Download Section 5.2: Trigonometric Functions of Angles

Section 5.2: Trigonometric Functions of Angles
Objectives: Upon completion of this lesson, you will be able to:







find the values of the six trigonometric functions for an angle  with given conditions
given an angle and the length of one side of a right triangle, find the length of any missing side
solve right-triangle application problems
know the exact values of the six trigonometric functions for 30, 45, 60 and the quadrantal angles
(angles also in radian form)
find the trigonometric function of any given angle in radians or degrees
find the quadrant containing an angle  with given parameters
use the reciprocal, tangent and cotangent, and Pythagorean identities to simplify expressions, verify
identities, or write trigonometric expressions in terms of a different trigonometric function.
Required Reading, Video, Tutorial





Read Swokowski/Cole: Section 5.2
Watch the Trig Introduction Video
Watch the Special Angles Video
Watch the Section 5.2 Video
Complete the Section 5.2 Before Class WA assignment
Discussion
Section 5.2 is packed with material, and this material is fundamental to understanding any future trigonometric
concepts. Please read the section thoroughly and pay close attention to all 12 examples in the text.
Analytic geometry; which is the use of algebra to study geometric properties and operations on symbols defined
in a coordinate system; is fundamental to the study of trigonometry. Following is a right triangle on a
coordinate grid in a circle of radius r. The circle and the triangle have the point P  x, y  in common. Also
pictured is the same right triangle without the grid or circle.
P(x, y)
hypotenuse
r
y


x
adjacent side
opposite side
Math130: Section 5.2 – Trigonometric Functions of Angles
The six trigonometric functions - sine, cosine, tangent, cotangent, secant,
and cosecant - are defined as ratios of the lengths of the sides of a right
triangle.

c = 17
Trigonometric Ratios/Functions

a=8

b
Using x, y  notation from the figure above, the six trigonometric ratios
or functions are defined as follows.
sin  
y
opposite

r hypotenuse
csc 
1
r

sin y
cos 
x
adjacent

r hypotenuse
sec 
1
r

cos x
tan  
y sin  opposite


x cos  adjacent
cot  
1
cos x


tan sin y
Note the relationships between the trigonometric ratios.
 cosecant is the reciprocal of sine
 secant is the reciprocal of cosine
 tangent is the ratio of sine divided by cosine
 cotangent is the reciprocal of tangent, which is also cosine
divided by sine
Notation: The common labeling of right triangles is to identify the
right angle as angle  (gamma) with the two acute angles labeled
 (alpha) and  (beta). The hypotenuse, the side opposite the right
angle  , shall be labeled c. The side opposite angle  shall be
labeled a, and the side opposite angle  shall be labeled b.

c

b
a

Example 1: For the given right triangle, find the 6 trig functions for angle  and angle  .
Solution: Before setting up the ratios, it is necessary to find the length of side b. This is accomplished by using
the Pythagorean Theorem.
a2  b2  c2
82  b 2  17 2
b 2  225
b  15
b  15
Choose +15 since b represents a length.
©2014;Dr. B. Shryock
Page 2 of 7
Math130: Section 5.2 – Trigonometric Functions of Angles
For angle 
opp 8
sin  

hyp 17
adj 15
cos 

hyp 17
opp 8
tan  

adj 15
For angle 
opp 15
sin  

hyp 17
adj
8
cos  

hyp 17
opp 15
tan  

adj
8
17
8
17
sec  
15
15
cot  
8
csc  
17
15
17
sec  
8
8
cot  
15
csc  
Notice that the definition of the trigonometric function remains constant regardless of the angle used; however,
the opposite and adjacent sides are redefined relative to the chosen angle. Also notice that sin   cos  and
that sin   cos  . This is due to the fact that  and  are complementary angles and we are dealing with
cofunctions. The topic of cofunction will be covered in Section 6.3.
It is helpful to set up the six trigonometric functions in table form as in the above example for easy computation.
An expedient method to use when working with many problems in trigonometry is to draw a picture of the
situation. It is critical to draw the picture correctly, especially when locating angles in a particular quadrant to
construct a triangle. The text summarizes the signs of the trigonometric functions on page 344 and in Figure 16
on page 345. It helps to remember that the cosine function is directly related to the x  coordinate and will
therefore have the same sign as the x  coordinate has when plotting points in the four quadrants. The same
holds true for the sine function, it is directly related to the y  coordinate.
Since each function shares the same sign as its reciprocal function (sine
and cosecant, cosine and secant, and tangent and cotangent), by
remembering the signs for sine and cosine in the four quadrants and that
tangent is the ratio of sine divided by cosine, you will automatically
know the corresponding sign for secant, cosecant, and cotangent.
Instead of using the Pythagorean identity sin 2   cos2   1, the
problems can also be worked by drawing a picture of the situation. Then
from the picture, it is a matter of using the trigonometric definitions and
reading the values from the picture.
Quadrant II
(x, y)  (-, +)
Quadrant I
(x, y)  (+, +)
Quadrant III
(x, y)  (-, -)
Quadrant IV
(x, y)  (+, -)
cos   0
sin   0
cos   0
sin   0
cos   0
sin   0
cos   0
sin   0
The Pythagorean Identities
The equation of the circle pictured at the bottom of page 1 is x 2  y 2  r 2 .
We know by the definition for sine and cosine that
x
y
cos  
or x  r cos  and sin  
or y  r sin  .
r
r
Substituting into the circle equation, we have
2
2
 r cos    r sin    r 2 cos2   r 2 sin 2   r 2 cos2   sin 2   r 2 .

©2014;Dr. B. Shryock

Page 3 of 7
Math130: Section 5.2 – Trigonometric Functions of Angles
Since r  0, we can divide by r 2 , and the result is the Pythagorean identity sin 2   cos2   1. Notice that if
we solve for either trigonometric term, this identity can be written as sin    1  cos2  or
cos    1  sin 2  .
Dividing sin 2   cos2   1 by sin 2  , the result is as follows.
sin 2   cos 2 
1

2
sin 
sin 2 
1  cot 2   csc 2 
Dividing sin 2   cos2   1 by cos 2  , the result is as follows.
sin 2   cos 2 
1

2
cos 
cos 2 
.
tan   1  sec 
2
2
In summary, the Pythagorean Identities are given below.
cos 2   sin 2   1
1  tan 2   sec 2 
cot 2   1  csc2 
We can work with factoring trigonometric expressions as in algebra.
Example 2: Factor and simplify.
sin 4 x  cos 4 x
Solution: Factoring sin 4 x  cos 4 x is like factoring a 4  b4 in algebra with just a few slight differences.
a 4  b4  (a 2  b 2 )(a 2  b2 )
 (a 2  b 2 )(a  b)(a  b)
sin 4 x  cos 4 x  (sin 2 x  cos 2 x)(sin 2 x  cos 2 x)
 (1)(sin 2 x  cos 2 x)
 sin 2 x  cos 2 x
 (sin x  cos x)(sin x  cos x)
Trigonometric Identities
 Establishing an identity is basically proving that a trigonometric equation is true for all values of the
argument.
 You may work on only one side of the identity to achieve the other given side. Even though you may
not work on both sides, you may certainly look to see what you are trying to prove which may provide you
with clues as to the next step in reaching your goal.
©2014;Dr. B. Shryock
Page 4 of 7
Math130: Section 5.2 – Trigonometric Functions of Angles
 In general, begin by working on the more difficult side.
 Convert to the sine and cosine functions if you do not see another technique to use.
 Be familiar with the basic identities listed on page 338; these are the main tools for verifying other, more
complex identities. You will want to memorize these since they will NOT be provided on the formula
sheet for the test or the final exam.
 Be Patient and Persistent! If things are not working out, sometimes it is necessary to start over with a
different plan, or begin again by working on the other side.
Example 2 on page 336 shows the derivation for the 30, 45, and 60 angles. A chart summarizing the
findings is at the bottom of the same page. Also refer to the Unit Circle you printed for Section 5.1. The sooner
you familiarize yourself with these values, the easier you will find this course. Please watch/re-watch the
Special Angles Video. Try to visualize this information instead of simply memorizing.
Example 3: Evaluate cos 45  sin 30  tan 2 60  .
Solution:


cos 45 sin 30  tan 2 60 
21
 
2 2
 3  

21

  3
2 2 

2  5
 
2  2

2
5 2
4
Pay close attention to your use of notation. tan   tan 2   tan 2
2
Many applications of right triangles involve the angle of elevation and the angle of depression. The angle of
elevation (see Figure 1 below) refers to the angle measured from the horizontal upward to the object in
question. The angle of depression (see Figure 2 below) refers to the angle measured from the horizontal
downward to the object.
object
observer
horizontal
angle of depresession
angle of elevation
observer
Figure 1
horizontal
Figure 2
object
Example 4: A person is standing 60 m from the base of a tower. The angle of elevation to the top of the tower
is 72. To the nearest tenth, how high is the tower?
©2014;Dr. B. Shryock
Page 5 of 7
Math130: Section 5.2 – Trigonometric Functions of Angles
Solution: The right triangle depicting the application is shown below. The height of the tower is labeled x, so
the equation to use is as follows.
x
tan 72 
60
T
x  60 tan 72  184.7
O
x W
E
R
The tower is approximately 184.7 m high.
72°
60 m
Example 5: A shoot from a 3rd floor apartment renovation has an angle of depression of 60. If the top of the
shoot is 11 meters above level ground, how long is the shoot?
Solution: The 60 angle of depression is the top angle in the depiction below. Recall from geometry that if
two parallel lines are cut by a transversal, alternate interior angles are congruent. Since the angle of
depression is 60, the base angle is also 60 because they are alternate interior angles.
11
L
11
L
sin 60
11 22
L

3
3
2
sin 60 
60°
11m
L
60°
The shoot is
22
m long.
3
Notice that this question did not say “round to the nearest …”, so we did not use our calculator and the answer is
given in exact form. Also notice that it is fine to leave answers with a radical in the denominator provided the
6
2 6 2
6
fraction is simplified. For instance,
is not simplified since


3 2.
2
2 2
2
Example 6: Find the exact value of sec if cot   3 and sin   0.
Solution Begin by determining the quadrant of the terminal side of  .
If
cot  is positive, then tan  is positive, and tan  is positive in
x  3
quadrants I and III. sin  is negative in quadrants III and IV.
Hence, the terminal side of  is in quadrant III.
y  1
1 1 y
10
 .
Since  is in quadrant III and if cot   3 , then tan   
3 3 x
Sketch a right triangle in quadrant III as shown below labeling the sides of the triangle. Now find the length
of the hypotenuse and insert that value into your sketch.
©2014;Dr. B. Shryock
Page 6 of 7
Math130: Section 5.2 – Trigonometric Functions of Angles
(3) 2  (1) 2  r 2
9 1  r2
r 2  10
r  10
Recall that sec is the reciprocal of cos  . From the diagram, we see that cos  
sec   
x 3

, hence
r
10
10
.
3
Practice Problems
Be sure to work the Not on WA in the list below, but do not submit them for grading. They could appear on
tests and the final exam. Answers to odd-numbered problems can be found at the end of your text.
Section Problems on WA
5.2
6, 9, 12, 16, 20, 21, 24, 25, 26, 31, 37, 39, 43, 46, 75, 77, 86, 88, 90, 92, 94
©2014;Dr. B. Shryock
Not on WA
55, 57, 59, 63, 65, 67
Page 7 of 7