theory of trigonometry
TRANSCRIPT
THEORY OF TRIGONOMETRY
1. INVESTIGATING TRIGONOMETRY
1.1 Six trigonometry ratios
If one angle of a triangle is 90 degrees and one of the other angles is known, the
third is thereby fixed, because the three angles of any triangle add up to 180 degrees.
The two acute angles therefore add up to 90 degrees: they are complementary angles.
The shape of a right triangle is completely determined, up to similarity, by the angles.
This means that once one of the other angles is known, the ratios of the various sides
are always the same regardless of the overall size of the triangle. The-ratios are defined
as follows:
Table 19-1.-Trigonometric ratios.
Figure 19-1.-Relationship of sides and angles in a right triangle. (A) Names of the sides;
(B) symbols used to designate the sides.
The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the
longest side of the triangle, and one of the two sides adjacent to angle A. The adjacent
leg is the other side that is adjacent to angle A. The opposite side is the side that is
opposite to angle A.
The reciprocals of these functions are named the cosecant (csc or cosec),
secant (sec) and cotangent (cot), respectively. The inverse functions are called the
arcsine, arccosine, and arctangent, respectively. There are arithmetic relations
between these functions, which are known as trigonometric identities. The secant,
cosecant, and cotangent are calculated, when needed, by using their relationships to
the three principal ratios. These relationships are as follows:
Negative Angles
We will just state:
sin (−θ) = −sin θ
cos (−θ) = cos θ
tan (−θ) = −tan θ
1.2 SOLUTION OF TRIANGLE
Sine Rule
The sine rule states that:
The triangle does not have to have a right angle in it for this rule to work.
The sine rule states that:
This can also be interpreted as three equations:
sin B
b =
sin C
c,
sin A
a =
sin C
c, and
sin A
a =
sin B
B
Since the three versions differ only in the labeling of the triangle, it is enough to verify
one just one of them, so we'll just consider the version stated first.
An explanation of the law of sine is fairly easy to follow, but in some cases we'll have to
consider sine of obtuse angles.
Example
We might want to find out what the angle opposite the side of length 20 is.
We can use the sine rule to solve for this angle, as follows:
Let a = the side of length 50
Let A = the angle opposite ‘a’ (100°)
Let b = the side of length 20
Let B = the angle opposite ‘b’
Inverting both sides gives:
Multiplying both sides by 20 gives:
Ambigous case
By definition, the word ambiguous means open to two or more interpretations.
Such is the case for certain solutions when working with the Law of Sines.
• If you are given two angles and one side (ASA or AAS),
the Law of Sines will nicely provide you with ONE solution
for a missing side.
• Unfortunately, the Law of Sines has a problem dealing with SSA.
If you are given two sides and one angle (where you
must find an angle), the Law of Sines could possibly provide you
with one or more solutions, or even no solution.
Before we investigate this situation, there are a few facts we need to remember.
Facts we need to remember:
1. In a triangle, the sum of the interior angles is
180º.
2. No triangles can have two obtuse angles.
3. The sine function has a range of
.
4. If the = positive decimal < 1, the can
lie in the first quadrant (acute <) or in the second
quadrant (obtuse <).
Let's look at some cases. In each example, decide whether the given information points
to the existence of one triangle, two triangles or no triangles.
Example 1: In , a = 20, c = 16, and m<A = 30º. How many distinct triangles can
be drawn given these measurements?
Use the Law of Sines:
C = sin-1 (0.4) = 24º (to the nearest degree) - in Quadrant I.
Sine is also positive in Quadrant II. If we use the reference angle 24º in Quadrant II,
the angle C is 156º.
But, with m<A = 30º and m<C = 156º the sum of the angles would exceed 180º.
Not possible!!!!
Therefore, m<C = 24º, m<A = 30º, and m<B = 126º and only ONE triangle is possible.
Example 2: In , a = 10, b = 16, and m<A = 30º. How many distinct triangles can
be drawn given these measurements?
Use the Law of Sines:
B = sin-1(.8) = 53.13010 = 53º.
Angles could be 30º, 53º, and 97º : sum 180º
The angle from Quadrant II could create angles 30º, 127º, and 23º : sum 180º
TWO triangles possible.
Cosine rule
c2 = a2 + b2 – 2ab cos C
There are two other versions of the law of cosines,
a2 = b2 + c2 – 2bc cos A
and
b2 = a2 + c2 – 2ac cos B.
Since the three versions differ only in the labeling of the triangle, it is enough to verify
one just one of them. We'll consider the version stated first.
Example :
So for this triangle…
You are given angles A and C and side c,
so use
B=180-(60+45)=75 and now
The cosine rule is also used when the three sides are given.
Using these methods, any triangle can be ‘solved’. That is all of the angles and sides of
a triangle can be worked out, if enough information is given to start with.
.
Area of Triangle
It is well known that to work out the
area of a rectangle you multiply the
length by the width.
Almost as well known is the formula
for the area of a triangle.
That is, the area of a triangle is half of
the area that the rectangle that the
triangle fits into.
But this formula requires the length of the base and the height of the triangle. There is
another formula working out the area of a triangle uses the lengths of two sides and the
angle between them.
As CB = b, and letting the area of
triangle ABC represented by T, we
have
T=12×base ×height
12×b×h ……… equation 1
From ∆ BNA ,
sin B=¿ hc¿
h=c sinB
substitute h=c sinB into equation 1,
T=12×b×c sin B=1
2bc sin B
By similar procedure above, we can also show that 𝑇= 1
2bc sin A and T=1
2ab sinC
So for this triangle:-
Of course, the way that the sides and
angles are labelled will not affect the
formula, as long as two sides and the
angle between them is used.
There is another formula, which is used when the lengths of the triangle's three sides
are known. It is Herons formula
In mathematics, Hero’s formula is the formula used to find the area of triangle in
geometry. The hero’s formula is defined by the mathematician named Heron of
Alexandria. The Pythagoras theorem is used to prove the hero’s formula which is found
in his book, Metrica. Hero’s formula is also known as Heron area formula. Let as
assume the sides of triangle as a, b and c.
The semi perimeter of the hero’s formula is given as,
s=a+b+c2
The hero’s formula is given as,
Area= √s (s−a ) ( s−b ) (s−c )
Given that area ∆=12ab sinC ……………(1)
Trigonometry identity sin2C+cos2C=1 sinC=√(1¿−cos2C)¿………….(2)
Cosine rule : c2 = a2 + b2 – 2ab cos C
cosC=a ²+b ²+c ²2ab
…………….(3)
Substitute (3) into (2) into (1)
12ab¿
√ ¿¿
14
√[(2ab−a¿¿2¿−b2+c2)(2ab+a2+b2−c2)]¿¿
14√ [c2−(a−b )2] [(a+b )2−c2]
14 √ [ (c−a+b ) (c+a−b ) ] [ (a+b−c ) (a+b+c ) ]
s=a+b+c2
,2 s=a+b+c
14
√ (2 s−a−b−a+b ) (2 s−a−b−a+b ) (a+b−2 s+a+b ) (2 s )
14
√16 (s−a ) ( s−b ) ( s−c ) ( s)
√164
√ (s−a ) ( s−b ) (s−c ) ( s)
√s (s−a ) ( s−b ) (s−c )
Example :
Find the area of a triangle using hero’s formula where every side is 8m long?
Given:
a = b = c = 8
Solution:
The semi perimeter of the hero’s formula is given as,
S = (a + b + c)/2 or perimeter/2.
S = (8+8+8)/2
S = (24)/2
S =12
The hero’s formula is given as,
Area=√ (s(s-a) (s-b) (s-c)),
Area= √ (12(12-8) (12-8) (12-8))
Area= √ (12(4) (4) (4))
Area = √ (768)
Area = 27.71
Therefore, the area of triangle is 27.71
1.3 PROBLEMS IN THREE DIMENSIONS
Normal to a Plane
In 3D graphics, an imaginary line that is perpendicular to the surface of a
polygon. It may be computed at the vertex of a triangle, in which case it is the
average of all the vertices of adjoining triangles. Or, it may be computed for
each pixel in the triangle as in Phong shading. Surface normals are used to
derive the reflectivity of a light source shining onto an object.
A straight line is normal to a plane if the line is perpendicular to all straight line
drawn on the plane.
diagram 1
In diagram 1, the straight line VO, drawn from a point V to meet the plane at
O, is a normal to the plane ABCD, as VO is perpendicular to the lines AO and
OE and other lines lying on the plane.
1.4 Angle between a Line and a Plane
The angle between a line AB and a plane π can be determined by making an
orthogonal projection from B to the plane π. Diagram 2 shows this orthogonal
projection from B meets the plane π at a point C. note that angle ACB=90⁰
Diagram 2
Diagram 3
Angle between two Planes
Two non-parallel planes π1 and π2, will meet at a common line. In diagram 4,
this common line is PQ. To find the angle between these two planes, take a
point, say A, on this common line. From this point A, draw two perpendicular
lines, AB and AC, to PQ with one of these lines lying on the plane π1and the
other line lying on the plane π2. Diagram 3 shows these two lines AB and AC,
with <BAP = <CAP =90⁰. Hence, from finding the angle between two planes,
we reduce this to find the angle between two lines, that is angle between the
line AB(representing plane π1) and the line AC(representing the plane π2).
The angle between two planes is the angle BAC.
Diagram 3
Problem in three dimensions
Question 1
Sally and Kate stand some distance away from a building (B) which is 10m
high. Sally is on a bearing of 030° from the building. From where she is
standing, the angle of elevation of the top of the building is 35°. Kate is on a
bearing of 090° from the building. From where she is standing, the angle of
elevation of the top of the building is 50°. This information is shown in the
diagram.
What is the distance between Sally and Kate?
Answer
Using trigonometry on the right-angled triangles OBS
so = 14.3 m (3 s.f.)
In the same way we can find OK = 8.39m (3 s.f.)
Applying the cosine rule to triangle KOS, we get:
KS2 = 8.392 + 14.32 - ( 2 × 8.39 × 14.3 × cos60°)
KS2 = 154.534
KS = 12.4m (3 s.f.)
1.4 ANGLE OF ELEVATION AND ANGLE OF DEPRESSION
Angle of Elevation
The word “elevation” means “rise” or “move up”.
Angle of elevation is the angle between the horizontal and the line of sight to an object
above the horizontal.
Example :
The angle of elevation of an object as seen by an observer is the angle between the
horizontal and the line from the object to the observer's eye (the line of sight).
Angle of Depression
The angle below horizontal that an observer must look to see an object that is lower
than the observer. (this assumes the object is close enough to the observer so that the
horizontals for the observer and the object are effectively parallel; this would not be the
case for an astronaut in orbit around the earth observing an object on the ground).
If the object is below the level of the observer, then the angle between the horizontal
and the observer's line of sight is called the angle of depression.
Note: The angle of depression is congruent to the angle of elevation
The picture below illustrates an example of an angle of depression and an angle of
elevation
Example of calculation :
The angle of elevation of the top of a pole measures 48° from a point on the ground 18
ft away from its base. Find the height of the flagpole.
Solution:
Step 1: Let’s first visualize the situation.
Step 2: Let ‘x’ be the height of the flagpole.
Step 4: x = 18 × tan 48° = 18 × 1.11061… = 19.99102… 20 Step 5: So, the flagpole is about 20 ft high.
2. EXTENDING TRIGONOMETRY
2.1 TRIGONOMETRY FORMULA
Basic identities
The picture shows a point on the unit circle. Since we know that any point on the unit
circle can be described by cos θ, sin θ. It is possible to draw the triangle that describes
this point. As you can see from the picture, the length of one side is cos θ and the
length of the other side is sin θ and, by definition, the radius of the unit circle is 1.
From these facts, the primary Pythagorean identity can be shown. sin2θ + cos2θ = 1.
This identity is just an application of the Pythagorean Theorem to the unit circle.
Formula for the Pythagorean Identities.
I. sin² θ + cos² θ = 1
II. tan² θ + 1 = sec² θ
III. 1 + cot² θ = csc² θ
Pictured below is the formula for the aptly named Pythagorean identity:
sin²θ + cos²θ = 1
Compound angle formula
The following are important trigonometric relationships:
sin(A + B) = sinAcosB + cosAsinB
cos(A + B) = cosAcosB - sinAsinB
tan(A + B) = tanA + tanB
1 - tanAtanB
To find sin(A - B), cos(A - B) and tan(A - B), just change the + signs in the above
identities to - signs and vice-versa:
sin(A - B) = sinAcosB - cosAsinB
cos(A - B) = cosAcosB + sinAsinB
tan(A - B) = tanA - tanB
1 + tanAtanB
Compound Angle Proof
The compound angle formula is: [4.1]
We construct a triangle, ABC, with CX being perpendicular to AB, and of length h. The line CS divides the angle C into two angles α and β.
We recall that the area of triangle ABC is:
[4.2]
Also, the area of triangle ACX is:
[4.3]
And the area of triangle XBC is:
[4.4]
We also note that h=b·cos α=a·cos β [4.5]
And for the set up, we note that the area of ABC is equal to the sum of the areas of triangles AXC and XBC:
[4.6]There are choices in substituting for h from Equation 4.5, and choosing the appropriate one to give 1/2ab throughout gives:
[4.7]Choosing an inappropriate value leads to more algebra, but the same result! A similar method is used to prove Pythagoras' Theorem
On division throughout by 1/2·a·b gives us the compound angle formula:
Double-angle formulae
sin(A + B) = sin A cos B + cos A sin B
Replacing B by A in the above formula becomes:
sin(2A) = sin A cos A + cos A sin A
so: sin 2A = 2 sin A cos A
similarly:
cos 2A = cos2 A - sin2 A
Replacing cos2A by 1 - sin2A (see Pythagorean identities) in the above formula gives:
cos 2A = 1 - 2sin2A
Replacing sin2 A by 1 - cos2 A gives:
cos2 A = 2 cos2 A - 1
It can also be shown that:
tan 2A = 2 tan A
1 - tan2A
Example. Check the identities
Answer. We will check the first one. the second one is left to the reader as an exercise. We have
Hence
which implies
Half-angle formulae
We start with the formula for the cosine of a double angle
cos 2θ = 1− 2sin2 θ
Now, if we let
then 2θ = α and our formula becomes:
cos α = 1 − 2sin2 (α/2)
We now solve for
(That is, we get sin(α/2) on the left of the equation and everything else on the right):
2sin2(α/2) = 1 − cos α
sin2(α/2) = (1 − cos α)/2
Solving gives us the following sine of a half-angle identity:
The sign of depends on the quadrant in which α/2 lies.
If α/2 is in the first or second quadrants, the formula uses the positive case:
If α/2 is in the third or fourth quadrants, the formula uses the negative case:
Using a similar process, with the same substitution of (so 2θ = α) we subsitute
into the identity
cos 2θ = 2cos2 θ − 1 (cosine of a double angle)
We obtain