√
2 – √4 = 0
2
0
Angle
cosinesine
Angle
72√2 – Phi
2
18
![Page 44: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/44.jpg)
75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
Here are two simple triangles which give us the formulae for the trig values of these three angles:-
![Page 45: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/45.jpg)
This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
![Page 46: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/46.jpg)
a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
![Page 47: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/47.jpg)
If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
![Page 48: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/48.jpg)
Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
![Page 49: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/49.jpg)
The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
![Page 50: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/50.jpg)
By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
![Page 51: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/51.jpg)
(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.
Contents of this PageThe icon means there is a Things to do investigation at the end of the section.The sections marked with have an online interactive calculator.
A Table of Exact Trig values Converter for trig functions of angles <0 or >90°
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Degress <-> Radians Converter Patterns
The Simple Square-Root pattern The √(2 ± Phi) pattern The √(2 ± √u) pattern
Proofs 30° 45° and 60° 36° and 54°, 18° and 72° 15° and 75°
o Ailles Rectangle Trig Formulae A diagram to relate many angles and Phi Other angles with exact trig expressions involving square-roots
Halving Superimposing Do we know all the angles?
A Table of Exact Trig values
that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
π24
7.5√6 – √3 + √2 – 2(√2 – 1)(√3 – √2)
82.5
11 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
π10
185
55
8 84cos2(18°) = [0; 1, 9, [2, 8]]
1 –
2√5
5
tan2(18°) = [0; 9, [2,8]]
72 2 π 5
![Page 53: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/53.jpg)
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36
√5 +
1
4 4
=
3 +
√5
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
√5 – 2√5
tan2(36°) = [0; 1, 1, [8, 2]]54 3 π
10
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π4
45
2
4
=
1
√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
1 45 π2
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
3 π 10
54
10 – 2
√5
=
5 –
√5
8 8
4cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
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π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72
√5 –
1
4 4
=
3 –
√5
8 8cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
√5 + 2√5
tan2(72°) = [9; [2, 8]]
18 π10
5 π 12
75
√6 –
√2
4 4
=
1 –
√3
2 4
=
4 –
√12
8 8cos2(75°) = [0; 14, [1, 12]]
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]15 π
12
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d
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+ ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
1. enter the angle as a number in one box leaving the other empty2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60 √3 30
Select a trig function:
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2
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos(9°)
=
1
2
2 + 2 + Φ
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cos(18°)
=
1
2
2 +2 + φ
=
1
2
2 + Φ
cos(27°)
=
1
2
2 + 2 – φ
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cos(36°)
=
1
2
2 +2 – Φ
=
1
2
2 + φ
=
Φ
2
cos(54°)
=
1
2
2 –2 – Φ
=
1
22 – φ
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cos(63°)
=
1
2
2 – 2 – φ
cos(72°)
=
1
2
2 –2 + φ
=
1
22 – Φ
=
φ
2
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cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
90√
2 – √4 = 0
2
0
Angle
cosinesine
Angle
72√2 – Phi
2
18
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75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
Here are two simple triangles which give us the formulae for the trig values of these three angles:-
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This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
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a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
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If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
![Page 65: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/65.jpg)
Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
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The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
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By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
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(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988).What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.
Contents of this PageThe icon means there is a Things to do investigation at the end of the section.The sections marked with have an online interactive calculator.
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A Table of Exact Trig values Converter for trig functions of angles <0 or >90°
Degress <-> Radians Converter Patterns
The Simple Square-Root pattern The √(2 ± Phi) pattern The √(2 ± √u) pattern
Proofs 30° 45° and 60° 36° and 54°, 18° and 72° 15° and 75°
o Ailles Rectangle Trig Formulae A diagram to relate many angles and Phi Other angles with exact trig expressions involving square-roots
Halving Superimposing Do we know all the angles?
A Table of Exact Trig values
that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
π24
7.5√6 – √3 + √2 – 2(√2 – 1)(√3 – √2)
82.5
11 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
π10
1810 +
= 5
+√5
1 –
2√572 2 π
5
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8 84
cos2(18°) = [0; 1, 9, [2, 8]]
5
tan2(18°) = [0; 9, [2,8]]
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36
√5 +
1
4 4
=
3 +
√5
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
√5 – 2√5
tan2(36°) = [0; 1, 1, [8, 2]]54 3 π
10
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π4
45
2
4
=
1
√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
1 45 π2
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
3 π 10
545
55
8 84cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
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π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72
√5 –
1
4 4
=
3 –
√5
8 8cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
√5 + 2√5
tan2(72°) = [9; [2, 8]]
18 π10
5 π 12
75
√6 –
√2
4 4
=
1 –
√3
2 4
=
4 –
√12
8 8cos2(75°) = [0; 14, [1, 12]]
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]15 π
12
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d
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+ ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
1. enter the angle as a number in one box leaving the other empty2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60 √3 30
Select a trig function:
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2
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos(9°)
=
1
2
2 + 2 + Φ
![Page 74: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/74.jpg)
cos(18°)
=
1
2
2 +2 + φ
=
1
2
2 + Φ
cos(27°)
=
1
2
2 + 2 – φ
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cos(36°)
=
1
2
2 +2 – Φ
=
1
2
2 + φ
=
Φ
2
cos(54°)
=
1
2
2 –2 – Φ
=
1
22 – φ
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cos(63°)
=
1
2
2 – 2 – φ
cos(72°)
=
1
2
2 –2 + φ
=
1
22 – Φ
=
φ
2
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cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
90√
2 – √4 = 0
2
0
Angle
cosinesine
Angle
72√2 – Phi
2
18
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75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
Here are two simple triangles which give us the formulae for the trig values of these three angles:-
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This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
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a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
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If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
![Page 82: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/82.jpg)
Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
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The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
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By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
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(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988).
If you are printing this page and the horizontal lines of fractions or the square-roots do not appear, make sure you have the "Print Images" and "Print Background" options checked when your browser's printing window appears.What angles have an exact expression for their sines, cosines and tangents? You might know that cos(60°)=1/2 and sin(60°)=√3/2 as well as tan(45°)=1, but are 30, 45 and 60 the only angles up to 90° with a formula for their trig values? No! There are lots more but not all angles have exact epxressions. Which angles do? What patterns are there in these expressions? This page shows expressions for many angles and even solves the complete problem of which angles do and which don't have exact trig expressions.
Contents of this Page
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The icon means there is a Things to do investigation at the end of the section.The sections marked with have an online interactive calculator.
A Table of Exact Trig values Converter for trig functions of angles <0 or >90°
Degress <-> Radians Converter Patterns
The Simple Square-Root pattern The √(2 ± Phi) pattern The √(2 ± √u) pattern
Proofs 30° 45° and 60° 36° and 54°, 18° and 72° 15° and 75°
o Ailles Rectangle Trig Formulae A diagram to relate many angles and Phi Other angles with exact trig expressions involving square-roots
Halving Superimposing Do we know all the angles?
A Table of Exact Trig values
that are expressible as simple terms involving square-roots.
acos(a)sin(b)
tan(a)cot(b)
b
radians
degrees
degrees
radians
0 0 1 0 90 π2
π24
7.5√6 – √3 + √2 – 2(√2 – 1)(√3 – √2)
82.5
11 π 24
π12
15
√6 +
√2
4 4
=
1 +
√3
2 4
=
4 +
√12
8 8cos2(15°) = [0; 1, 13, [1, 12]]
2 – √3tan(15°) = [0; 3, [1, 2]]
tan2(15°) = [0; 13, [1, 12]]75 5 π
12
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π10
18
10 + 2
√5
=
5 +
√5
8 8
4cos2(18°) = [0; 1, 9, [2, 8]]
1 –
2√5
5
tan2(18°) = [0; 9, [2,8]]
72 2 π 5
π8
22·5
2 + √2
2
=
4 +
√8
8 8cos2(22·5°) = [0; 1, 5, [1, 4]]
√2 – 1tan(22·5) = [0; [2]]
tan2(22·5) = [0; 5, [1,4]]
67·5
3 π 8
π6
30
3
4
=
√3
2
cos(30°) = [0; 1, 6, [2, 6]]cos2(30°) = [0; 1, 3]
√3
3
tan(30°) = [0; 1, [1,2]]tan2(30°) = [0; 3]
60 π3
π5
36
√5 +
1
4 4
=
3 +
√5
8 8cos(36°) = [0; 1, [4]]
cos2(36°) = [0; 1, 1, 1, [8, 2]]
√5 – 2√5
tan2(36°) = [0; 1, 1, [8, 2]]54 3 π
10
5 π 24
37.5
√6 + √3 – √2 – 2(√2 + 1)(√3 – √2)
52.5
7 π 24
π4
45
2
4
=
1
√2
cos(45°) = [0; 1, [2]]cos2(45°) = [0; 2]
1 45 π2
7 π 24
52.5
√6 – √3 – √2 + 2(√2 – 1)(√3 + √2)
52.5
5 π24
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3 π 10
54
10 – 2
√5
=
5 –
√5
8 8
4cos2(54°) = [0; 2, 1, [8, 2]]
1 +
2√5
5
tan2(54°) = [1; 1, [8, 2]]
36 π5
π3
60
1
4
=
1
2
cos(60°) = [0; 2]cos2(60°) = [0; 4]
√3tan(60°) = [1; [1,2]]
tan2(60°) = 330 π
6
3 π 8
67·5
2 – √2
2
=
4 –
√8
8 8cos2(67·5) = [0; 6, [1, 4]]
1 + √2tan(67·5) = [2; [2]]
tan2(67·5) = [5; [1,4]]
22·5
π8
2 π 5
72
√5 –
1
4 4
=
3 –
√5
8 8cos(72°) = [0; 3, [4]]
cos2(72°) = [0; 10, [2, 8] ]
√5 + 2√5
tan2(72°) = [9; [2, 8]]
18 π10
5 π 12
75
√6 –
√2
4 4
=
1 –
√3
2 4
=
4 –
√12
8 8cos2(75°) = [0; 14, [1, 12]]
2 + √3tan(75°) = [3; [1, 2]]
tan2(75°) = [13; [1,12]]15 π
12
11 π 24
82.5
√6 + √3 + √2 + 2(√2 + 1)(√3 + √2)
7.5 π24
π2
90 0 Infinity 0 0
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The values in the table are those angles of the form n° or n/2 or n/3 for a whole number n, between 0 and 90° whose sin or cosine is rational, or whose continued fraction is periodic or the square of the trig value has a periodic continued fraction.
Continued fraction [a; b,c,d,...] means a +
1
b +1
c +
1
d + ...
and the periodic continued fraction [a;b, c, d, e, d, e, d, e, d, e,...] is written as [a; b, c, [d, e]].
Trig functions of Angles <0 or >90°
To find the trig. values of all angles including those bigger than 90 degrees and negative angles:
1. select a trig function2. type the angle in the box and then3. click on the button
to find which angle in the range 0-90° has the same value:
( °)
Degrees-Radians ConverterTo convert between DEGREES and RADIANS:
1. enter the angle as a number in one box leaving the other empty2. then click the button to do the conversion
You can use Pi in the radians box and * for multiplication e.g. 3*Pi/2:
degrees radians
Patterns
Select a trig function:
![Page 90: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/90.jpg)
The Simple Square-Root pattern
Ernesto La Orden of Madrid pointed out the following neat way to connect and remember the easiest of the sines (cosines):
Angle
sinecosine
Angle
90√4
= 12
0
60√3
230
45√2
=
1
2 √245
30√1
=
1
2 260
0√0
= 02
90
The √(2 ± Phi) pattern
cos(9°)
=
1
2
2 + 2 + Φ
![Page 91: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/91.jpg)
cos(18°)
=
1
2
2 +2 + φ
=
1
2
2 + Φ
cos(27°)
=
1
2
2 + 2 – φ
![Page 92: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/92.jpg)
cos(36°)
=
1
2
2 +2 – Φ
=
1
2
2 + φ
=
Φ
2
cos(54°)
=
1
2
2 –2 – Φ
=
1
22 – φ
![Page 93: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/93.jpg)
cos(63°)
=
1
2
2 – 2 – φ
cos(72°)
=
1
2
2 –2 + φ
=
1
22 – Φ
=
φ
2
![Page 94: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/94.jpg)
cos(81°)
=
1
2
2 – 2 + Φ
This pattern uses the identitiesphi = φ = √2 – Φ and Phi = Φ = √2 + φ
together with the half-angle formula for cos(A/2) (see below) starting from cos(36)=Phi/2 and cos(72)=phi/2. The pattern continues with the cosines of 4.5°, 13.5°, etc.
The √(2 ± √u) pattern
Ernesto La Orden also put many angles into this pattern:
Angle
cosinesine
Angle
90√
2 – √4 = 0
2
0
Angle
cosinesine
Angle
72√2 – Phi
2
18
![Page 95: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/95.jpg)
75√
2 – √3
2
15
67.5
√2 – √2
2
22.5
60√2 – √1
=
1
2 2
30
45√2 – √0
=
1
2 √2
45
30√2 + √1
=
√3
2 2
60
22.5
√2 + √2
2
67.5
15√
2 + √3
2
75
0√
2 + √4 = 1
2
90
54√2 – phi
2
36
36√2 + phi
2
54
18√2 + Phi
2
72
The table on the right has values of u that are Phi2 = 2.618033.. and phi2 = 0.381966..
Proofs
30° 45° and 60°
Here are two simple triangles which give us the formulae for the trig values of these three angles:-
![Page 96: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/96.jpg)
This triangle is just a square cut along a diagonal. If the sides are of length 1, the diagonal is length √2. This gives the sin, cos and tan of 45°.
Here is an equilateral triangle where all sides and all angles are equal (to 60°). If the sides are of length 2, then when we cut it in half as shown, the two triangles have 60°, 30° and 90° angles with a side of length 1 and a hypotenuse of length 2. The other side is therefore of length √3. So we can read off the sin cos and tan of both 30° and 60°.
36° and 54°, 18° and 72°
For 36° and 72° we need some further work based on the geometry of a regular pentagon which has angles of 36° and 72°. If the sides of the pentagon are of length 1, the diagonals are of the golden section number in length Phi where:
Phi = = 1.618033988.. =
1 + √5 = 1 +
1
2 Phi
The upper triangle with angles 72°, 72° and 36° and sides of lengths 1, Phi and Phi shows the trig values for 18° and 72°.
The lower triangle with angles of 36°, 36° and 108° and sides of lengths 1, 1 and Phi shows the trig values of 36° and 54°.
15° and 75°
If we take the triangle on the left, we can calculate the length of the third side using the Cosine Formula. If, in a triangle with sides a, b and c, we know both sides b and c and also the angle A between sides b and c then we can compute the length of third side, a, as follows:
a2 = b2 + c2 – 2 b c cos(A)For our triangle on the left, the known sides are b=2 and c=2 and the angle between them is A=30°. The length of the third side, the base a, is therefore:
![Page 97: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/97.jpg)
a2 = 22 + 22 – 2 x 2 x 2 x cos(30°)= 8 – 4 √3= 2 (4 – 2 √3)
But (√3 – 1)2 = 3 + 1 – 2 √3 = 4 – 2 √3 and soa2 = 2 (√3 – 1)2
Taking the square-root:a= √2 (√3 – 1) which we can also write as
= 2 (√3 – 1) / √2Using this expressions for a, we can expand the triangle by a factor of √2, to get rid of the denominator. Finally, we put in a line from the top of the triangle to the centre of the base a to make two right-angled triangles. This will halve the side a and cut the triangle into two and gets rid of the factor 2 also. We then arrive at the triangle on the right which shows the sines and cosines of 75° and 15°:
Ailles Rectangle
An alternative (easier) method for sine and cosine of 15° and 75° is found in Ailles Rectangle (named after an Ontario high school teacher). It is easy to remember because it is two (green) 45° right-angled triangles stuck onto the sides of a (white) 30-60-90 triangle and the rectangle completed with a (yellow) 15-75-90 triangle on the hypotenuse of the 30-60-90 triangle as shown here.The 30-60-90 sides are "as usual", namely 1, 2 and √3. From the two 45-45-90 triangles, it is quite easy to see that x is √3/√2 and y is 1/√2 from which we can read off the sines and cosines of 15° and 75°.
Trig Formulae
Many symmetries and patterns are apparent in the table. They reflect some underlying identities such as:
sin(x) = a / hcos(x) = b /
htan(x) = a / bcot(x) = b / a
sin(x) = cos(90° – x) sin2(x) + cos2(x) = 1tan(x) =
sin(x)
cos(x)
cot(x) =
1 =
cos(x)
tan(x) sin(x)tan2(x) + 1 =
1
cos2(x)
cot2(x) + 1 =
1
sin2(x)
If we know the value of a trig function on two angles A and B, we can determine the trig function values of their sum and difference using the following identities:
sin( A + B ) = sin(A)cos(B) + cos(A)sin(B)
tan(A + B) =tan(A) + tan(B)
1 – tan(A) tan(B)sin( A – B ) = sin(A)cos(B) – cos(A)sin(B)
cos( A + B ) = cos(A)cos(B) – sin(A)sin(B)tan(A – B) =
tan(A) – tan(B)
1 + tan(A) tan(B)cos( A – B ) = cos(A)cos(B) + sin(A)sin(B)
![Page 98: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/98.jpg)
If the two angles are the same (i.e. A=B) we get the sines and cosines of double the angle. Rearranging those formulae gives the formula for the sin or cosine of half an angle:
sin( 2A ) = 2 sin(A) cos(A)cos( 2A ) = 1 – 2 sin2(A)cos( 2A ) = cos2(A) – sin2(A)cos( 2A ) = 2 cos2(A) – 1
sin
A
=
1 – cos(A)
22
cos
A
=
1 + cos(A)
22
tan( 2A ) =
2 tan(A)
1 – tan2(A)
tan
A
2
=
sin(A)
1 + cos(A)
=
1 – cos(A)
A diagram to relate many angles and Phi
![Page 99: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/99.jpg)
Robert Gray's page on Coordinates for many regular solids has an amazing diagram at the bottom which relates Phi to the angles of 18°, 30°, 36°, 45°, 54°, 60° and 72° according to their 3D coordinates in the solids.
Each of those angles is measured from the top most point of the circle when a vertical line is turned through that angle. Each line from the base point meets the circle at a point whose a height is 1 (72°), 1+Phi (60°), 2+Phi (54°), 2+2 Phi (45°), 2+3 Phi (36°), 3+3 Phi (30°) or 3+4 Phi (18°).
Do look at his pages for more fascinating information on 120 3D solids, of which we will also explore the most symmetrical 5 on our next
Things to do
1. Suppose the origin of the circle is the lowest point and its radius is 2 + 2 Phi. Find the equation of the circle.
2. Use your answer to the previous question to find the coordinates of each of the points on the circle with the angles shown.
3. Compute the lengths of each of the red lines from the lowest point to the points shown on the circle.
4.
From any two points A and B on a circle, the angle AOB at the centre of a circle, O, is twice the angle at any point on the circumference in the same sector.
In the diagram,all the red angles at the circumference are equal;the red angles are twice the blue angle AOB at the centre;
the red angles are to a point in the same sector of the circle as is the centre of the circle so they cannot be in the grey sector.
Use the above theorem to find three points on the circle ABOVE this Things To Do section where a line from the centre makes an angle with the vertical ofi. 2×18=36°
ii. 2×30=60°
iii. 2×36=72°
Other angles with exact trig expressions involving square-roots
Are there other angles with a simple exact expression for their cosine or sine?Well it all depends upon what you mean by simple!
Carl Friedrich GAUSS (177 - 1855) looked at a similar problem which answers this question. He investigated if there was a method of constructing a regular polygon of n sides using only a pair of compasses (to draw circles) and a straight-edge (a ruler with no markings). We know we can construct a regular polygon for all of the values of n=3, 4, 5, 6, 8 and 10.
Halving
There is a simple geometrical way to use compasses to divide an angle into two (angle bisection). So all the angles in a regular n-gon can be split into two to make a regular 2n-gon. We can repeat the process to get a 4n-gon, 8n-gon and in general a 2kn-gon for any k once we have a method of constructing a regular n-gon.
![Page 100: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/100.jpg)
The Trig Formula section above contains a formula for the cosine of half an angle in terms of the cosine of the (whole) angle:
cos
A
2
=
1 + cos(A)
2
=
2 + 2cos(A)
2As Mitch Wyatt pointed out to me, since we know that cos(90°) is 0 and 90° is /2 radians, we can use it to find the cosine of half that angle (45° or /4 radians) and then halve that angle again and so on. Each time we introduce another square root so we get a cascading or nested sequence of square roots:
cos
4
=
√2
2
cos
8
=
√2 +
√2
2
cos
16
=
√
2 +
√2 +
√2
2
cos
32
=
√
2 +
√
2 +
√2 +
√2
2However, this page is about sines and cosines which have simpler expressions, so we will not expand on this except to say that it shows how we can always find an exact expression for the sine (or cosine) of 1/2, 1/4, 1/8, ..., 1/2
n, ... of any angle for which we have an exact sine (or cosine) expression.
Superimposing
If we construct a regular triangle (3 sides) and with the same circle centre, construct three regular pentagons (5 sides) with each having one vertex in common with the triangle, we will have the 15 vertices of a regular 15-gon. This is shown on the right with the 3 pentagons in blue on the same circle, each having a vertex in common with the red triangle and the regular 15-gon appears in yellow.
![Page 101: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/101.jpg)
By superimposing two regular polygons like this, we can construct a regular P×Q-gon (if P and Q have no factors in common otherwise more than one vertex of each will coincide).
Do we know all the angles?
All this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have
![Page 102: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/102.jpg)
(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
Tom Ace pointed out that there is more about this in chapter 15 of Oystein Ore's Number Theory and Its History from 1948 and now available as a Dover book(1988).
If you are printing this page and the horizontal lines of fractions or the square-roots do not appear, make sure you have the "Print Images" and "Print Background" options checked when your browser's printing window appears.ll this was known in Euclid's time, around the year 300 BC. So what about 7ths and 9ths? Is it possible to find sines and cosines of all the multiples of 1/7 and 1/9 of a turn in exact terms (using square roots)? What about 11ths and 12ths etc.?
In the next 2000 years no one found an exact geometric method for 7-gons or 9-gons but also no one had proved it was impossible to construct such regular polygons.
![Page 103: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/103.jpg)
Then C F Gauss completely solved the problem while he was a student at Göttingen between 1795 and 1798. Gauss found the conditions on n and its the prime factors to solve two equivalent problems:
drawing a regular n-sided polygon using only a straight edge and compass and expressing the cos and sin of 360/n° using only square roots.
If we factor n as 2ap1bp2
c..., i.e. a, b, c, ... are the powers of n's prime factors: 2, p1, p2, ... (the prime's power is 0 if it is not a factor of n) then both of the problems are solvable when
b,c,... and all the powers except a, the power of 2, must be 1, and the primes>2 that are factors of n (that is p1, p2, ...) must be of the form 22k+1 for some number k.
Both problems are solvable for these values of n and only for these values.
Prime numbers of the form 22k+1 are called Fermat primes. The series of numbers of the form 22k+1 begins
220 + 1 = 3, 221 + 1 = 5, 222 + 1 = 17, 223 + 1 = 257, 224 + 1 = 65537, ...However not every number of the form 22k + 1 is prime -- and it is only the prime ones that we must have as factors of n. The next one, 225 + 1 is 4294967297 and has a factor of 641 so it is not prime. In fact, we do not know if there are any more primes of this form except the first 5 listed above.
Such numbers, n, of the form Gauss describes are as follows, one per row, each a product of some of the Fermat primes (but each prime at most once) followed by its multiples of two. For any number in the table, its double is also in the table:
24 8 16 32 ...36 12 24 48 ...510 20 40 80 ...
3×5=1530 60 120 240 ...1734 68 136 ...
3×17=51102 204 408 ...5×17=75150 300 600 ...
3×5×17=225 450 900 1800 ...257514 1028 2056 ...
When put in order, we have(1), 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, ...
which is Sloane's A003401.Here, for instance, is the cosine of an angle involving 17ths:
cos
8 =
√17 – 1 + S –
√68 + 12√17 + 2(√17 – 1)S – 16 T
17 16
![Page 104: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/104.jpg)
where S =
√34 – 2√17
and T =
√34 + 2√17
Things to do
1. From the five 'starting values' above: 3, 5, 17, 257 and 65537 we can multiply these and also double any number to get a constructible polygon or an angle with a sine which involves nothing more than square-roots. Don't forget that we can double 2 to get 4 (a square), 8 (octagon), 16, etc too but we cannot use any of the 5 odd primes more than once in any product (so 3x3=9 and 5x5=25 are not in the list but 3x5x7 is). A complete list will involve these five numbers of course together with their products 3x5=15, 3x17=51, ... and you can double any of these any number of times: 2x3=6, 2x15=30 as well as 2x6=12, 2x12=24,... .
a. What are the first 12 values in the list that starts 3, 4, 5, ...?
b. Check that there are 24 values (excluding 1 and 2) less than 100.
c. Is 100 in the list?
2. From the five known values: 3, 5, 17, 257 and 65537, there are a finite number of odd numbers n for which sin(360/n°) can be written with square-roots alone i.e. products of these 5 numbers where no number can be used more than once. How many odd numbers (excluding 1) can you make using these 5 no more than once in each product? This is the total number of known polygons we can construct with ruler and compass or which have a sine (cosine) formed from nothing more than square-roots. With thanks to Richard Duffy for suggesting this puzzle
3. Pythagorean Identities
4.5.
6. Addition Formulas
7.8.
9. Subtraction formulas
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10.11.
12.Double Angle Formulas
13.14.
15.Half-Angle Formulas
16.17.
18.Product Formulas
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19.20.
21.Factoring Formulas
22.23.The following two formulas are of only limited use:
24.
Trigonometric Addition Formulas
Angle addition formulas express trigonometric functions of sums of angles in terms of functions of and . The fundamental formulas of angle addition in trigonometry are given by
(1
)
![Page 107: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/107.jpg)
(2
)
(3
)
(4
)
(5
)
(6
)
The first four of these are known as the prosthaphaeresis formulas, or sometimes as Simpson's formulas.
The sine and cosine angle addition identities can be compactly summarized by the matrix equation
(7
)
These formulas can be simply derived using complex exponentials and the Euler formula as follows.
(8)
(9)
(10)
(11)
Equating real and imaginary parts then gives (1) and (3), and (2) and (4) follow immediately by substituting for .
Taking the ratio of (1) and (3) gives the tangent angle addition formula
(12)
(13)
(14)
(15)
The double-angle formulas are
(16)
(17)
(18)
(19)
![Page 108: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/108.jpg)
(20)
Multiple-angle formulas are given by
(21)
(22)
and can also be written using the recurrence relations
(23)
(24)
(25)
The angle addition formulas can also be derived purely algebraically without the use of complex numbers. Consider the small right triangle in the figure above, which gives
(26)
(27)
Now, the usual trigonometric definitions applied to the large right triangle give
(28)
(29)
(30)
(31)
Solving these two equations simultaneously for the variables and then immediately gives
![Page 109: Summary of Trigonometric Identities](https://reader030.vdocuments.net/reader030/viewer/2022013102/54752405b4af9f7f5f8b45f4/html5/thumbnails/109.jpg)
(32)
(33)
These can be put into the familiar forms with the aid of the trigonometric identities
(34)
and
(35)
(36)
(37)
(38)
which can be verified by direct multiplication. Plugging (◇) into (◇) and (38) into (◇) then gives
(39)
(40)
as before.
A similar proof due to Smiley and Smiley uses the left figure above to obtain
(41)
from which it follows that
(42)
Similarly, from the right figure,
(43)
so
(44)
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Similar diagrams can be used to prove the angle subtraction formulas (Smiley 1999, Smiley and Smiley). In the figure at left,
(45)
(46)
(47)
giving
(48)
Similarly, in the figure at right,
(49)
(50)
(51)
giving
(52)
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A more complex diagram can be used to obtain a proof from the identity (Ren 1999). In the above figure, let . Then
(53)
An interesting identity relating the sum and difference tangent formulas is given by
(54)
(55)