trigonometric functions - wordpress.com · trigonometric functions similar right-angled triangles...
TRANSCRIPT
Trigonometric Functions
Similar Right-Angled Triangles
We plan to introduce trigonometric functions as certain ratios determined by similar right-angled trian-gles. By de�nition, two given geometric �gures are similar if they have the same shape and di¤er only insize. Pairs of similar right-angled triangles are shown in the examples below.
Example 1
A B
C
P Q
R
Two similar right-angled triangles
Because the two triangles have the same shape, the angle at A is equal to the angle at P , the angle at B isequal to the angle at Q, (they are both right angles), and the angle at C is equal to the angle at R. AB andPQ are called corresponding sides. BC and PQ are also corresponding sides as are AC and PQ.
Example 2
U V
W
X Y
Z
Another pair of similar right-angled triangles
In this pair of similar right-angled triangles, the angle at U is equal to the angle at X, the angle at V isequal to the angle at Y and the angle at W is equal to the angle at Z. UV and XY are corresponding sidesas are VWand Y Z. WU and ZX are also corresponding sides.
As we can see from the above examples, corresponding sides in a pair of similar triangles need not havethe same length. What is the same are their ratios. Thus in Example (1)
Length of ABLength of PQ
=Length of BCLength of QR
. We may also write this asLength of ABLength of BC
=Length of PQLength of QR
Likewise
Length of BCLength of QR
=Length of CALength of RP
which we may also write asLength of BCLength of CA
=Length of QRLength of RP
1
The last one is
Length of CALength of RP
=Length of ABLength of PQ
which may be written asLength of CALength of AB
=Length of RPLength of PQ
In Example (2),Length of UVLength of XY
=Length of VWLength of Y Z
=Length of WULength of ZX
which is a short form for the following three:
Length of UVLength of XY
=Length of VWLength of Y Z
Length of VWLength of Y Z
=Length of WULength of ZX
Length of WULength of ZX
=Length of UVLength of XY
Of course they may be re-arranged at will.The above ratios may be used to calculate unknown lengths in similar triangles if su¢ cient information
is furnished about some of the lengths.
Example 3 Say we are given that the lengths, in centimeters, of the sides AB, BC and PQ in Example 1
are 8, 15 and 20 respectively. Since we know the ratioLength of PQLength of AB
, (it is20
8), we may use it to calculate
the length of QR. We simply use the fact fact that
Length of QRLength of BC
=Length of PQLength of AB
This translates intoLength of QR
15=20
8. Therefore the length of QR is
20� 158
=75
2centimeters.
Exercise 4
1. Suppose, in Example 1, the lengths, in inches, of QR, RP and BC are 18, 20:4 and 15 respectively.Calculate the length of CA.
2. Suppose, in Example 2, the lengths, in feet, of ZY , XY and UV are 13, 12 and 21 respectively.Calculate the length of VW .
3. A 6 feet tall man walks away from the base of a house as shown in the �gure below. When he is 18 feetaway from the house, his shadow starts forming 8 feet ahead of him. How tall is the building?
house shadow
house
2
Trigonometric Functions
As a prelude to trigonometric functions, consider the following problem: A student was assigned the task ofestimating the height of an upright tree in her home back yard. She proceeded as follows:She measured the length of the tree�s shadow and found it to be 22 feet. She then erected an upright stick
and measured its height and the length of its shadow. She found them to be 4 feet and 1.7 feet respectively.
She then argued that the two �gures above are similar. Therefore
The height of the treeThe length of the tree�s shadow
=The height of the stick
The length of the stick�s shadow(1)
Substituting the known lengths gives
The height of the tree22
=41.7
Therefore the height of the tree is4� 221:7
= 51:8 feet, rounded to 1 decimal place.
As we have said before, we will de�ne trigonometric functions as ratios of right-angled triangles like theones used by the student to determine the height of the tree in her backyard. There are six of them, denotedby tanx, sinx, cosx, cotx, cscx and cotx. We start with tanx. To introduce it, consider the angle x shownbelow in the �rst quadrant. We denoted the origin by A and took the positive horizontal axis as the initialray. We then then dropped a perpendicular to the horizontal axis from a point C on the terminal ray for x.Thus ABC is a right-angled triangle and the right angle is at B.
3
The longest side AC is called the hypotenuse of the triangle. It is common practice to call BC the "sideopposite angle x" and AB the "side adjacent to the angle x", then de�ne
tanx =Length of opposite sideLength of adjacent side
However, because opposite sides and adjacent sides are not so obvious for angles in the other quadrants, itis more practical to use the coordinates of the point C. Clearly, the length of the opposite side BC is thevertical coordinate of C and the length of the adjacent side AB is the horizontal coordinate of C,therefore
tanx =Vertical coordinate of CHorizontal coordinate of C
Of course this ratio depends on the value of x, not on the size of the triangle, (since ratios of correspondingsides in similar �gures are equal). It turns out to be a very useful computational tool, therefore it has beenevaluated accurately for di¤erent angles x and recorded in tables and the common calculators. For someangles x, it can be calculated using the geometry of triangles. Among such angles are 30�, 45�, and 60�: Ofthese three, tan 45� is the easiest one to calculate because when x is 45�, the opposite side and the adjacentside are equal in length, (see the left �gure below). This implies that the vertical and horizontal componentsof C are equal, therefore tan 45� = 1:
45A B
C
An angle of 45� An angle of 30�
To determine tan 30�, consider the equilateral triangle ACD, (i.e. a triangle whose sides are all equal inlength) shown in the right �gure above. Actually, all we need is the the upper half of the triangle. Thereis no harm assuming that each of the sides AC; CD and AD has length 2. Then BC has length 1, (sinceit is half the length of CD). If we now use the Pythagorean theorem we easily obtain the length of AB tobep22 � 12 =
p3. Now the coordinates of C can be read o¤ the �gure. The horizontal one is
p3 and the
vertical one is 1. Therefore
tan 30� =Vertical coordinate of CHorizontal coordinate of C
=1p3=
p3
3
4
To determine tan 60�, consider the same equilateral triangle ACD but drawn di¤erently as shown below.
All we need is the left half of the triangle. If we assume that each of the sides AC; AD, and CD has length2 then AB must have length 1 and BC must have length
p22 � 12 =
p3. Therefore C has horizontal
coordinate 1 and vertical coordinatep3, so that
tan 60� =Vertical coordinate of CHorizontal coordinate of C
=
p3
1=p3
Angles Bigger Than 90�
The same formula is used to de�ne the tangent of an angle bigger than 90�. For an example, take theangle 135�, shown in the �gure below, which the line segment AC makes with the horizontal axis. Drop aperpendicular from C to the horizontal axis as shown in the next �gure. We have denoted the point whereit intersects with the horizontal axis by B.
Since angle BAC is 45�, (i.e. 180� � 135�), angle BCA must also be 45�, therefore the line segments BAand BC must have the same length. If, for pure convenience, we assume that the length of BC is one unitthen so must be the length of BA. Therefore the point C has coordinates (�1; 1) and so
tan 135� =Vertical coordinate of CHorizontal coordinate of C
=1
�1 = �1
5
For another example, we calculate tan 120�. The 120� is drawn in the �gure below.
The coordinates of B are (�1; 0) and BC has lengthp3. Therefore the coordinates of C are
��1;
p3�and
so
tan 120� =Vertical coordinate of CHorizontal coordinate of C
=
p3
�1 = �p3
Use the �gure below to calculate the exact value of tan 150�:
For most angles x, we use a calculator to evaluate tanx. Say you want to �nd tan 23�. Make sure thecalculator is in degree mode. Press the button labelled tan then enter the number 23. If you now press the= symbol, a number which we may round o¤ to 0:424 47, should appear.Use a calculator to complete the table below. Round o¤ the values to 3 decimal places.
x� 21� 48� 79� 126� 180� 235� 260� 300� 331�
tanx�
Soon you will be required to draw an angle whose tangent is a given number. Here are two examples:
Example 5 To draw an angle u that satis�es the following conditions: (i) it is in standard position, (ii) itis in the �rst quadrant and (iii) its tangent is 5
6 , (i.e. tanu =56).
Solution We have to take the positive horizontal axis as the initial ray. To draw a terminal ray for u, itsu¢ ces to draw a line segment that originates from (0; 0) and passes through a point C with verticalcoordinate 5 and horizontal coordinate 6 so that
tanu =Vertical coordinate of CHorizontal coordinate of C
=5
6
6
2 4 6
2
4
6 .C
u
Example 6 To draw an angle w which, (i) is in standard position, (ii) is in the third quadrant, and (ii) hastangent 43 , (i.e. tanw =
43).
Solution We have to take the positive horizontal axis as the initial ray. We have to draw a terminal raywith the property that if C is a point on the ray then
Vertical coordinate of CHorizontal coordinate of C
=4
3
Since the horizontal and vertical coordinates of points in the third quadrant are negative, it must be thecase that
Vertical coordinate of CHorizontal coordinate of C
=�4�3
The two negative signs happen to cancel out to give the result tanw = 43 . Therefore, it su¢ ces to draw
a line segment that originates from (0; 0) and passes through (�3;�4). The angle is drawn in the �gurebelow.
Example 7 In the given triangle, angle A is 39�, the side AB has length 32 cm and angle ABC is a rightangle. Calculate the length of the side BC and round o¤ to 1 decimal place.
Solution We use the de�nition
tan 39� =Length of opposite sideLength of adjacent side
=Length of BC
32
Therefore Length of BC = 32 tan 39�. We use a calculator to determine tan 39� then multiply it by32 to get 25.9 cm, to 1 decimal place.
7
Exercise 8
1. Draw an angle u, in the second quadrant, whose tangent is � 38 , (i.e. tanu = �
38) and an angle w in
the third quadrant whose tangent is 45 .
Angle u Angle w
2. In triangle ABC, the angle at B is a right angle, the angle at A is 43� and the side AB has length 28cm.
The length of the side BC is:
A) 25:88 cm B) 26:11 cm. C) 26:74 cm D) 27:56 cm
Practice Problems Set 5, v1
1. The shadow of an electric pole is found to be 8 feet long. At the same time and place, the shadow of
8
a 6 foot man is found to be 2.7 feet long. What is the height of the electric pole, to the nearest foot?
2. Draw an angle u, in the fourth quadrant, whose tangent is � 512 , (i.e. tanu = �
512 ) and an angle w in
the third quadrant whose tangent is 75 .
Angle u Angle w
9
3. In triangle ABC, the angle at B is a right angle, the angle at A is 32� and the side BC has length 43ft. Calculate the length of the side AB.
4. To determine the height of a building and the communication tower on top of the building, the anglesof elevation of the bottom and top of the tower were measured from a point 47 meters from the bottomof the building and the results are as shown in Figure 2. If we remove the un-necessary details, we get�gure 3. The angle at A is 90�. Use triangle OAB to calculate the height AB, then use triangle OACto calculate the height AC.
Building
Tower
Figure 1 Figure 2 Figure 3
Practice Problems Set 5, v2
1. The shadow of an electric pole is found to be 8 feet long. At the same time and place, the shadow of
10
a 6 foot man is found to be 2.7 feet long. What is the height of the electric pole, to the nearest foot?
2. Draw an angle u, in the fourth quadrant, whose tangent is � 59 , (i.e. tanu = �
59 ) and an angle w in
the third quadrant whose tangent is 87 .
Angle u Angle w
3. In triangle ABC, the angle at B is a right angle, the angle at A is 35� and the side BC has length 48
11
ft. Calculate the length of the side AB.
A B
C
35°
48 ft
4. To determine the height of a building and the communication tower on top of the building, the anglesof elevation of the bottom and top of the tower were measured from a point 47 meters from the bottomof the building and the results are as shown in Figure 2. If we remove the un-necessary details, we get�gure 3. The angle at A is 90�. Use triangle OAB to calculate the height AB, then use triangle OACto calculate the height AC.
Building
Tower
Figure 1 Figure 2 Figure 3
12
General De�nition of the Basic Trigonometric Functions.
Consider an angle x in standard position. As we have done above, we the origin by A and take positivehorizontal axis, (starting at A), as the initial ray. Take any C on the terminal ray for x. Then the linesegment AC makes an angle x with the positive horizontal axis. Drop a perpendicular CB from C to meetthe horizontal axis at B and use it to determine the coordinates of C. In the example shown below, weconsidered an angle x in the third quadrant, but that need not be the case.
The line segment AC is called the hypotenuse of the right triangle ABC. The trigonometric functions tanx,sinx and cosx are de�ned as follows:
tanx =Vertical coordinate of CHorizontal coordinate of C
, sinx =Vertical coordinate of CLength of hypotenuse AC
, cosx =Horizontal coordinate of CLength of hypotenuse AC
For angles x in the �rst quadrant and in standard position, we may de�ne sinx and cosx by
sinx =Length of opposite sideLength of hypotenuse
cosx =Length of adjacent sideLength of hypotenuse
It turns out that tanx =sinx
cosx. Verifying this is easy:
sinx
cosx=
�Vertical coordinate of CLength of hypotenuse AC
��Horizontal coordinate of CLength of hypotenuse AC
� =Vertical coordinate of CLength of hypotenuse AC
� Length of hypotenuse ACHorizontal coordinate of C
=Vertical coordinate of CHorizontal coordinate of C
= tanx
The following table gives a few values of these functions. The angles x may be given in degrees or radians.
13
We have given them in both units.
x indegrees
0� 30� 45� 60� 90� 120� 135� 150� 180�
x inradians
0 rad�
6rad
�
4rad
�
3rad
�
2rad
2�
3rad
3�
4rad
5�
6rad � rad
tanx 01p3
1p3
Un-de�ned
�p3 �1 � 1p
30
sinx 01
2
1p2
p3
21
p3
2
1p2
1
20
cosx 1
p3
2
1p2
1
20 �1
2� 1p
2�p3
2�1
x indegrees
210� 225� 240� 270� 300� 315� 330� 360� 390�
x inradians
7�
6rad
5�
4rad
4�
3rad
3�
2rad
5�
3rad
7�
4rad
11�
6rad 2�
13�
6rad
tanx1p3
1p3
Un-de�ned
�p3 �1 � 1p
30
1p3
sinx �12
� 1p2
�p3
2�1 �
p3
2� 1p
2�12
01
2
cosx �p3
2� 1p
2�12
01
2
1p2
p3
21
p3
2
More Trigonometric Functions
There are three more trigonometric functions denoted by cotx, cscx and secx.
Using the same �gure above, they are de�ned by
cotx =1
tanx=Horizontal coordinate of CVertical coordinate of C
cscx =1
sinx=Length of hypotenuse ACVertical coordinate of C
secx =1
cosx=Length of hypotenuse ACHorizontal coordinate of C
Most calculators give only the values of sinx, cosx and tanx. The other functions are evaluated using the
above formulas. For example, sec 60� =1
cos 60�= 2.
14
Example 9 (�5; 12) is a point on the terminal ray of an angle x . To �nd the exact value of: sin x , cos x ,tan x , cot x , csc x and sec x .
Solution: The angle is shown in the �gure below
The horizontal and vertical coordinates of the point (�5; 12) on the terminal ray are �5 and 12 respec-tively. The hypotenuse has length p
5 + 122 =p169 = 13
Therefore sin x = 1213 , cos x = �
513 , tan x = �
125 , cot x = �
512 , csc x =
1312 and sec x = �
135 .
Example 10 To �nd the exact values of tan x , sin x and cos x given that sec x = 32 and x is in quadrant IV.
Solution: The angle is shown in the �gure below. Since sec x = 32 , we may take the horizontal coordinate
of a point on the terminal ray to be 2 and the corresponding hypotenuse to have length 3.
Then the corresponding vertical coordinate should be �p32 � 22 = �
p5. It has to be negative because
the vertical coordinates of points in the fourth quadrant cannot be positive. Thus�2;�
p5�is a point
on the terminal ray for the given angle x. Therefore tan x = �p52 , sin x = �
p53 and cos x = 2
3 .
Exercise 11
1. (�3;�4) is a point on the terminal ray of an angle �. Find the exact value of sec �.
A) 54 B) � 5
3 C) � 35 D) 4
3
2. You are given that � is an acute angle and sin � =
p5
3. Find the exact value of sec �.
A) sec � = �2p5
5B) sec � =
3p5
5C) sec � =
3
2D) sec � =
p5
2
15
3. The angles x, y and z are shown in the �gure below.
Determine the exact value of:
(a) sin z (b) cos z (c) tan z (d) sin y (e) cos y (f) tan y
(g) sinx (h) cosx (i) tanx (j) sec z (k) csc z (l) cot z
4. You are given that u is an angle in the second quadrant with sinu =5
13and v is an angle in the fourth
quadrant with tan v = �43
(a) Draw the angles in standard position on the axes below.
(b) Determine the exact value of:
(a) tanu (b) cosu (c) cotu (d) secu (e) cscu
(g) cos v (h) cot v (i) sec v (j) csc v (k) sin v
16
5. You are given that x is an angle in the second quadrant and sinx = 34 . Draw the angle in standard
position then calculate cosx, tanx, cotx, secx, and cscx.
6. You are given that � is an angle in the third quadrant and tan � = 512 . Draw the angle in standard
position then calculate cos �, sin �, cot �, sec �, and csc �.
The signs of the Sine, Cosine and Tangent functions
� Since the horizontal and vertical coordinates of the points in the �rst quadrant are positive, it followsthat ALL the three functions sinx, tanx and cosx are positive in the �rst quadrant.
� The vertical coordinates of points in the second quadrant are positive but the horizontal coordinates arenegative. It follows that, of the three, it is the SINE function that is positive in the second quadrant.The other two are negative.
� The vertical and horizontal coordinates of points in the third quadrant are negative. It follows that ofthe three, it is the TANGENT function that is positive in the third quadrant, (because it is a quotientof two negative numbers). The other two are negative.
� The horizontal coordinates of points in the fourth quadrant are positive but the vertical coordinatesare negative. It follows that of the three, it is the COSINE function that is positive in the fourthquadrant. The other two are negative.
� The above observations are summarized in the following table
Quadrant 1 2 3 4
Positive function All Sine Tangent Cosine
We may condense this further to
1 2 3 4
A S T C
Or simply ASTC
Reference Angles
Consider the angles 30�, 150�, 210�, 330�, 390�. The non-acute angles 150�, 210�, 330�, 390�, 510�, . . . areall related to the single acute angle 30� as follows:
� 150� is short of 180� by 30� therefore its terminal ray makes an angle of 30� with the x-axis.
Terminal ray makes an angle of 30� with the x-axis
17
� 210� exceeds 180� by 30� therefore its terminal ray makes an angle of 30� with the x-axis.
Terminal ray makes an angle of 30� with the x-axis
� 330� is short of 360� by 30� therefore its terminal ray makes an angle of 30� with the x-axis.
Terminal ray makes an angle of 30� with the x-axis
The pattern should be clear. We say that 30� is the reference angle for the 150�, 210�, 330�, etc. Ingeneral, the reference angle of a given angle x is the acute angle that the terminal ray for x makes withthe x-axis, assuming that one has taken the positive x-axis as the initial ray.
Note that sin 150� = sin 30�, cos 150� = � cos 30�, tan 150� = � tan 30�, thus the values of the trigono-metric functions at 150� are either the same or they di¤er from the corresponding values at 30� by just asign. Before calculators were introduced, the values of sine, cosine and tangent functions were tabulated foracute angles only. If, say, one wanted to determine sin 290�, which is not an acute angle, one would
(a) Determine the reference angle for 290�, which is 70�;
(b) Look up sin 70� from a table of the sine function. This would be a number rounded to 4 decimal places,most probably 0:9397;
(c) Note that sinx is negative in the fourth quadrant, therefore sin 290� = � 0:9397:
Exercise 12
1. Find the reference angle for 115�
A) 35� B) 25� C) 65� D) 75�
2. Find the reference angle for �237�
3. Find the reference angle for �44�
18
Practice Problems Set 6, v1
1. You are given that � is an angle in the second quadrant and cos � = � 13 . Draw the angle, form an
appropriate right triangle, calculate its unknown length then determine the exact values of sin �, tan �,csc � and cot �. Where applicable, give your answer as a radical with a rational denominator.
Calculate the unknown length in this space
sin � =
tan � =
csc � =
cot � =
2. You are given that � with sin � = � 34 and tan � > 0, (i.e. the tangent of � is positive). Determine the
quadrant in which � lies, draw it then calculate the exact value of cos �
19
3. A guy wire is attached to a point 2 feet below the top of a vertical electric pole. The pole is 18 feettall and the guy wire makes an angle of 78� with the horizontal ground. Draw a diagram displayingthis information then calculate the length of the guy wire.
20
Practice Problems Set 6, v2
1. You are given that � is an angle in the second quadrant and sin � = 14 . Draw the angle, form an
appropriate right triangle, calculate its unknown length then determine the exact values of cos �, tan �,sec � and cot �. Where applicable, give your answer as a radical with a rational denominator.
Calculate the unknown length in this space
sin � =
tan � =
csc � =
cot � =
2. You are given that � with cos � = 23 and tan � < 0, (i.e. the tangent of � is negative). Determine the
quadrant in which � lies, draw it then calculate the exact value of sin �
21
3. A guy wire is attached to a point 2 feet below the top of a vertical electric pole. The pole is 17 feettall and the guy wire makes an angle of 75� with the horizontal ground. Draw a diagram displayingthis information then calculate the length of the guy wire.
22
Graphs of sin x and cosx
Sample values of cosx are given below.
x indegrees
0 30 45 60 90 120 135 150 180
cosx to2 dec. pl.
1 0:87 0:71 0:5 0 �0:5 �0:71 �0:87 �1
x indegrees
210 225 240 270 300 315 330 360
cosx to2 dec. pl.
�0:87 �0:71 �0:5 0 0:5 0:71 0:87 1
They are plotted on the axes below.
Joining them with a smooth curve gives the graph below.
0 180 36090 270| |
As shown in the �gure below, we also get this same shape between any two consecutive multiples of 360�.
23
For this reason, we say that cosx is periodic with period 360�.
Its values lie strictly between �1 and 1. We say that it has amplitude 1.
One cycle of sinx is given in the �gure below.
0 180 36090 270| |
1_
_1
It is this same shape that one gets between 360� and 720� and between �360� and 0�, (see the graph
24
below).
In general, we get the same shape between any two consecutive multiples of 360�. For this reason, we saythat sinx is periodic with period 360�. Its values are also strictly between �1 and 1, therefore its amplitudeis also 1.
Magnifying the graphs of sinx and cosx
For an example, consider the graph of 2 sinx. We get it by simply doubling the values of sinx. The resultis a graph with amplitude 2 drawn below on the same axes as the graph of sinx.
0 180 36090 270| |
1 _
_1
2 _
2_
graph of 2sinx
In general, if b is a positive number then the graph of b sinx will have the familiar shape of a sine functionbut with amplitude b. A similar statement with sine replaced by cosine is also true. In the �gure below, the
25
graph of 1:8 cosx is drawn on the same axes as the graph of cosx.
0 180 36090 270| |
1 _
_1
1.8_
1.8_
graph of 1.8cosx
The graph of � sinx is obtained by drawing the re�ection, in the horizontal axis, of the graph of sinx,(the positive values are made negative and the negative values are made positive).
0 180 36090 270| |
1_
_1
0 180 36090 270| |
1_
_1
Graph of sinx Graph of � sinx
The graph of �2 sinx is obtained by �rst magnifying the graph of sinx by a factor 2 then draw a re�ection,
26
in the horizontal axis, of the magni�ed graph.
0 180 36090 270| |
1_
_1
2_
2_
graph of 2sinx
0 180 36090 270| |
1_
_1
2_
2_
graph of 2sinx
The same procedure is followed to sketch a cycle of � cosx or �b cosx, (when b is positive). For example,to sketch a cycle of �1:8 cosx, �rst magnify the graph of cosx by a factor of 1:8 then draw a re�ection, inthe horizontal axis, of the magni�ed graph.
0 180 36090 270| |
1_
_1
1.8_
1.8_
graph of 1.8cosx
0 180 36090 270| |
1_
_1
1.8_
1.8_
graph of 1.8cosx
In general, to sketch a cycle of a sinx or a cosx where a is a given negative number, �rst magnify thegraph of sinx or cosx by a factor jaj then draw a re�ection, in the horizontal axis, of the magni�ed graph.
Translating the graphs of sinx and cosx horizontally
Say you are asked to sketch one cycle of the graph of cos (x� 60�). You simply take a cycle of cosx andslide it, (i.e. translate it), to the right through 60 degrees. The result is drawn below on the same axes asone cycle of cosx, (shown dotted). We say that the graph of cosx is shifted through 60� to the right. The
27
60� angle is called the horizontal shift.
Graphs of cosx, (dotted), and cos (x� 60�)Here is a way of visualizing what is going on: Take the largest value 1 of cosx. You get it when x is 0� or360�, (assuming you are restricting yourself to one cycle of cosx). It follows that cos(x � 60�) has value 1when the angle x� 60� is 0� or 360�, i.e. when x = 60� or when x = 420� as shown on the above graph.
If you are asked to sketch the graph of cos (x+ 60�), you would shift the graph of cosx to the left, (NOTTO THE RIGHT), through 60�.
Graphs of cosx, (dotted), and cos (x+ 60�)
In general, if b is a positive number then the graph of cos (x� b) is obtained by shifting the graph ofcosx through b degrees to the RIGHT and the graph of cos (x+ b) is obtained by shifting the graph of cosxthrough b degrees to the LEFT.
Changing the Period of sinx and cosx
We change the period of sinx or cosx when we multiply the variable x by a constant. For an example,consider the function sin 32x. We note that
32x is zero when x = 0
� and it is 360 when
x =36032
= 360� 23= 240�:
28
It follows that one cycle of sin 32x is between 0 and 240 degrees, suggesting that its period is 240�. This is
indeed the case as the graph below shows.
In general, the function sin bx has period360
b. The same applies to cos bx.
A Combination of All Three
Consider sketching one cycle of the graph of 2 sin 34 (x� 40�). The �rst step is to deduce useful information
about it from its formula.
� It has amplitude 2.
� Its period is 360��34
�= 360� 4
3= 480 degrees.
� It has a horizontal shift of 40� to the right.
Now draw coordinate axes and introduce an interval from 0� to 480� on the horizontal axis, (because thefunction has period 480�). Divide the interval into four equal parts corresponding to angles 0�, 120�, 240�;360� and 480�. The vertical axis should extend at least to 2 and �2 because the function has amplitude 2.
0 240 480120 360| |
2_
_2
| |
29
Without the shift, you have the function 2 sin 34x whose graph is shown dotted in the �gure below. Thegraph of 2 sin 34 (x� 40
�) is obtained by shifting the dotted graph through 40� to the right.
Another Example To sketch one cycle of y = �2 cos (3x+ 15�)
Solution We start by extracting useful information from the given formula. (i) Its amplitude is 2. (ii)
Its period is�360
3
��= 120�. (iii) To get its shift we �rst write it in the form �2 cos 3 (x+ 5).
Now the shift is clear; it is 5� to the left. Turning to the required sketch, we �rst sketch a cycleof cos 3x, shift it 5� to the left, magnify it by a factor 2 then re�ect, in the horizontal axis, themagni�ed graph. The various stages are shown in the �gures below.
0 60 12030 90| |
0 60 12030 90| |
Sketch of cos 3x Sketch of cos 3(x+ 5)
30
0 60 12030 90| |
0 60 12030 90| |
Sketch of 2 cos 3(x+ 5) Sketch of �2 cos 3(x+ 5)
Exercise 13
1. Determine the amplitude of y = � 14 sinx
A) � 14 B) �
4 C) 4 D) 14
2. Determine the period of y = 5 cos 4�x
A) �2 B) 4� C) 1
2 D) 2
3. Determine the phase shift of y = 5 sin�14x�
�4
�A) �
16 units to the left B) � units to the right C) �4 units to the left D) �
4 units to the right
4. The current I, in amperes, �owing through a particular ac (alternating current) circuit at time t secondsis
I = 120 cos�24�t� �
6
�(a) What is the period of the current?
A) �120 seconds B) 24� seconds C) 1
12 seconds D) 1144 seconds
(b) What is the phase shift of the current?
A) �6 seconds B) 1
6 seconds C) �144 seconds D) 1
144 seconds
5. You are given one cycle of the graph of f(x) = sinx.
31
Extend the coordinate axes as necessary then sketch the graph of g(x) = sin (x+ 30�).
6. You are given a sketch of the graph of h(x) = cosx:
Extend the coordinate axes as necessary then sketch the graph of u(x) = cos (x� 45�) :
7. You are given a sketch of the graph of h(x) = cosx:
Extend the coordinate axes as necessary then sketch the graph of v(x) = 3 cosx
8. Sketch, on the same axes, one cycle each for the graph of
(a) sinx and 2:5 sinx (b) cosx and = �2 cosx (c) sinx and 34 sinx
(d) cosx and 45 cosx (e) sinx and sin (x� 30�) (f) sinx and sin (x+ 40�)
(g) cosx and cos (x+ 45�) (h) cosx and cos (x� 28�) (i) sinx and sin (x+ 54�)
32
Practice Problems Set 7, v1
1. Sketch one cycle of the curve y = cosx and the curve y = 2 cosx on the same axes below.
| | | ||
_
_
_
_
_
_
_
_
2. Sketch one cycle of the curve y = 2 sinx and the curve y = 2 sin(x� 30�) on the same axes below.
| | | ||
_
_
_
_
_
_
_
_
33
3. Sketch one cycle of the curve y = sin 34x
| | | ||
_
_
_
_
_
_
_
_
4. Determine the amplitude, period and shift of the curve y = 3 sin 23 (x� 30�)
Amplitude =
Period =
Shift = (Specify whether it is to the left or right.)
Sketch one cycle of the curve y = 3 sin 23 (x� 30�)
| | | ||
_
_
_
_
_
_
_
_
34
Practice Problems Set 7, v2
1. Sketch one cycle of the curve y = sinx and the curve y = 3 cosx on the same axes below.
| | | ||
_
_
_
_
_
_
_
_
2. Sketch one cycle of the curve y = 2 cosx and the curve y = 2 cos(x+ 40�) on the same axes below.
| | | ||
_
_
_
_
_
_
_
_
35
3. Sketch one cycle of the curve y = cos 32x
| | | ||
_
_
_
_
_
_
_
_
4. Determine the amplitude, period and shift of the curve y = 4 sin 45 (x+ 30�)
Amplitude =
Period =
Shift = (Specify whether it is to the left or right.)
Sketch one cycle of the curve y = 3 sin 23 (x� 30�)
| | | ||
_
_
_
_
_
_
_
_
36
Exercise 14
1. Determine the period of the given function then sketch one cycle of its graph:
(a) sin 3x (b) cos 23x (c) sin 34x (d) cos 53x
(e) sin 35x (f) sin 43x (g) cos 56x (h) cos 45x
2. Give the period and amplitude of f(x) = 2 sin 3x2 then sketch one cycle of its graph on the given axes.
| | | ||____
____
3. Determine the period, amplitude and shift of the trigonometric function f(x) then sketch one cycle ofits graph on the given axes.
| | | ||____
____
4. Determine the period, amplitude and shift of the given function then sketch one cycle of its graph:
(a) 2 sin 3 (x� 10�) (b) 3 cos 23 (x+ 45�) (c) 4
3 sin32 (x� 20
�)
(d) 3:5 cos 45 (x+ 15�) (e) 1:8 cos 35 (x� 50
�) (f) 2:5 sin 34 (x+ 48�)
(g) 34 sin 2 (x+ 25
�) (h) 23 sin
32 (x� 35
�) (i) 35 cos
35 (x+ 90
�)
5. Determine the period, amplitude and shift of f(x) = � 43 sin
�34x+
3�8
�then sketch one cycle of its
graph. The angles are in radians.
6. You are given that x is in degrees. Determine the period, amplitude and shift of y = �3 cos�23x� 20
��then sketch one cycle of the curve.
7. Sketch one cycle of the curve y = 2 sin�34x� 15
��� 3. Assume that x is in degrees.
37
Inverse Trigonometric Functions
Here is a problem that may be solved using an inverse trigonometric function:A rectangle has sides of length 12 inches and 22 inches. What is the acute angle x, shown in �gure (i),
between its diagonals?
12
22
x 12
22
_ _ _ _ _ _ _ _ _ _y11
6
Figure (i) Figure (ii)
One approach is to to calculate angle y shown in �gure (ii) then double it to get x. Clearly, tan y = 611 .
Therefore we have to �nd an acute angle y whose tangent is 611 . This is the opposite of what we have done
so far, which was to calculate the tangent or sine or cosine of a given angle. Now we have to �nd an anglewhose tangent is a given number. This requires the use of an inverse trigonometric function. The inversesof tanx, sinx and cosx are introduced below.
It is convenient to view a function as a table with two rows, (or two columns), that shows how twovariable quantities are related. The table must satisfy one crucial condition: Every number in the �rst row,(column), is paired with exactly one number in the second row, (column).Take the function tanx: The table below shows a sample of its values.
x �80� �60� �50� �45� �30� �20� 0� 20� 30� 45� 54� 60� 72�
tanx �5:67 �p3 �1:19 �1 � 1p
3�0:36 0 0:36
1p3
1 1:38p3 3:08
The decimals are approximate values. We have restricted the angles x: they are between �90� and 90�.When we do this then di¤erent angles in the �rst row are paired with di¤erent numbers in the second row.Under these circumstances, we can de�ne an inverse function. To this end, swap the two rows. The result isa table that gives sample values for the inverse of tanx which is denoted by arctan y or tan�1 (y). (The useof y instead of x is purely for convenience.)
y �5:67 �p3 �1:19 �1 � 1p
3�0:36 0 0:36
1p3
1 1:38p3 3:08
tan�1 y �80� �60� �50� �45� �30� �20� 0� 20� 30� 45� 54� 60� 72�
Thus tan�1(�5:67) = �80�; tan�1(1) = 45�, tan�1(1:38) = 54�, etc. Clearly, if y is a given number thenarctan y, or tan�1 (y), is the angle between�90� and 90� whose tangent is y. This function is given on calcula-tors. Set your calculator to degree mode then use it to con�rm that, (to 1 decimal place), tan�1 (0:72) = 35:8�,tan�1 (�0:42) = �22:8�, then determine the following:
a) tan�1 (�0:567) b) tan�1 (0:333) c) tan�1 (0:157) d) tan�1 (4:7) e) tan�1 (�6)
38
To determine the acute angle between the diagonals of the above rectangles, we had to �nd an angle ywhose tangent is 6
11 . Now we know that we have to �nd tan�1 � 6
11
�. A calculator gives 28:61� to 2 decimal
places. Therefore x = 57:22� or 57� to the nearest degree.
The inverse of sinx is denoted by sin�1 y or arcsin y and it is obtained in a similar way. We �rst restrictthe angles x to values between �90� and 90�, (so that di¤erent angles have di¤erent sines). A sample ofvalues of sinx is given below
x �90� �80� �60� �50� �45� �30� 0� 30� 45� 54� 60� 72� 90�
sinx �1 �0:98 �p3
2�0:77 � 1p
2�12
01
2
1p2
0:81
p3
20:95 1
When we swap the two rows we get sample values for sin�1 y or arcsin y:
y �1 �0:98 �p3
2�0:77 � 1p
2�12
01
2
1p2
0:81
p3
20:95 1
sin�1 y �90� �80� �60� �50� �45� �30� 0� 30� 45� 54� 60� 72� 90�
As you would expect, sin�1 (y) is the angle between �90� and 90� whose sine is y. The values of sin�1 y arealso given on calculators. Use it to con�rm that, to one decimal place, sin�1(0:3) = 17:5�, sin�1(0:8732) =60:8� and sin�1(�0:721) = �46:1�.
To get an inverse for cosx, we restrict the angle to values between 0� and 180� to guarantee that di¤erentangles have di¤erent cosines. A sample of values of cosx is given in the table below.
x 0� 30� 45� 50� 60� 71� 90� 120� 135� 140� 150� 165� 180�
cosx 1
p3
2
1p2
0:641
20:33 0 �1
2� 1p
2�0:77 �
p3
2�0:97 �1
Sample values for its inverse are obtained by swapping the two rows.
y 1
p3
2
1p2
0:641
20:33 0 �1
2� 1p
2�0:77 �
p3
2�0:97 �1
cos�1 y 0� 30� 45� 50� 60� 71� 90� 120� 135� 140� 150� 165� 180�
If y is a number between �1 and 1 then cos�1 (y) is the angle between 0� and 180� whose cosine is y.Set your calculator to degree mode then use it to determine cos�1 (0:6), sin�1 (�0:42), cos�1 (�0:567) ;
sin�1 (0:333), cos�1 (0:157).
Example 15 To �nd all the angles x such that 4 sinx = 3:
Solution 16 Divide both sides by 4 to get sinx = 0:75. Therefore x is an angle whose sine is 0:75. In otherwords, x = arcsin 0:75. A calculator gives x = 48:6� to 1 dec. pl. But this is not the only angle whose sineis 0.75. Another one is 180� � 48:6� = 131:4�. Since the sine function is periodic with period 360�, we mayadd any multiple of 360� to 48:6� or 131:4� and the result will still be a solution of the equation 4 sinx = 3.Thus 48:6� + 360�, 131:4� + 360�; 48:6� + 720�, 131:4� + 720�; and many others are solutions of the givenequation. The expressions (48:6 + 360n)�, (131:4 + 360n)�, n = �1;�2; : : : capture all the solutions.
39
Example 17 To calculate the exact value of cos�sin�1( 35 )
�:
Solution 18 sin�1( 35 ) is an angle in the �rst quadrant whose sine is35 . We may name that angle u. Thus
u = sin�1( 35 ) and so sinu =35 . The �gure below shows the angle.
u
5
|
|
|
|
3
We use the Pythagorean theorem to calculate the length of the third side of the triangle. If its length is a,then
a2 + 32 = 52
When we solve for a we get a = 4. We are required to calculate cos�sin�1( 35 )
�which is actually cos(u).
Since the horizontal side of the triangle has length 4, it follows that cosu = 45 , therefore cos
�sin�1( 35 )
�= 4
5
Exercise 19
1. Solve the following equations for x between 0� and 360�. When necessary, round o¤ your answers to 1decimal place.
(a) 5 sinx� 3 = 0 (b) 5 tanx+ 9 = 0 (c) 3 cosx+ 1 = 2
(d) 2 sinx+ 3 = 1 (e) 4 cosx+ 1 = 3 7 tanx+ 6 = 4
2. Clearly, tan�tan�1
�p3��= tan (60�) =
p3. Explain why tan
�tan�1 y
�= y for any given number y.
What is tan�1 (tanx)?
3. A student was asked to determine tan�sin�1( 12 )
�and he proceeded as follows: We know from a table
of sample values of the sine function that sin�1( 12 ) = 30�. Therefore the problem requires one to�nd tan (30�) which is 1p
3. It follows that tan
�sin�1( 12 )
�= tan (30�) = 1p
3. Find the exact value of
cos�tan�1(1)
�and sin(cos�1(�
p32 )) in a similar way.
4. Follow the steps of Example 17 to calculate the exact value of:
(a) sin(tan�1( 125 )):
(b) tan�sin�1( 45
�):
(c) cos(tan�1( 724 )):
(d) cos(sin�1(x)), (hint: write x as x1 ).
5. Verify that sin�cos�1(� 3
4 )�=
p74
6. Use a sketch to �nd the exact value of cos�sin�1 45
�A) 1
5 B) � 45 C) � 3
5 D) 35
7. Assume that x is positive. Use a sketch to �nd sin�tan�1 x
�A) x
px2+1
x2+1 B) xpx2�1
x2+1 C) xpx2 � 1 D)
px2+1x2+1
40
8. The exact value of cos�tan�1 2x
�in terms of x is
(A)
p4� x2x
(B)
p4 + x2
x(C)
xp4 + x2
(D)xp4� x2
9. Evaluate the following expressions
(a) sin (arccos 2x) (b) tan�arcsin x2
�(c) cos
�arctan 3
x
�(d) sec
�arctan
px2
�10. Find all the angles x such that 6 sin2 x� 7 sinx+ 2 = 0 :
11. Find all the angles x such that tan2 x� 6 tanx+ 5 = 0.
41