Chapter 6 Trigonometric Identities and Equations
(picture of someone tuning a piano)
Ever wonder how a piano tuner does his or her job? The quick answer is that they listen for lower
beat frequencies created by what is being tuned and a tuning fork, and they try eliminating that sound
by adjusting the instrument being tuned. The long answer is that there is a lot of math behind what is
going on. If two instruments are making sound at the same time their sound waves are being added.
In math terms the combined amplitude (size of the sound) is represented by:
A  k sin(2 f1t )  k sin(2 f 2t )  k sin(2 f1t )  sin(2 f 2t ) 

 x  y   x  y 
 sin 
  , that you will be learning
 2   2 
By using the trigonometric identity sin x  sin y  2cos 

about in this chapter, we can alter the expression above into something that is equivalent to A.

 2 f1t  2 f 2t   2 f1t  2 f 2t  
A  k sin(2 f1t )  sin(2 f 2t )  k  2cos 
 sin 

2
2

 


  f f     f f  
 2k cos  2  1 2  t  sin  2  1 2  t 
  2     2  

 f1  f 2  
 t  represents the sound whose frequency is the average
 2  
The factor of the expression sin  2 


 f1  f 2  
 t  represents the lower beat frequency that the
 2  
of the two instruments and the factor cos 2 

tuner is trying to eliminate. By the use of a trigonometric identity one can show what is going on and
why, what the tuner is doing, works!
Chapter 6 Trigonometric Identities and Equations
Section 6.1 Trigonometric Identities
Objectives

Understanding the basic identities

Using identities to simplify
In this chapter we are going to look at a lot of identities used to manipulate trigonometric
expressions. As you learn to solve equations and evaluate trigonometric functions you will found that
you will need to change trigonometric functions by using identities in order to be capable of
answering questions.
THE BASIC IDENTITIES
You have seen some of the identities we will be looking at in this chapter. An identity is something
which is true for all inputs. For example, x + 7 = 7 + x is an example of an identity. It is true for all
real numbers. Here is a list of some of the basic trigonometric identities.
Reciprocal Identities
csc u =
1
sin u
sec u =
1
cos u
cot u =
1
tan u
sin u =
1
csc u
cos u =
1
sec u
tan u =
1
cot u
cot u =
cos u
sin u
Quotient Identities
tan u =
sin u
cos u
Pythagorean Identities
cos2 u  sin 2 u  1
1  tan 2 u  sec2 u
cot 2 u  1  csc2 u
Cofunction Identities


sin   u   cos u
2



cos   u   sin u
2



tan   u   cot u
2




cot   u   tan u
2




sec   u   csc u
2



csc   u   sec u
2

You did see all of these identities in chapter 5 but they just weren’t specifically pointed out as such.
They come out of the definitions of the trigonometric functions. Let’s look at where a few of these
come from.
Discussion 1: Proof of Some Identities
Section 6.1 Trigonometric Identities
Let’s investigate how we get some of our identities.
Unit circle definitions of trigonometric
sin u  y , cosu  x , tan u 
functions.
y
, the other three
x
are just the reciprocals of these three.
Let’s look at the quotient identity for tangent.
tan u 
y
by definition.
x
Since sin u  y and cos u  x we substitute and
get; tan u 
sin u
.
cos u
Let’s see where the Pythagorean identities
P (x, y)
come from. Remember that with the unit
1
circle you have a right triangle. If you look at
x
u
y
(1, 0)
the triangle notice that x 2  y 2  1 . Therefore,
use substitution again and we have the first
x2  y 2  1 ,
identity
so,  cos u    sin u   1 which is written as;
sin u  y ,
2
cos u  x
2
cos2 u  sin 2 u  1
To get the other two Pythagorean identities
simply divide the first identity found by
2
2
cos u and then by sin u. Notice
2
2
2

cos u
cos2 u
that 1  tan u  sec u is only true if cos u  0,
2
cos2 u
2
sin u
sin 2 u
2
cos u

sin 2 u
sin 2 u
1

 1  tan 2 u  sec2 u
2
cos u

1
sin 2 u

cot 2 u  1  csc2 u
and cot 2 u  1  csc2 u is only true if sin2u  0.
This is a start at showing you how mathematicians use the definitions and identities to arrive at new
identities and formulas.
Example 1: Two Ways to Answer
What are two ways of answering the question of finding sin θ given that cos θ =
4
and knowing
5
that θ is a third quadrant angle?
Solution:
Method 1: Make a triangle and use the
coordinate definitions (used in section 5.5
example 6).
x = −4
θ
y=?
r=5
cos  
x
y
, sin  
r
r
52 − (4)2 = y2  y2 = 25 − 16 = 9  y =  3
So, for this example y = −3 since we go down
to the point in the third quadrant. Our answer
Chapter 6 Trigonometric Identities and Equations
then is, sin  
Method 2: Use the Pythagorean identity
y 3

r
5
2
cos2   sin 2   1 , and substitute in the
 4 
2

  sin   1
 5 
value we know for cos  . Using this
sin 2   1 
method does leave us at the end wondering
if the answer is positive or negative. But, if
we have memorized ASTC we know that
only tangent and cotangent are positive in
sin 2  
sin  
16
25
9
3
 sin   
25
5
3
5
the third quadrant thus sine must be
negative in this example.
Question 1: What could cos θ be if given that sin θ =
2
?
3
Example 2: Finding Values Using Identities
What are the values of all six trigonometric functions given that cos θ =
9
10 19
and csc θ =
?
10
19
Use identities to calculate your answers.
Solution:
We need to think about
sin u =
1
sin u
cos u
1
, tan u =
, cot u =
, sec u =
csc u
cos u
sin u
cos u
cos θ =
9
,
10
which identities will help us.
Since we have csc θ we could
use the reciprocal identity to
find sin θ. We would then
sin  
know sine and cosine so we
then we could use the
sin 
tan  

cos
reciprocal identities to find
cot θ and sec θ.
cot  
sec 
SIMPLIFING WITH IDENTITIES
10 19
, so
19
1
1
19
 19



csc 10 19 10 19
10
19
could use the quotient
identity to find tan θ. And
csc θ =
 19
10   19  10   19
9
10
9
9
10
1
1
9
9 19



tan 
19
 19
19
9
1
1 10


9
cos
9
10
Section 6.1 Trigonometric Identities
Discussion 2: Simplifying Expressions
Simplify the following expression by using identities.
a) cot x sin x
b) cos 2 x (sec 2 x − 1)
c) (1 − sin 2 x) sec 2 x
a) Convert cot x into sines and cosines by a
cot x sin x =
quotient identity.
b) Convert (sec 2 x − 1) into tan 2 x by a
cos x
sin x = cos x
sin x
cos 2 x (sec 2 x − 1) = cos 2 x tan 2 x
Pythagorean identity. Then, convert
= cos2 x
tangent by using a quotient identity.
sin 2 x
cos2 x
= sin 2 x
c) Convert (1 − sin 2 x) into cos 2 x by a
(1 − sin 2 x) sec 2 x = cos 2 x sec 2 x
Pythagorean identity. Then, convert
= cos2 x
secant by using a reciprocal identity.
1
cos2 x
=1
Question 2: Can
1  cos 2 x
be simplified, and if so how?
1  cos x
In the last question you might have been tempted to replace (1 − cos 2 x) with sin 2 x but that doesn’t
lead you anywhere. Factoring the numerator as a difference of two squares is the key. We can’t
forget our algebra. We need it here in trigonometry too.
Example 3: Simplify Expressions
What are the simplified versions of the following?
a)

1
csc x  1
2

b) sec     cos θ
2

c) sec 2 x tan 2 x + sec 2 x
Solution:
a) Use a Pythagorean identity then
reciprocal identity
b) Use a cofunction identity then reciprocal
then quotient.
c) This time we need to factor then use a
Pythagorean identity.
1
csc2 x  1
=
1
cot 2 x
= tan 2 x


   cos θ = csc θ cos θ
2

sec 
=
1
cos θ
sin 
=
cos 
 cot 
sin 
sec 2 x tan 2 x + sec 2 x = sec 2 x (tan 2 x + 1)
= sec 2 x sec 2 x
= sec 4 x
Chapter 6 Trigonometric Identities and Equations
Section Summary:
Be sure to memorize your identities.
Reciprocal Identities
Answer Q1:
cos2   sin 2   1
2
 2 
cos 2   
 1
 3 
4
cos 2    1
9
5
cos 2  
9
cos   
5
9
5
,
3
which answer
will depend on
whether θ is in
the third or
fourth
quadrants.

csc u =
1
sin u
sec u =
1
cos u
cot u =
1
tan u
sin u =
1
csc u
cos u =
1
sec u
tan u =
1
cot u
cot u =
cos u
sin u
Quotient Identities
tan u =
sin u
cos u
Pythagorean Identities
cos2 u  sin 2 u  1
1  tan 2 u  sec2 u
Cofunction Identities


sin   u   cos u
2



cos   u   sin u
2



tan   u   cot u
2



cot   u   tan u
2



sec   u   csc u
2



csc   u   sec u
2

cot 2 u  1  csc2 u
Section 6.1 Trigonometric Identities
SECTION 6.1 PRACTICE SET
(1-- 4) Prove the following identities using the unit circle definition.
1. sec  
1
csc 
2. sec  
1
cos 
3. cot  
1
tan 
4. cot  
cos 
sin 
(530) (If no answers are possible explain why)
5
5. Given sin  
in the 1st quadrant find the other five trigonometric function values.
13
6. Given cos  
3
in the 1st quadrant find the other five trigonometric function values.
5
7. Given cos  
5
in the 3rd quadrant find the other five trigonometric function values.
13
8. Given sin  
3
in the 4th quadrant find the other five trigonometric function values.
5
9. Given tan  
5
in the 1st quadrant find the other five trigonometric function values.
12
10. Given cot  
4
in the 1st quadrant find the other five trigonometric function values.
3
11. Given cot  
5
in the 2nd quadrant find the other five trigonometric function values.
12
12. Given tan  
4
in the 4th quadrant find the other five trigonometric function values.
3
13. sec  
13
in the 1st quadrant find the other five trigonometric function values.
12
14. Given csc  
5
in the 1st quadrant find the other five trigonometric function values.
4
15. Given csc  
13
in the 3rd quadrant find the other five trigonometric function values.
12
16. Given sec 
5
in the 2nd quadrant find the other five trigonometric function values.
4
Answer Q2:
1  cos 2 x
=
1  cos x
(1  cos x)(1  cos x)
1  cos x
= (1  cos x)
The difference
of two squares.
Chapter 6 Trigonometric Identities and Equations
17. Given sin  
2
in the second quadrant find the other five trigonometric function values.
5
18. Given cos  
3
in the 4th quadrant find the other five trigonometric function values.
8
19. Given tan  
5
in the 3rd quadrant find the other five trigonometric function values.
8
20. Given cot  
5
in the 3rd quadrant find the other five trigonometric function values.
8
21. Given sec 
5
in the 3rd quadrant find the other five trigonometric function values.
2
22. Given csc  
6
in the 3rd quadrant find the other five trigonometric function values.
5
23. Given sin  
3
in the 3rd quadrant find the other five trigonometric function values.
8
24. Given cos  
3
in the 4th quadrant find the other five trigonometric function values.
8
25. Given tan  
3
in the 3rd quadrant find the other five trigonometric function values.
4
26. Given cot  
4
in the 2nd quadrant find the other five trigonometric function values.
5
27. Given sec  
4
in the 1st quadrant find the other five trigonometric function values.
5
28. Given csc  
5
in the second quadrant find the other five trigonometric function values.
13
29. Given sin  
13
in the 2nd quadrant find the other five trigonometric function values.
5
30. Given cos  
4
in the 4th quadrant find the other five trigonometric function values.
3
(3136) Using trigonometric identities find the values of the other four trigonometric
functions given the values of two of the trigonometric functions.
31. sin  
5
 11
and cos  
6
6
32. cos  
4
 5
and sin 
9
3
Section 6.1 Trigonometric Identities
33. sin  
5
12 119
and sec 
12
119
34. cos  
3
7 10
and csc 
7
20
35. cos  
3
8 55
and sec 
8
55
36. sin  
3
7 10
and sec 
7
20
(3758) Simplify each of the following:
37. tan  cos 
38. cot  sin 
39. cos  csc
40. sin  sec 
41. cos  sec
42. sin  csc 
43.
46.
49.
52.
sin  tan 
cos 
44.
30sin 3  cos 2 
10 cos  sin 
3
2
sin 2   1
sin   1
sin 4   cos4 
sin 2   cos2 


55. sin     csc 
2



58. tan     tan 
2

47.
50.
cos  cot 
sin 
sin 2 
cos 
2
 sin  csc 
cos 2   1
cos   1
2
45.
48.
51.
sin 2  cos 
cos 2  sin 
1
tan 2 
 cot  tan 
cos4   sin 4 
cos2   sin 2 
2
1
 cos  
53. 
  2
sin


 sin 
1
 sin  
54. 
 
cos


 cos 2 


56. cos     sec 
2



57. cot     cot 
2

Chapter 6 Trigonometric Identities and Equations
Section 6.2 Verifying Identities
Objectives

Understanding how to verify identities
In this section you will get a chance to practice your problem solving skills and you will get a better
understanding of how the trigonometric functions relate to each other. Verifying identities will
require that you know all of your identities well. When you verify an identity you are checking to see
if two different expressions are equivalent.
Caution, this is different from solving equations. You do not want to move terms from one side of
the equal sign to the other or multiply both sides of the equation by something. You must work with
only one side of the equation at a time.
IDENTITY VERIFICATION
Let’s begin with an example.
Discussion 1: Verifying Identities
Verify the identity
sec2 x  1
sec2 x
 sin 2 x .
We might begin by working with the left side since it is
the more complicated looking of the two. When verifying
sec2 x  1
sec2 x
 sin 2 x
identities there can be different ways of proving them
equivalent.
One way of starting this example might be to change the
tan 2 x
sec 2 x − 1 by using a Pythagorean identity.
sec2 x
Next we might eliminate the fraction by changing sec 2 x
 sin 2 x
tan 2 x cos2 x  sin 2 x
by using a reciprocal identity.
Now, change tan 2 x into sines and cosines and simplify
sin 2 x
(quotient identity).
cos2 x
cos2 x  sin 2 x
sin 2 x  sin 2 x
This isn’t the only way to approach this example. Here is another way to work the problem.
As above we begin by working with the left side since it is
sec2 x  1
2
the more complicated looking one of the two.
We now break the fraction up into two fractions.
 sin 2 x
sec x
sec2 x
sec2 x

1
sec2 x
 sin 2 x
Section 6.1 Trigonometric Identities
Next we simplify each fraction.
1  cos2 x  sin 2 x
Now, convert 1  cos2 x , by using a Pythagorean identity.
sin 2 x  sin 2 x
Let’s talk about some guidelines that can be helpful when trying to verify identities.
1) You must have your basic identities memorized!
2) You should work with the more complicated looking side first. Remember that you
can’t move terms from one side to the other or multiply both sides by something.
3) Typically, you will want to add fractions together, simplify fractions so that they have
monomials in the denominator, and/or factor when possible.
4) Look for opportunities to use trigonometric identities to get functions that are the same
or that are paired up like sine and cosine, or tangent and secant, or cotangent and
cosecant or that are paired up with the other side of the identity.
5) Another strategy might be to convert everything to sines and cosines.
6) You may want to multiply the numerators and denominators of fractions by something
in order to create the difference of two squares like multiplying (1 + sin x) by (1 − sin x)
to get (1 − sin 2 x) which equals cos 2 x.
7) If nothing comes to mind just try something. It may lead somewhere or it might not but
either way you will gain some insight about how to verify the identity.
Question 1: What would you want to try first when verifying
tan 2 x  1
 sec x ?
sec x
Example 1: Verifying Identities
Verify that the following are identities.
a)
tan 2 x  1
 sec x
sec x
b)
sin 2 
 sec  cos 
cos 
c)
cot 2 t  1
1  cot 2 t
Solution:
a) Here we would like to change tan 2 x + 1 so that
everything will be in secants.
Now canceling out the like factor secant, and
you are done.
b) Here it is hard to say which is more complicated
but let’s try putting the right side into one term
by subtracting them. First, we would need to
change secant by the reciprocal identity.
tan 2 x  1
 sec x
sec x
sec 2 x
 sec x
sec x
sec x  sec x
sin 2 
 sec  cos 
cos 
sin 2 
1

 cos 
cos 
cos 
 1  2sin 2 t
Chapter 6 Trigonometric Identities and Equations
Next, get common denominators and subtract.
sin 2 
1
cos 2  1  cos 2 



cos
cos
cos
cos
Now convert the numerator by a Pythagorean
sin 2  sin 2 

cos
cos
identity.
c) On this problem let’s try manipulating the left
cot 2 t  1
side. We should get a single term in the
1  cot 2 t
denominator by replacing (1 + cot2 t) with csc2t.
cot 2 t  1
csc2 t
cot
Now let’s use a reciprocal identity to change
2
2
 1  2sin 2 t
 1  2sin 2 t

t  1 sin 2 t  1  2sin 2 t
csc t.
Next, distribute and replace cot 2 t with
cos 2 t
sin 2 t
cos2 t
sin 2 t
.
sin 2 t  sin 2 t  1  2sin 2 t
cos2 t  sin 2 t  1  2sin 2 t
The sin 2t canceled out
1  sin t   sin
Lastly use a Pythagorean identity to convert
2
cos 2 t to (1 − sin 2 t) and then simplify.
2
t  1  2sin 2 t
1  2sin 2 t  1  2sin 2 t
Example 2: Working with both sides
Verify that
sin 
sin  cos

 csc (1  cos 2  ) is an identity.
1  cos
1  cos
Solution:
It might be easier if we change all
trigonometric functions to sines and
sin 
sin  cos

 csc θ(1 + cos 2 θ)
1  cos
1  cos
cosines.
Thus, replace csc θ with
1
.
sin 
Next, let’s work with the left side.
Let’s use hint number 6.
(1  cos 2  )
sin 
sin  cos


sin 
1  cos
1  cos
 1  cos  sin 
 1  cos



 1  cos  1  cos  1  cos
sin   sin  cos
1  cos2 

 sin  cos 1  cos 2 


sin 
 1  cos
sin  cos  sin  cos2 
1  cos2 
Combine the two fractions.
sin   sin  cos  sin  cos  sin  cos2 
Simplify.
sin   sin  cos2 
1  cos2 
1  cos 
2
Factor and replace (1−cos2θ) with
sin 2 θ.
Now cancel the sines.

1  cos2 
sin 
sin  (1  cos 2  ) 1  cos 2 

sin 
sin 2 


1  cos2 
sin 
1  cos 2 
sin 
Section 6.1 Trigonometric Identities
(1  cos 2  ) 1  cos 2 

sin 
sin 
Here is an example of one area where trigonometric functions show up in electronics.
Example 3: Capacitors and Inductors
The energy stored in a capacitor is given by the function C(t) = k cos 2 (2πFt) and the energy stored in
an inductor is given by I(t) = k sin 2 (2πFt) where t is time and F is the frequency. The total energy in
a circuit is given by E(t) = C(t) + I(t). Prove that E(t) must be a constant.
Solution:
Let’s begin by substituting into the formula for E(t).
E(t) =
C(t)
+
I(t)
E(t) = k cos 2 (2πFt) + k sin 2 (2πFt)
Now factor out the k.
E(t) = k [cos 2 (2πFt) + sin 2 (2πFt)]
cos 2 (u) + sin 2 (u ) = 1, so;
E(t) = k (1) = k
a constant.
Section Summary:
Hints for verifying identities:
1) You must have your basic identities memorized!
2) You should work with the more complicated looking side first. Remember that you
can’t move terms from one side to the other or multiply both sides by something.
3) Typically, you will want to add fractions together, simplify fractions so that they have
monomials in the denominator, and factor when possible.
4) Look for opportunities to use trigonometric identities to get functions that are the same
or that are paired up like sine and cosine, or tangent and secant, or cotangent and
cosecant or that are paired up with the other side of the identity.
5) Another strategy might be to convert everything to sines and cosines.
6) You may want to multiply the numerators and denominators of fractions by something
in order to create the difference of two squares like multiplying (1 + sin x) by (1 − sin x)
to get (1 − sin 2 x) which equals cos 2 x.
7) If nothing comes to mind just try something. It may lead somewhere or it might not but
either way you will gain some insight about how to verify the identity.
Answer Q1:
You would
want to
change
tan 2x + 1 into
sec2x.
Chapter 6 Trigonometric Identities and Equations
SECTION 6.2 PRACTICE SET
(1– 64) Prove the following identities are true:
1. sec cos   cot 
2. csc  sin   tan 
3. cos  tan   cot    csc
4. sin   cot   tan    sec
5.
cot 
 cos 
csc 
6.
tan 
 sin 
sec 
7. sin 2   cot 2   cos2   csc2 
8. sin 2   tan 2   cos2   sec2 
9. tan   cos  cot    sin   1
10. cot  sin   tan    cos  1
11.
1  tan 2 
 sec 
sec 
12.
1  cot 2 
 csc 
csc 
13.
1  tan  cot   1

1  tan  cot   1
14.
csc   1 1  sin 

csc   1 1  sin 
15.
1  sin  csc   1

1  sin  csc   1
16.
cos   1 1  sec 

cos   1 1  sec 
17.
1
1

 2 csc 2 
1  cos  1  cos 
18.
1
1

 2sec 2 
1  sin  1  sin 
19.
21.
1  sin  2
cos 
2
cos   1
tan 
2


1  sin 
1  sin 
cos 
sec   1
20.
22.
sin 2 
1  cos  
2
sin   1
cot 
2


1  cos 
1  cos 
sin 
csc   1
23.
1  cos 2 
 sin 
1  sin 
24.
1  sin 2 
  cos 
1  cos 
25.
1
 sec   tan 
sec   tan 
26.
1
 csc   cot 
csc   cot 
27.
1
1

 2sec 2 
1  sin  1  sin 
28.
1
1

 2 csc 2 
1  cos  1  cos 
29.
1  sin 4 
1  sin 
2
 cos 2 
31. sec4   sec2   tan 4   tan 2 
30.
1  cos 4 
1  cos 2 
 sin 2 
32. csc4   csc2   cot 4   cot 2 
Section 6.1 Trigonometric Identities
33. sec4   tan 4  
39.
41.
cos 
2
34. csc4   cot 4  
cos 
 sec 
1  sin 
36.
tan   cot 
 sin 2   cos 2 
tan   cot 
38.
35. tan  
37.
1  sin 2 
1  tan 2 
1  tan 
2
 1  2cos 2 
sin 2   tan 
cos 2   cot 
 tan 2 
sin  cos 
cos 2   sin 2 
sin 2 
tan 
1  tan 2 
tan   cot 
 1  2sin 2 
tan   cot 
40. 1 
42.

1  cos 2 
1  cot 2 
1  cot 
2
 2cos 2 
cos 2   cot 
sin 2   tan 
 cot 2 
43.
1  sin 
 (sec   tan  ) 2
1  sin 
44.
1  cos 
2
  cos   cot  
1  cos 
45.
1  cos  1  cos 

 4 cot  csc 
1  cos  1  cos 
46.
1  sin  1  sin 

 4 tan  sec 
1  sin  1  sin 
47.
1
1

 2 tan 
tan   sec  tan   sec 
48.
1
1

 2 cot 
cot   csc  cot   csc 
49.
sin 3   cos3 
 1  sin  cos 
sin   cos 
50.
sin 3   cos3 
 sin   cos 
1  sin  cos 
51. sin 4   cos4   1  2cos2 
52. cos4   sin 4   1  2sin 2 
53.
1  sin   cos 
 csc   cot 
1  sin   cos 
54.
1  sin   cos 
 sec   tan 
1  cos   sin 
55.
1  2sin 2 
 cot   tan 
sin  cos 
56.
1  2cos 2 
 tan   cot 
sin  cos 
57.
tan   tan x
 tan  tan x
cot   cot x
58.
cot   cot x
 cot  cot x
tan   tan x
59. sin 2  sec2   sin 2  csc2   sec2 
61.
63.
cot 2   1
60. cos2  sec2   cos2  csc2   csc2 
 1  2sin 2 
62.
sin 
cos 

 sin   cos 
1  cot  1  tan 
64.
1  cot 
2
tan 2   1
1  tan 
2
 2sin 2   1
sin 
cos 
1


1  cot  1  tan  sin   cos 
(65−70) Graph each of the following to show they are identities, then prove they are identities.
Chapter 6 Trigonometric Identities and Equations
65.
 csc  1 csc  1  cot 2 
67. csc   sin   cot  cos 
69.
1
1

 2sec 2 
1  sin  1  sin 
66.
sec  1sec  1  tan 2 
68. sec   cos   tan  sin 
70.
1
1

 2 csc 2 
1  cos  1  cos 
(71−76) Show each of the following are not identities by finding an angle value for  that makes the
statement false. (Check by using the graphing calculator)
71. sin   cos   1
73. sin  
75.
1
sec 
sin 2   cos 2   sin   cos 
72. tan   1  sec
74. cos  
76.
1
csc 
1  sin 2   cos 
(77−82) Show each of the following are not true by finding an angle values for x and y so
the statement is false.
77. sin  x  y   sin x  sin y
78. cos  x  y   cos x  cos y
79. sin  x  y   sin x  sin y
80. cos  x  y   cos x  cos y
81. tan  x  y   tan x  tan y
82. tan  x  y   tan x  tan y
Section 6.1 Trigonometric Identities
Section 6.3 More Identities
Objectives

Understanding more complicated identities
In this section we are going to look at more trigonometric identities. Some are used more frequently
than others, but you will need them as you study more mathematics.
MORE COMPLICATED IDENTITIES
First we will begin with identities involving the addition and subtraction of angles.
Sum and Difference Identities
sin(u  v)  sin u cos v  cos u sin v
sin(u  v)  sin u cos v  cos u sin v
cos(u  v)  cos u cos v  sin u sin v
cos(u  v)  cos u cos v  sin u sin v
tan(u  v) 
tan u  tan v
1  tan u tan v
tan(u  v) 
tan u  tan v
1  tan u tan v
Let’s take some time here to prove one of the above identities.
C(cos (u−v), sin (u−v))
Since arc AC is equivalent to arc BD (both arcs have the same
angle (u − v)), line segment AC is equal in length to BD . Therefore,
u
D(cos u, sin u)
B(cos v, sin v)
u−v
v
A(1, 0)
we can create the following equation and result.
Distance
formula
(cos(u  v)  1) 2  (sin(u  v)  0) 2  (cos u  cos v) 2  (sin u  sin v) 2
Square both
sides
(cos(u  v)  1)2  (sin(u  v)  0) 2  (cos u  cos v) 2  (sin u  sin v) 2
Square terms
sin2x+cos2x=1
subtract 2
divide by −2
cos 2 (u  v)  2cos(u  v)  1  sin 2 (u  v)  cos 2 u  2cos u cos v  cos 2 v  sin 2 u  2sin u sin v  sin 2 v
2cos(u  v)  1  1  2cos u cos v  2sin u sin v  1  1
2cos(u  v)  2cos u cos v  2sin u sin v
cos(u  v)  cos u cos v  sin u sin v
The other formulas can be proven as well but we are not going to do that here. Here is an example of
one way in which these were used before calculators came on the scene.
Example 1: Finding Trigonometric Values
Chapter 6 Trigonometric Identities and Equations
Find the cos

.
12
Solution:
We need to convert this to angles we know
like
   
, , , and any multiple of these.
6 4 3 2
cos
Now, we can use a difference identity and

  
= cos   
12
3 4
   
   
 3 4 12 
  
 
 
   
  = cos   cos    sin   sin  
3
4
3
4


 
 
3 4
cos 

determine the value of cos .
12
 1   2   3  2 
=  



 2   2   2 
 2 
 2  6

 4   4 

 

=

2 6
 0.9659
4
Example 2: Algebraic Use


What would sin cos 1 x  sin 1 x simplify too? Notice that this example fits here because inverse
trigonometric functions yield angle answers. This is the sine of the sum of two angles.
Solution:
sin(u  v)  sin u cos v  cos u sin v
Sum identity for sine.
    
x)  sin  cos x  cos  sin x   x  x 
 
We now have:
sin(cos1 x  sin 1 x)  sin cos1 x cos sin 1 x  cos cos1 x sin sin 1 x
Simplify the inverses.
sin(cos1 x  sin 1
Now we need a triangle to
help us with the other two


parts. Thus, sin cos 1 x =
sin θ = 1  x2 and

1
r=1
θ
y=
1
r=1
1  x2
α
x=x
cos −1 x = θ
x=
sin −1 x = α

cos sin 1 x = cos α
= 1  x2 .
sin(cos1 x  sin 1 x)  1  x2 1  x2  x  x 
Perform the multiplication.
sin(cos 1 x  sin 1 x)  1  x 2  x 2
Simplify.
sin(cos 1 x  sin 1 x)  1
Here are some more identities that show up in calculus and other courses.
y=x
1  x2

Section 6.1 Trigonometric Identities
Double Angle
and
Power Reducing Identities
sin 2 u 
sin(2u )  2sin u cos u
cos(2u )  cos 2 u  sin 2 u
 2cos 2 u  1
cos 2 u 
1  cos  2u 
2
1  cos  2u 
2
 1  2sin 2 u
tan(2u) 
2tan u
tan 2 u 
1  tan u
2
1  cos  2u 
1  cos  2u 
Discussion 1: Proving Some Identities
Let’s show how some of the above identities are derived.
sin(2u )  2sin u cos u
Let’s start by looking at sin 2u. We will
verify the identity.
sin(u  u )  2sin u cos u
First, let’s rewrite the left side.
Second, use the sum identity.
sin u cos u  cos u sin u  2sin u cos u
2sin u cos u  2sin u cos u
Lastly, combine like terms.
cos 2u is verified similarly, so let’s assume
tan(2u) 
it’s an identity and verify tan 2u.
First, let’s replace tan 2u by a quotient
identities.
Divide the numerator and the denominator
1  tan 2 u
sin(2u )
2 tan u

cos  2u  1  tan 2 u
identity.
Now, substitute with a double angle
2tan u
2sin u cos u
cos u  sin u
2
2

2tan u
1  tan 2 u
2sin u cos u
by cos 2 u.
2 tan u
cos 2 u

2
cos u sin u 1  tan 2 u

cos 2 u cos 2 u
Simplify.
2sin u
cos u  2 tan u
sin 2 u 1  tan 2 u
1
cos 2 u
Lastly,
2
sin u
 tan u
cos u
2tan u
1  tan u
2
Let’s do one more and verify the identity
cos 2 u 
1  cos  2u 
2
cos 2 u 
2tan u
1  tan 2 u
1  cos  2u 
2
.
To start we will use the cos 2u identity on
the right side.

cos2 u 

1  cos2 u  sin 2 u
2

Chapter 6 Trigonometric Identities and Equations
Now, replace sin 2 u with a Pythagorean
cos2 u 
identity.
cos 2 u 
Simplify the numerator.
Finally, we are done.

1  cos2 u  1  cos2 u

2
1  cos 2 u  1  cos 2 u 2cos 2 u

2
2
cos2 u  cos2 u
Question 1: Verify the identity tan 2 u 
1  cos  2u 
1  cos  2u 
.
Example 3: Simplifying a Double Angle
cos (2cos −1 x)
Find the following.
Solution:
Double angle identity.
cos(2u )  cos 2 u  sin 2 u





cos(2cos1 x)  cos2 cos1 x  sin 2 cos1 x
We now have:

2


cos(2cos1 x)  cos cos1 x   sin cos1 x 

 


Now we need a triangle to

1

2
cos(2cos1 x)  x  sin cos1 x 


Simplify the inverses.

r=1
θ
help us with, sin cos x =
y=
2
2
1  x2
x=x
sin θ = 1  x2 .
cos −1 x = θ
Perform the multiplication.
2
cos(2cos 1 x)   x    1  x 2 



cos(2cos 1 x)  x 2  1  x 2
Simplify.
2

cos(2cos 1 x)  2 x 2  1
Of course, we have more identities. Let’s look at what are called half angle identities.
Question 2: What do you think could be the right side of this identity
(Hint: The power reduction formula is a good start.)
u
sin   = _______?
2
 
Section 6.1 Trigonometric Identities
 u
Discussion 2: Discovering the Identity for sin  
 2
Let’s look at how we could have answered question 2.
We will start with the power reduction
2
sin 2 u 
1  cos  2u 
identity for sin u.
Let’s take the square root of both sides.
We’re close, now just replace u with
u
.
2
Simplify and we’re done.
sin u  
u
sin    
2
2
1  cos  2u 
2
 u
1  cos  2 
 2
2
1  cos u
u
sin    
2
2
The other half angle identities can be verified easily. Here are all three of them.
Half Angle Identities
1  cos u
u
sin    
2
2
1  cos u
u
cos    
2
2
sin u
 u  1  cos u
tan   

sin u
1  cos u
2
With the sine or cosine of half an angle, the + or − symbol is determined by which quadrant the angle
u
is located. Here are a couple of examples.
2
Example 4: Using the Half Angle Identities
 
 7 
 , and cos 
 exactly?
 12 
 12 
What are the values of sin 
Solution:
 
6
 
The sin   = sin   which means that
2
 12 
the half angle identity can help us.

6
We evaluate cos 

.

Simplify the square root.
1  cos u
u
sin    
2
2
 
 
 
1  cos  
 
6
6
sin      
 2 
2




 
sin    
 12 
3
2
1
2
2 3
 
sin    
12
4
 
2 3
 
sin    
2
 12 
Chapter 6 Trigonometric Identities and Equations
 
 is
 12 
The final result is positive because 
Answer Q1:
1  cos  2u 
tan 2 u 
1  cos  2u 
sin 2 u
 
sin   
 12 
2 3
2
in the first quadrant where sine is positive.
 7 
 6 
 7 
The cos   = cos   which means
2
 12 

cos 2 u
1  cos(2u )
2

1  cos(2u )
2
1  cos(2u )

1  cos(2u )
1  cos u
u
cos    
2
2
  7

6
cos  
 2


that the half angle identity can help us.
 7
 6
We evaluate cos 

 7 
1  cos 



 6 

2


 7 
cos 

 12 

.

Simplify the square root.
 7 
The final result is negative because   is
 12 
in the second quadrant where cosine is
1
 3
2
2
 7
cos 
 12
2 3


4

 7
cos 
 12
2 3


2

 7
cos 
 12
2 3


2

negative.
Question 3: Why did the two answers to example 4 differ by only a negative sign?
Example 5: Double and Half Angles
Given sin θ =
 
 
3
3
and that  < θ <
, find sin 2θ, cos 2θ, sin   , and cos   .
5
2
2
2
Solution:
For finding sin 2θ, we will use a
x = 4
double angle formula along with a
Answer Q2:
 v
1  cos  2 
v
 2
sin 2  
2
2
sin 2
1  cos  v 
v

2
2
picture of what is given and our
knowledge that θ is in the third
quadrant.
y = −3
θ
r=5
sin 2θ = 2 sin θ cos θ
24
 3  4 
= 2    =
5
5
25
  
As in the first part we will find cos 2θ
cos 2θ  1  2sin 2 
 3 
= 1  2 
 5 
2
Section 6.1 Trigonometric Identities
=1
 
Now we find sin   . Since θ is in
2
1  cos
 
sin    
2
2
 
the third quadrant half of θ will be in
either the first or second quadrants.
=
Sine yields positive values in those
two quadrants.
 
 
As with sin   we find cos   .
2
2
18 7
=
25 25
=
(sign depends on quadrant)
 4 
9
1  
 
5
  = 5
2
2
9
3
3 10
or

10
10
10
1  cos
 
cos    
2
2
 4 
1
1  
 
5
  = 5
=
2
2
Since the reference angle in quadrant
three is less than 45° (the magnitude
of x is bigger than the magnitude of y)
half of θ will put the angle in the first
=
1
1
10

or
10
10
10
quadrant where cosine yields positive
values.
Lastly, we have these identities.
Product to Sum
sin u sin v 
1
 cos(u  v)  cos(u  v) 
2
cos u cos v 
1
 cos(u  v)  cos(u  v) 
2
sin u cos v 
1
 sin(u  v)  sin(u  v) 
2
cos u sin v 
1
 sin(u  v)  sin(u  v) 
2
Sum to Product
uv
u v
sin u  sin v  2sin 
 cos 

 2 
 2 
uv uv
sin u  sin v  2cos 
 sin 

 2   2 
uv
u v
cos u  cos v  2cos 
 cos 

 2 
 2 
u v u v
cos u  cos v  2sin 
 sin 

 2   2 
Example 6: Using Product Sum Identities
Convert the following products to sums and sums to products.
a) sin θ cos 2θ
b) cos 3θ + cos θ
Solution:
a) We need the formula for the product of
sin u cos v 
sine and cosine.
Substitute in our angles.
sin  cos 2 
1
 sin(u  v)  sin(u  v) 
2
1
 sin(  2 )  sin(  2 ) 
2
Chapter 6 Trigonometric Identities and Equations
Simplify the right side.
sin  cos 2 
1
1
sin(3 )  sin(  )
2
2
Sine is an odd function.
sin  cos 2 
1
1
sin(3 )  sin( )
2
2
b) We need the formula for the sum of two
cosines
Substitute in our angles.
 3   
 3   
cos3  cos  2cos 
 cos 

2


 2 
Simplify the right side.
 4
cos3  cos  2cos 
 2
Example 7: Using Identities
Simplify the following by using an identity.
a) cos5x cos2 x  sin5x sin 2 x
 5 
 12  are
Solution:


complementary.
a) This looks like one of the sum or
 
Thus sin  
difference identities. In fact it is the
 12 
 5 
and cos 

 12 
are equal. Now
 7 
 12  has a


reference angle
 5 
of 
 so
 12 
that is why we
get the same
answers just
opposite signs.

 2 
 cos 


 2 
cos3  cos  2cos  2  cos  
Fractions simplify.
Answer Q3:
The angles
 
  and
 12 
uv
u v
cos u  cos v  2cos 
 cos 

 2 
 2 
cosine of the difference of two angles,
 7x   x 
 sin  
 2  2
b) 2sin 
cos5x cos2 x  sin5x sin 2 x
cos(5 x  2 x)
cos3x
thus we get:
b) This looks like one of the sum to product
identities. In fact it is the cosine minus
 7x   x 
2sin 
 sin  
 2  2
 4 x  3x   4 x  3x 
2sin 
 sin 

 2   2 
cosine, thus we get:
cos 4 x  cos3x
Section Summary:
Identities:
Sum and Difference Identities
sin(u  v)  sin u cos v  cos u sin v
sin(u  v)  sin u cos v  cos u sin v
cos(u  v)  cos u cos v  sin u sin v
cos(u  v)  cos u cos v  sin u sin v
tan(u  v) 
tan u  tan v
1  tan u tan v
Double Angle
sin(2u )  2sin u cos u
tan(u  v) 
and
tan u  tan v
1  tan u tan v
Power Reducing Identities
sin 2 u 
1  cos  2u 
2
Section 6.1 Trigonometric Identities
cos(2u )  cos 2 u  sin 2 u
 2cos 2 u  1
cos 2 u 
1  cos  2u 
2
 1  2sin 2 u
tan(2u) 
2tan u
tan 2 u 
1  tan u
2
1  cos  2u 
1  cos  2u 
Half Angle Identities
1  cos u
u
sin    
2
2
1  cos u
u
cos    
2
2
sin u
 u  1  cos u
tan   

sin u
1  cos u
2
Product to Sum
sin u sin v 
1
 cos(u  v)  cos(u  v) 
2
cos u cos v 
1
 cos(u  v)  cos(u  v) 
2
sin u cos v 
1
 sin(u  v)  sin(u  v) 
2
cos u sin v 
1
 sin(u  v)  sin(u  v) 
2
Sum to Product
uv
u v
sin u  sin v  2sin 
 cos 

 2 
 2 
uv uv
sin u  sin v  2cos 
 sin 

 2   2 
uv
u v
cos u  cos v  2cos 
 cos 

2


 2 
u v u v
cos u  cos v  2sin 
 sin 

 2   2 
Chapter 6 Trigonometric Identities and Equations
SECTION 6.3 PRACTICE SET
(1–12) Use the sum and difference identities to evaluate exactly the following:
 5 
3. cos 

 12 
1. sin150o
 
2. sin  
 12 
4. cos 75o
 13 
5. sin 

 12 
7. cos 285o
 19 
8. cos 

 12 
10. tan105o
11. tan195o
r
r
r
6. sin195o
 7 
9. tan 

 12 
r
r
 13 
12. tan 

 12 
r
(13–26) Show each of the following is true using sum and difference identities.




13. cos(  2 )  cos 
14. sin   360  sin 
15. sin   90o   cos 


16. cos      sin 
2



17. sin      cos 
2

18. cos   90o   sin 
21. cos       cos


20. sin       sin 
22. sin 180o    sin 


23. sin 270o     cos 
3

25. sin   
2

26. cos   270o   sin 
19. cos   180o   cos 

  cos 





 3

24. cos 
     sin 
 2

(27–36) Use the sum and difference identities in reverse to rewrite each of the following as a single
trigonometric function.
27. sin 5 x cos 3x  cos 5 x sin 3 x
28. sin 3x cos 7 x  cos 3x sin 7 x
29. cos 4 x cos 7 x  sin 4 x sin 3x
30. cos 9 x cos 2 x  sin 9 x sin 2 x
31. sin 5 x cos 3x  cos 5 x sin 3 x
32. sin 3x cos 7 x  cos 3x sin 7 x
33. cos 4 x cos 3x  sin 4 x sin 3x
34. cos 9 x cos 2 x  sin 9 x sin 2 x
35.
tan 5 x  tan 3 x
1  tan 5 x tan 3 x
36.
tan 5 x  tan 3 x
1  tan 5 x tan 3 x
Section 6.1 Trigonometric Identities
(37–44) Use the given information to find exact values of:
 
 
a.) sin 2 b.) cos 2 c.) sin   d.) cos  
2
2
37. sin  
4
5
39. cos  
5
13
41. sin  
12
13
43. cos  
3
5
0  

2
0  

2
  

2
3
2
 

38. sin  
4
5
40. cos  
5
13
42. sin  
12
13
44. cos  
3
5
2
 
3
   2
2
3
   2
2
  
3
2
(45–48) Use the given information to find exact values of:
 
a.) tan 2 b. tan  
2
45. tan  
3
4
47. tan  
5
12
0  

2

2
 
46. tan  
4
3
48. tan  
12
5
  
3
2
3
   2
2
(49–70) Prove the following identities.
50.
 sin   cos  2  1  sin 2
51. cos4   sin 4   cos 2
52.
 cos  sin   cos  sin    cos 2
53. cos 4  cos2 2  sin 2 2
54. sin 4  4sin  cos   8sin 3  cos 
49.
 sin   cos  2  sin 2  1
55.
sin 2
 cot 
1  cos 2
56.
1  cos 2
 cot 
sin 2
57.
cos 2
 cot   tan 
sin  cos 
58.
cos 2
cot   1

1  sin 2 cot   1
59. cos 2 
1  tan 2 
1  tan 
2
   tan   sin 
61. sin 2   
2 tan 
2
63.
sin 2
 
 
 cos2    sin 2  
2sin 
2
2
60. cos 2 
cot   tan 
cot   tan 
   csc  cot 
62. sin 2   
2csc
2
   tan   sin 
64. csc2   
2 tan 
2
Chapter 6 Trigonometric Identities and Equations
 
65. tan    csc  cot 
2
sec
 
66. tan   
 2  sec csc  csc
2
 
67. tan 2    1 
1  cos 
2
   2cos 
68. 1  tan 2   
 2  1  cos 
 
1  tan 2  
 2   cos 
69.
 
1  tan 2  
2
70.
cos   sin  cos   sin 

 2 tan 2
cos   sin  cos   sin 
(71–76) Convert the following products to sums or differences.
71.
sin 4 sin 2 
72.
sin 3 sin 5 
73.
sin 4  cos 2 
74.
sin 5  cos3 
75.
 cos5 sin 3 
76.
 cos 4 sin 2 
(77–84) Convert the following sums or differences to products.
77. sin 5  sin 3
78. sin 4  sin 2
80. sin 5  sin 3
80. sin 4  sin 2
81. cos5  cos3
82. cos 4  cos 2
83. cos5  cos3
84. cos 4  cos 2
Section 6.1 Trigonometric Identities
Section 6.4 Solving Trigonometric Equations
Objectives

Understanding how to solve trigonometric equations
It is now time to solve equations that have trigonometric functions. To solve equations of this type
you will need to get the equations into factors equal to zero with each factor containing only one
trigonometric function.
General Strategy:
1.
Get the equation equal to zero
2.
Convert all trigonometric functions into the same function by using identities or if
that isn’t possible, then factor the equation into factors where each has only one
type of trigonometric function.
3.
Set each factor equal to zero and solve for the trigonometric function.
4.
Lastly identify which angles make each equation true.
SOLVING TRIGONOMETRIC EQUATIONS
Discussion 1: Solving a Simple Equation
Let’s find all the angles that make 2sin θ − 1 = 0.
As we think about our strategy for solving this we see that it is equal to zero already and that there is
only one function.
Therefore, we will begin by getting sin θ
2sin θ − 1 = 0
2sin θ = 1
alone.
sin θ =
Now, we need to think about what angle
input would cause the output of the sine
function to be
1
.
2
We know that

is one angle. Now we
6
need to think about every other angle
whose reference angle is

that would
6
1
2
(added one)
(divided by two)
sin θ = y on the unit circle. An angle that has a
point on the unit circle with y-value of
1 
is .
2 6
Sine is positive in the 1st and 2nd quadrants. So,
we have

5
and
(whose reference angle in the
6
6
2nd quadrant is

). In addition to these two we
6
cause the output of the sine function to
also have every 2π revolutions around the unit
be positive.
circle from these two angles, which will also
Chapter 6 Trigonometric Identities and Equations
make sine equal
1
, since sine is periodic with a
2
period of 2π.
Final answer.
θ=

6
 2 n
and θ =
5
 2 n
6
(n any integer)
We could have used the calculator to help us with this problem as long as we are comfortable with
non-exact solutions.
2sin θ − 1 = 0
We would still need to solve for sine.
2sin θ = 1
sin θ =
1
2
(added one)
(divided by two)
But now we could use the calculator and
1
find the sin -1   =θ.
2
Notice that 0.52359… is the same as

which we found earlier as one of the many answers.
6
But since we are using the inverse sine function to assist us in solving our problem we only
get one answer on the calculator even though there are many answers to sin θ =
1
. We still
2
need to think in order to arrive at all of the solutions to the equation. The answer 0.52359…
is in radians and is the reference angle so in the second quadrant the angle would be π 
0.52359… = 2.61799…
Final answer.
θ = 0.52359...  2 n and θ = 2.61799...  2 n (n any integer)
Remember that the trigonometric functions are not one-to-one and thus we have many inputs that
yield the same outputs. Specifically we have an infinite number of them. In “real life” though, it may
very well be that all we need are the answers that exist in just the first revolution (in the domain of
0, 2  ).
Question 1: If we have cos x =
 2
what angles, x, would make this equation true?
2
Let’s look at some more examples.
Example 1: Solving Equations
Solve the following equations.
a)
3 cos   cos tan 
b) sec x csc x  2csc x  0
c) csc2   cot 2   3
Section 6.1 Trigonometric Identities
Solution:
3 cos   cos tan 
a) Get the equation equal to zero
3 cos  cos tan   0

cos
Factor
Set each factor equal to zero and solve.

3  tan   0
cos  0
3  tan   0
cos  0
We now look for angles that will satisfy

each equation on the domain of 0, 2  ,

then we extend to the whole domain.

or
2

tan    3
3
2

n
2
2
5
or
3
3

2
n
3
Notice the πn in both answers. This happened because our two answers in each case were
half a revolution apart, so we could combine them into just one. To finally answer the
question, what is the solution to
3 cos   cos tan  , we need to make one more
observation. Tan θ is not defined for

2
or
3
2
n .
, so our final solution is  
2
3
b) This is already set equal to zero, so we
sec x csc x  2csc x  0
csc x  sec x  2   0
begin by factoring.
Now set each factor equal to zero and
csc x  0
 sec x  2   0
1
0
sin x
sec x  2
solve.
We now look for angles that will satisfy
1
 0 is impossible so there isn’t a
sin x
each equation on the domain of 0, 2  ,
csc x 
then we extend to the whole domain.
solution from this equation. With the
second, think what makes cos x 
1
(secant
2
and cosine are reciprocals). Two angles that
work are
x=

3

5
and
.
3
3
5
 2 n
3
 2 n or
Now the final answer.
c) Get the equation equal to zero
csc2   cot 2   3
csc2   cot 2   3  0
Use a Pythagorean identity to convert
cosecant squared to cotangent squared.
Solve for cotangent.
cot   1  cot   3  0
2
2
Chapter 6 Trigonometric Identities and Equations
cot 2   1  cot 2   3  0
cot 2   2  0
cot 2   2
A more convenient way to write
cot   2 
2 when looking for exact answers.
2
2
We now look for angles that will satisfy
this equation on the domain of 0, 2  ,
It might be easiest to think what angles
then we extend to the whole domain.
cause tangent to equal


Now the final answer.

4

4
or
2
.
2
5
4
n
Let’s look at another example.
Discussion 2: Multiple of an Angle
Let’s find the solution, on the domain of 0, 2  , for 5sin3x  10  0 .
Just as with our previous equations we
10sin3x  5  0
10sin3x  5
need to get the sine alone.
Next, we need to think what angles yield
1
as an output (write all the answers).
2
sin 3 x 
1
2
sin  
1
2

7
6

7
11
 2 n or
 2 n
6
6
Notice that we are not concerned about
Answer Q1:
The reference
angle that has
2
as an
2

answer is
.
4
Since cosine is
negative we
want angles in
the 2nd and 3rd
quadrants.
Thus
3
 2 n
x=
4
or
5
 2 n
x=
4
the 3x at the moment.
Now we can solve for x.
(multiplied by
1
)
3
To get our final answer remember that
sin 3x, has a period that is
or
11
6
3x 
7
 2 n
6
x
7 2 n

18
3
x
7 12 n

18
18
x
7 19 31
,
,
18 18 18
x
7 11 19 23 31 35
,
,
,
,
,
18 18 18 18 18 18
we were looking for answers on the
domain of 0, 2  . Since this function,
(θ = 3x)
or
3x 
11
 2 n
6
x
11 2 n

18
3
or
or
or
x
11 12 n

18
18
x
11 23 35
,
,
18 18 18
1
the usual
3
period we need to look for 3 times as
many answers as usual.
The final answer is…
Section 6.1 Trigonometric Identities
It is easy to forget to find all the answers to an equation like the one we just solved. Make sure that
once you think you have finished a problem check to see that you have found all of the possible
solutions. Let’s do another example.
Example 2: Fraction of an angle
x
x
 
 
Find all of the solutions to tan    cos  2 x  tan    0 on the domain of 0, 2  .
2
2
Solution:
It looks as though we will need to
factor and then set each factor
equal to zero.
Set each factor equal to zero and
solve for the trigonometric
function.
As was done in discussion 2,
x
x
tan    cos  2 x  tan    0
2
2
x
tan   1  cos  2 x    0
2
x
tan    0
2
or
x
tan    0
2
or
tan  0
1  cos  2 x   0
cos  2 x   1
cos  1
or
(θ =
think what angles will make each
of our two equations true.
Now we can solve for x.
  0   n,
x
 n
2
    2 n
x0
2 x    2 n
or
x  2 n
To get our final answer remember
x
,   2x )
2
x
or
or x 
that we were looking for answers

2
n
 3
2
,
2
on the domain of 0, 2  .
Section Summary:
Here are the guidelines for solving trigonometric equations.
1.
Get the equation equal to zero
2.
Convert all trigonometric functions into the same function by using identities or if that isn’t
possible, then factor the equation into factors where each has only one type of trigonometric
function.
3.
Set each factor equal to zero and solve for the trigonometric function.
4.
Lastly identify which angles make each equation true.
Remember that when you are solving equations where the trigonometric functions have multiples of
angles we need to be very careful about finding all of the solutions.
Chapter 6 Trigonometric Identities and Equations
SECTION 6.4 PRACTICE SET
(1–22) Solve each equation for , with 0<<2. (Give the answers in radians)
1.
3 sec  2
2.
2 sec  2
3. 2sin   1  0
4.
2 sin   1  0
5. 2 2 cos   5  10
6. 4cos   5  3
7. tan 2   3  0
8. tan 2   2  3
9. 2sin 2   1  0
10. 4sin 2   3  0
11. 4cos2   3  5
12. 12cos2   8  11
13. tan(2 )  1
 
14. tan    3
2
3
 
15. cos   
2
2
 
1
2
17. sin 3   1
18. sin  3   0
20. 2sin  2   3  0
 
21. 5cos    3  2
2
16. cos  2  
19.
2 sin  3   1  0
 
22. 4cos    3  3
2
(23–44) Solve each equation for , with 0<<360o. (Give the answers in degrees)
1
4
23. sin 2  cos   0
24. 2sin   sin 2  0
25. sin  cos  
26. 4sin  cos   2
 
 
27. sec    cos  
2
2
28. csc 2  2sin 2
29. sin   3 cos 
30. cos    3 sin 
 
 
31. sin    cos  
2
2
 
 
32. sin     3 cos  
2
2
33. 2cos2   cos   0
34. sin 2   sin   0
35. 2sin 2   sin   1  0
36. 2cos2   cos   1  0
37. 2cos  sec  1  0
38. 2sin   csc   1  0
39. sin 2   cos2   1  cos 
40. cos2   sin 2    sin 
41. sin 2   6  cos   1
42. 2sin 2   3 1  cos  
43. cos 2  6sin 2   4
44. cos 2  2  2sin 2 
Section 6.1 Trigonometric Identities
(45–54) Each of the following has an infinite number of solutions, write an expression
the represents all the possible solutions. (Give the expression in radians)
45. 2sin   1
46. 2cos   1
47. 4sin 2   3
48. 4cos2   1
49. 4 tan   4 3
50. 5cot   5
51. 8 2 cos   3  11
52. 5 2 sin   1  4
53. 7 sin   3  4
54. 5cos   7  2
(55–64) Use the calculator to solve each equation for , with 0<<2.
(Give the answer in radians, approximated to two decimal places, if there is no answer
explain why)
55. 3sin   2  4
56. 5sin   7  3
57. 5cos   8  4
58. 3cos   2  4
59. 16sin 2   3
60. 4cos 2   3
61. 3sin   10  3
62. 2cos  5  3
63. 5sin 2   2sin   1  0 (Hint: Use the quadratic formula)
64. 7cos2   cos   1  0 (Hint: Use the quadratic formula)
(65–70) Use the calculator to solve each equation for , with 0<<360o.
(Give the answer in degrees, approximated to two decimal places, if there is no answer
explain why)
65. 8sin   3  4
66. 9cos  11  3
68. 5cos   11  2
69. 3sin 2   5sin   1  0 (Hint: Use the quadratic formula)
67. 3sin   2  3
70. 2cos2   8cos   3  0 (Hint: Use the quadratic formula)
(71–76) Graph each of the following with the graphing calculator and use the graphing
calculator to find all solution(s) for , with 0<x<2.
(Give the answer(s) in radians approximated to two decimal places, if there is no solution
explain how the graph tells you this)
71. sin x  x3  1  0
 x
72. 3cos    x  2  0
2
73. x  5cos x  0
74. x  4sin x  0
75. 3sin x  2 cos x  x
76. 2sin x  cos x  x
77. sin x  cos x  x  0
78. sin x  cos x  x  0
(79– ) Mathematical modeling needed
Chapter 6 Trigonometric Identities and Equations
Chapter 6 Review
Topic
Section
Key Points




CHAPTER 6 REVIEW HOMEWORK
Section 6.1
1. Given sin  
12
is in the 1st quadrant, find the other five trigonometric functions.
13
2. Given cos  
4
is in the 2nd quadrant, find the other five trigonometric functions.
5
3. Given tan  
12
is in the 3rd quadrant, find the other five trigonometric functions.
5
4. Given sec 
5
is in the 4th quadrant, find the other five trigonometric functions.
4
(5−6) Given the value of two trigonometric functions, use the trigonometric identities to find
the values of the other four trigonometric functions.
5. sin  
3
8
cos  
 55
8
6. cos  
5
9
csc  
9 14
28
(710) Simplify each of the following:
7. sin  cot 
9.
1  sin  1  sin  
cos 

8. cos2  1  tan 2 
10.

sin 2 
cos 2   1  tan 2 
Section 6.2
(1118) Prove the following identities are true.
11. cos tan  csc  1
13.
cos   sin 
 1  cot 
sin 
15. tan 2   sin 2   tan 2  sin 2 
17.
sec   1 1  cos 

sec   1 1  cos 
12. sec   cos   tan  sin 
14.
1  tan 2 
 cot   tan 
tan 
16. cos4   sin 4   cos2   sin 2 
18.
1  sin 
cos 

cos 
1  sin 
(1920) Show each of the following are not identities by finding an angle  that makes the statement
false.
Chapter 5 Trigonometric Functions
19. sin   cos  tan   2
20. 2sin  cos   sin   0
(2122) Graph each of the following to show they are identities, then prove they are identities.
21. csc  cot  sec
22.
1
1

 2sec 2 
1  sin  1  sin 
Section 6.3
(2326) Use the sum and difference identities to evaluate exactly the following:
23. sin 15o
 13 
24. cos 

 12 
 19 
25. tan 

 12 
26. tan 285o
 


(2730) Use the sum and difference identities in reverse to rewrite each of the following as
a single trigonometric function.
27. sin 3x cos 7 x  cos 3x sin 7 x
29.
tan 8 x  tan 5 x
1  tan 8 x tan 5 x
28. sin 7 x cos 5 x  cos 7 x sin 5 x
30.
tan 9 x  tan 6 x
1  tan 9 x tan 6 x
(3134) Use the given information to find the exact values of:
 
 
a. sin 2
b. cos 2
c. sin  
d. cos 
2
2
31. sin  
5
,
13
33. cos  
12
,
13
0  

2

2
 
3
32. cos   ,
5
34. sin  
4
,
5
3
   2
2
  
3
2
(3536) Use the given information to find the exact values of:
 
a. tan 2
b. tan  
2
35. tan  
12
,
5
0  

2
36. tan  
4
,
3

2
 
(3746) Prove the following identities.
37. tan  
sin 2
1  cos 
38. tan   csc 2  cot 2
Section 5.1 Geometry Review

39. 4csc2 2  sec2  1  cot 2 
41.

sin   cos 2  sec2   tan2   sin 2
40.
sin 2 cos 2

 sec 
sin 
sin 
42. sin 2 
2 tan 
1  tan 2 
 
43. 1  cos   2 cos  
2
   sec   1
44. tan   
tan 
2
 
 
45. sin   2sin   cos  
2
2
 
2 tan  
2
46. sin  
 
1  tan 2  
2
(4748) Convert the following products to sums or differences.
47. sin 5 sin 3
48. sin 4 cos 2
(4952) Convert the following sums or differences to products.
49. sin 8  sin 6
50. sin 7  sin 4
51. cos5  cos3
52. cos 7  cos5
Section 6.4
(5360) Solve each equation for , with 0<  < 2. (Give the answers in radians)
53. 2sin   1  0
54. 2cos2   1  0
55. 2cos2   cos   1  0
56. sin 3  0
57. cos 2  sin   0
58. cos 4  1
 
59. 6 cos    3  3
2
 
60. tan 2    2  1
2
(6168) Solve each equation for , with 0<  < 360o. (Given the answers in degrees)
61. 2cos   1  1
62. 4sin 2   3
63. 2sin 2   cos   1  0
64. 2cos2   3sin   3  0
65. tan   cot   2  0
66. 2sin   csc   3  0
67. tan  tan 2  1
  1
 
68. sin 2    sin    0
4 2
4
(6972) Each of the following has an infinite number of solutions, write an expression that
represents all the possible solutions. (Given the answer in radians)
Chapter 5 Trigonometric Functions
69. 2cos   1
70. 2sin   3
71. 4sin 2   1
72. 4cos 2   3
(7378) Use the calculator to solve each equation for , with 0<  < 2.
(Give the answers in radians approximated to 2 decimal places)
73. 8cos  3  2
74. 5sin   7  4
75. 25sin 2   2
76. 16cos2   3
77. 3sin 2   2sin   3  0
78. 2cos2   3cos   4  0
(7984) Use the calculator to solve each equation for . with 0<  < 360o/
(Give the answers in degrees approximated to 2 decimal places)
79. 9sin   1  3
80. 7 cos   5  2
81. 36cos2   7
82. 49sin 2   3
83. 3cos2   cos   3  0
84. 2sin 2   3sin   3  0
(8588) Graph each of the following with the graphing calculator and use the graphing calculator
to find all the solutions of x, with 0 < x < 2.
(Give the answer(s) in degrees approximated to 2 decimal places , if there is no solutions
explain how the graph tells you this)
85. cos x  x2  1  0
87. cos x  sin x  2 x 2  0
x
86. 2sin    x  2  0
2
88. sin x  cos x  x2  0
Section 5.1 Geometry Review
CHAPTER 6 EXAM
1. Using the unit circle definition prove: cos  
1
sec 
2. Given sin  
12
in the 2nd quadrant, find the other five trigonometric function values.
13
3. Given sin  
2
 7
find the values of the other four trigonometric functions.
and cos  
3
3
(45) Simplify:
4.
cos 2   1
cos   1
5.
1
cot 2 
 tan  cot 
(613) Prove the following identities are true
6. sec cos   sin 2   cos2 
8.
1  tan  cos   sin 

1  tan  cos   sin 
10. cos 2 
12.
7.
sec2   1
 tan 
tan 
9. sin 4   cos4   2sin 2   1
cot 2   1
11. cos 2 
1  cot 
2
 cos   sin  2  1  sin 2
cot   tan 
tan   cot 
13. sin 4   cos4    cos 2
(1416) Use the sum and difference identities in reverse to rewrite each of the following as a single
trigonometric function.
14. sin 7 x cos 5 x  cos 7 x sin 5 x
16.
15. sin 5 x cos8 x  cos8 x sin 5 x
tan 7 x  tan 5 x
1  tan 7 x tan 5 x
(1719) Use the sum and difference identities to evaluate each of the following exactly.
17. sin
5
12

18. cos 195o

19. tan
13
12
(2122) Use the given information to find exact values of:
 
 
a. sin 2
b. cos 2
c. sin  
d. cos 
2
 
2
21. sin  
4
,
5
3
   2
2
22. cos  
12
,
13
0  

2
Chapter 5 Trigonometric Functions
23. Given tan  
4
,
3

2
    find: a. tan2
 
b. tan  
2
(2425) Convert the following to a sum or difference.
24. sin 6 sin 4
25. sin 3 cos 
(2627) Convert the following sums to a product.
26. sin 9  sin 7
27. cos 7  cos5
(2833) Solve each of the following equations for , with 0    2 .
(Given the answers in radians and degrees)
28. 2cos  1  0
29. 16sin 2   3  5
30. 2sin 4  1  0
 
31. 3cos    1  5
2
32. 2sin 2   sin   1  0
33. cos 2  sin 2   0
(3435) Use the graphing calculator to solve each equation for , with 0    2 .
(Give the answers in radians, approximated to two decimal places)
34. 5sin   3  2
35. 16cos2   5