Download Section 6.1 Verifying Trigonometric Identities

Section 6.1
Verifying Trigonometric Identities
Do you Enjoy Solving Puzzles?
We have already established some basic relationships among the trigonometric
functions. The reciprocal, quotient, and Pythagorean identities follow easily from
the definitions of the trigonometric functions. The even-odd identities are
established using properties of the Cartesian coordinate system.
In this section, we see how new identities can be verified using identities that we
already know. The process will involve some trial and error and may remind you
of solving a puzzle.
Objective #1: Use fundamental trigonometric identities to verify identities.
Solved Problem #1
1a. Verify the identity: csc x tan x  sec x.
In many cases, it is helpful to rewrite all the trigonometric functions on one side in terms of sines and cosines
using reciprocal and quotient identities. In general, we start by working with the more complicated side, so we
1
begin with the left side and rewrite it in terms of sines and cosines using the reciprocal identity csc x 
sin x
sin x
and the quotient identity tan x 
.
cos x
csc x tan x 
1 sin x
1
sin x
1


 sec x


sin x cos x sin x cos x cos x
1
in the last step. Also note how we started with the left side
cos x
of the given identity, csc x tan x  sec x, and ended with the right side.
Note the use of the reciprocal identity sec x 
Pencil Problem #1
1a. Verify the identity: sin x sec x  tan x.
Copyright © 2014 Pearson Education Inc.
217
Algebra and Trigonometry 5e
1b. Verify the identity: cos x cot x  sin x  csc x.
Once again we will rewrite all the trigonometric
functions on one side in terms of sines and cosines.
We start by working with the more complicated side,
the left side, and rewrite it using the quotient identity
cos x
cot x 
.
sin x
cos x cot x  sin x 

1b. Verify the identity: csc θ  sin θ  cot θ cos θ .
[Hints: The Pythagorean identity
cos2 θ  sin 2 θ  1 can also be used in the form
1  sin 2 θ  cos2 θ and a fraction of the form
can be rewritten as
a2
b
a a
 .]
b 1
cos x cos x

 sin x
1
sin x
cos2 x sin x sin x


sin x
1 sin x
cos2 x sin 2 x

sin x
sin x
2
cos x  sin 2 x

sin x
1

sin x
 csc x

Note how we expressed both terms on the left side as
fractions with a common denominator and then added;
this technique is often helpful when one side of the
given identity has two terms and the other side has
only one term. Also note the use of the Pythagorean
identity cos2 x  sin 2 x  1.
1c. Verify the identity: sin x  sin x cos 2 x  sin 3 x.
Sometimes factoring out a common factor is helpful.
We start by working with the more complicated side,
the left side, and factor out the common factor sin x.
1c. Verify the identity by using factoring first:
sec x  sec x sin 2 x  cos x. [Hint: After factoring
out the common factor on the left side and using a
Pythagorean identity as we did in Solved Problem
#1c, rewrite secant in terms of cosine.]
Then we will use the Pythagorean identity
cos2 x  sin 2 x  1 in the form 1  cos2 x  sin 2 x.
sin x  sin x cos2 x  sin x (1  cos2 x )
 sin x  sin 2 x  sin 3 x
218
Copyright © 2014 Pearson Education Inc.
Section 6.1
1d. Verify the identity:
1  cos θ
 csc θ  cot θ .
sin θ
There is usually more than one way to verify an
identity. Here we will verify the identity in two ways.
In the first method, we will start with the left side and
work toward the right. In the second method, we will
start with the right side and work toward the left.
1d. Verify the identity in two different ways as in
1  sin θ
Solved Problem #1d:
 sec θ  tan θ .
cos θ
Method 1: Start with the left side and write the single
fraction as the sum of two fractions. Then use
appropriate reciprocal and quotient identities.
1  cos θ
1
cos θ


 csc θ  cot θ
sin θ
sin θ sin θ
Method 2: Start with the right side and rewrite the
given functions in terms of sines and cosines. Then
add the resulting fractions.
csc θ  cot θ 
1
cos θ 1  cos θ


sin θ sin θ
sin θ
Although the equation in Method 2 could be obtained
by reversing the equation in Method 1, the thought
processes are different. Perhaps one direction is more
obvious to you than the other.
1e. Verify the identity:
sin x
1  cos x

 2 csc x.
1  cos x
sin x
1e. Verify the identity:
cos x
1  sin x

 2 sec x.
1  sin x
cos x
We will work with the left side and begin by getting a
common denominator so that we can combine the two
fractions into one fraction. The LCD is
(1  cos x )sin x. Watch for opportunities to factor and
simplify within the fraction and to apply basic
identities. The Pythagorean identity
cos2 x  sin 2 x  1 will be used to replace
cos2 x  sin 2 x with 1 in a key step along the way.
(continued on next page)
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219
Algebra and Trigonometry 5e
sin x
1  cos x
sin x sin x 1  cos x 1  cos x





1  cos x
sin x
1  cos x sin x
sin x 1  cos x
sin 2 x
(1  cos x ) 2


(1  cos x ) sin x (1  cos x ) sin x

sin 2 x  (1  cos x )2
(1  cos x )sin x

sin 2 x  1  2 cos x  cos 2 x
(1  cos x )sin x
(sin 2 x  cos2 x )  1  2 cos x
(1  cos x )sin x
1  1  2 cos x

(1  cos x ) sin x
2  2 cos x

(1  cos x ) sin x
2(1  cos x )

(1  cos x ) sin x

2( 1  cos x )
( 1  cos x ) sin x
2

sin x
 2 csc x

Answers for Pencil Problems (Textbook Exercise references in parentheses):
sin x
1
sin x


 tan x (6.1 #1)
1 cos x cos x
1
sin θ sin θ
1
sin 2 θ 1  sin 2 θ cos2 θ cos θ cos θ
csc θ  sin θ 








 cot θ cos θ . (6.1 #11)
sin θ
1 sin θ sin θ sin θ
sin θ
sin θ
sin θ
1
1
sec x  sec x sin2 x  sec x(1  sin 2 x)  sec x cos2 x 
 cos2 x  cos x (6.1 #7)
cos x
1  sin θ
1
sin θ
1
sin θ 1  sin θ
(6.1 #24)


 sec θ  tan θ ; sec θ  tan θ 


cos θ
cos θ cos θ
cos θ cos θ
cos θ
cos x
1  sin x
cos x  cos x
(1  sin x )(1  sin x ) cos 2 x  (1  sin x ) 2 cos 2 x  1  2 sin x  sin 2 x





1  sin x
cos x
(1  sin x ) cos x
(1  sin x ) cos x
(1  sin x ) cos x
(1  sin x ) cos x
1a. sin x sec x 
1b.
1c.
1d.
1e.

220
(cos 2 x  sin 2 x )  1  2 sin x
1  1  2 sin x
2  2 sin x
2(1  sin x )
2




 2 sec x
(1  sin x ) cos x
(1  sin x ) cos x (1  sin x ) cos x (1  sin x ) cos x cos x
Copyright © 2014 Pearson Education Inc.
6.1 #31
Section 6.2
Sum and Difference Formulas
More Identities?
In this section, you will learn to use another set of identities, known as the sum
and difference formulas. The first of these is established using the definitions of
the trigonometric functions and the properties of the Cartesian plane. Once the
first formula is established, the other three can be verified using it and other
basic identities.
Once verified, these formulas can be used to find function values as well as
verify other identities. In the Exercise Set, you will see how these formulas can
be used to simplify functions that model sound vibrations.
Objective #1: Use the formula for the cosine of the difference of two angles.
Solved Problem #1
3
. Use the formula for
2
the cosine of the difference of two angles to find the
exact value of cos 30  cos(90  60).
1a. We know that cos 30 
Pencil Problem #1
1a. Use the formula for the cosine of the difference of
two angles to find the exact value of
cos15  cos(45  30).
The formula is
cos(α  β )  cos α cos β  sin α sin β.
In cos(90  60), α  90 and β  60. Apply
the formula above, substitute exact values for the
four resulting trigonometric expressions, and
simplify.
cos(90  60)  cos 90 cos 60  sin 90 sin 60
 0
1
3
3
3
 1
 0

2
2
2
2
Note that this result is the same as the known value
of cos 30 
3
, as it should be.
2
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Algebra and Trigonometry 5e
1b. Use the formula for the cosine of the difference of
two angles to find the exact value of
cos 70 cos 40  sin 70 sin 40.
1b. Use the formula for the cosine of the difference of
two angles to find the exact value of
cos50 cos 20  sin 50 sin 20.
The formula is
cos(α  β )  cos α cos β  sin α sin β.
Notice that cos 70 cos 40  sin 70 sin 40 looks
like the right side of the formula with α  70 and
β  40.
cos 70 cos 40  sin 70 sin 40  sin(70  40)
 sin 30
1

2
1c. Verify the identity:
cos(α  β )
 1  tan α tan β.
cos α cos β
1c. Verify the identity:
We begin with the left side and apply the formula
for the cosine of the difference of two angles in the
numerator. We then rewrite the fraction as the sum
of two fractions and simplify.
cos(α  β ) cos α cos β  sin α sin β

cos α cos β
cos α cos β
cos α cos β sin α sin β


cos α cos β cos α cos β
sin α sin β
 1

cos α cos β
 1  tan α tan β
222
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cos(α  β )
 tan α  cot β.
cos α sin β
Section 6.2
Objective #2: Use sum and difference formulas for cosines and sines.
Solved Problem #2
2.
We are given that sin α 
4
for a quadrant II angle
5
1
for a quadrant I angle β. Use this
2
information in Solved Problems #2a−d.
α and sin β 
2a. Find the exact value of cos α .
Pencil Problem #2
2.
We are given that sin α 
3
for a quadrant I angle
5
5
for a quadrant II angle β. Use this
13
information in Pencil Problems #2a−d.
α and sin β 
2a. Find the exact value of cos α .
The value of cos α is negative, since α is a quadrant
II angle. Use the Pythagorean identity
sin 2 α  cos2 α  1 to find cos α , choosing the
negative value.
2
4
2
   cos α  1
5
2
25 16 9
4
cos2 α  1    


5
25 25 25
cos α  
9
3

25
5
2b. Find the exact value of cos β.
2b. Find the exact value of cos β.
The value of cos β is positive, since β is a quadrant
I angle. Use the Pythagorean identity
sin 2 β  cos2 β  1 to find cos β , choosing the
positive value.
2
1
2
   cos β  1
2
2
4 1 3
1
cos2 β  1      
2
4 4 4
cos β 
3
3

4
2
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Algebra and Trigonometry 5e
2c. Find the exact value of cos(α  β ).
2c. Find the exact value of cos(α  β ).
We use the sum formula for cosine and substitute
4
1
the given values, sin α  and sin β  , as well
5
2
3
as the values we just found, cos α   and
5
3
cos β 
.
2
cos(α  β )  cos α cos β  sin α sin β
3 3 4 1
 
 
5 2
5 2

3 3 4 3 3  4


10 10
10
2d. Find the exact value of sin(α  β ).
2d. Find the exact value of sin(α  β ).
We use the sum formula for sine and substitute the
4
1
3
values sin α  , sin β  , cos α   , and
5
2
5
3
cos β 
.
2
sin(α  β )  sin α cos β  cos α sin β
224

4 3  3 1

   
5 2  5 2

4 3 3 4 33


10 10
10
Copyright © 2014 Pearson Education Inc.
Section 6.2
Objective #3: Use sum and difference formulas for tangents.
Solved Problem #3
3.
Verify the identity: tan( x  π )  tan x.
Pencil Problem #3
3.
Verify the identity: tan(2π  x )   tan x.
Start with the left side and apply the sum formula for
tangent. Use the fact that tan π  0.
tan x  tan π
1  tan x tan π
tan x  0

1  tan x  0
tan x

1
 tan x
tan( x  π ) 
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Algebra and Trigonometry 5e
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1a.
1b.
1c.
6 2
(6.2 #1)
4
3
(6.2 #5)
2
cos(α  β ) cos α cos β  sin α sin β
cos α cos β sin α sin β

 cot β  tan α  tan α  cot β


cos α sin β
cos α sin β
cos α sin β cos α sin β
12
63
16
2c. 
(6.2 #57a) 2d. 
(6.2 #57b)
13
65
65
tan 2π  tan x
0  tan x
 tan x
3. tan(2π  x ) 


  tan x (6.2 #37)
1  tan 2π tan x 1  0  tan x
1
2a.
226
4
5
2b. 
Copyright © 2014 Pearson Education Inc.
(6.2 #9)
Section 6.3
Double Angle, Power-Reducing, and Half-Angle Formulas
How Far Can You Throw That?
When you throw an object, the distance that the object will travel before hitting
the ground depends on the initial speed of the object as well as the angle at
which the object leaves your hand. This distance can be modeled by a formula
that involves both sines and cosines.
In the Exercise Set, you will see how a trigonometric identity introduced in this
section can be used to rewrite a formula involving two trigonometric functions
as a formula involving only one such function. The simpler form can be used to
determine an angle that maximizes throwing distance.
Objective #1: Use the double-angle formulas.
Solved Problem #1
4
and θ lies in quadrant II, find the exact
5
value of sin 2θ , cos 2θ , and tan 2θ .
1a. If sin θ 
Pencil Problem #1
15
and θ lies in quadrant II, find the
17
exact value of sin 2θ , cos 2θ , and tan 2θ .
1a. If sin θ 
The formulas sin 2θ  2 sin θ cos θ and
cos 2θ  cos2 θ  sin 2 θ require that we know both
4
sin θ and cos θ . We are given that sin θ  . We
5
need to find cos θ . Since θ lies in quadrant II, the
value of cos θ is negative. Use the Pythagorean
identity sin 2 θ  cos 2 θ  1 to find cos θ , choosing
the negative value.
2
4
2
   cos θ  1
5
2
25 16
9
4
cos2 θ  1    


5
25 25 25
cos θ  
9
3

25
5
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Algebra and Trigonometry 5e
Now substitute the values into the formulas.
24
 4   3
sin 2θ  2 sin θ cos θ  2       
 5  5
25
cos 2θ  cos2 θ  sin 2 θ
2
2
9 16
7
 3
4
      


 5
5
25 25
25
The formula tan 2θ 
2 tan θ
1  tan 2 θ
requires that we
know tan θ . Since we know sin θ 
4
and
5
3
cos θ   , we can use a quotient identity to find
5
tan θ .
4
sin θ
4  5
4
 5      
tan θ 
3


cos θ
5
3
3

5
Now we can use this value in the formula.
tan 2θ 
2 tan θ
1  tan 2 θ

 4
2  
 3
 4
1   
 3
2
8
8

8  9  24
3

 3      
16
7
3  7 7
1

9
9

Note that we can also find tan 2θ using the values
of sin 2θ and cos 2θ and a quotient identity.
sin 2θ

tan 2θ 
cos 2θ
228
24
25  24
7
7

25

Copyright © 2014 Pearson Education Inc.
Section 6.3
1b. Find the exact value of 2 sin15 cos15.
1b. Find the exact value of cos2 15  sin 2 15.
The given expression looks like the right side of the
double-angle formula cos 2θ  cos2 θ  sin 2 θ with
θ = 15°. We use the formula to rewrite the
expression and then evaluate.
cos2 15  sin 2 15  cos(2  15)  cos 30 
3
2
1c. Verify the identity: sin 3θ  3sin θ  4 sin 3 θ .
Notice that the sine on the left side has 3θ as its
argument while the ones on the right have just θ, so
we need to use one or more identities that allow us
to change the argument. The right side may look
more complicated, but we will start with the left
side because it contains the more complicated
argument and we see more opportunities to apply
identities.
1c. Verify the identity:
sin 4t  4 sin t cos3 t  4 sin 3 t cos t.
We first rewrite sin 3θ as sin(2θ  θ ) and use the
sum formula for sine. Then we use double-angle
formulas to simplify the occurrences of sin 2θ and
cos 2θ that result from the first step. Since the right
side contains only sine, we choose to use the form
of the double-angle formula for cosine that
expresses cos 2θ in terms of sine only. Still keeping
in mind that the right side contains only sine, we
replace an occurrence of cos2 θ with 1  sin 2 θ ,
using a form of the Pythagorean identity
sin 2 θ  cos 2 θ  1.
sin 3θ  sin(2θ  θ )
 sin 2θ cos θ  cos 2θ sin θ
 (2 sin θ cos θ ) cos θ  (1  2 sin 2 θ ) sin θ
 2 sin θ cos2 θ  sin θ  2 sin 3 θ
 2 sin θ (1  sin 2 θ )  sin θ  2 sin 3 θ
 2 sin θ  2 sin 3 θ  sin θ  2 sin 3 θ
 3sin θ  4 sin 3 θ
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Algebra and Trigonometry 5e
Objective #2: Use the power-reducing formulas.
Solved Problem #2
2.
Write an equivalent expression for sin 4 x that does
not contain powers of trigonometric functions
greater than 1.
Pencil Problem #2
2.
Write an equivalent expression for sin 2 x cos2 x that
does not contain powers of trigonometric functions
greater than 1.
We begin by writing sin 4 x as (sin 2 x ) 2 and
applying the power-reducing formula for sine. After
some simplification, we will use the power-reducing
formula for cosine.
sin 4 x  (sin 2 x )2
 1  cos 2 x 



2
2
1  2 cos 2 x  cos2 2 x
4
1 1
1
  cos 2 x  cos 2 2 x
4 2
4
1 1
1  1  cos 2(2 x ) 
  cos 2 x  

4 2
4
2

1 1
1 1
  cos 2 x   cos 4 x
4 2
8 8
3 1
1
  cos 2 x  cos 4 x
8 2
8

230
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Section 6.3
Objective #3: Use the half-angle formulas.
Solved Problem #3
3.
3
and a half-angle formula to
2
find the exact value of cos105.
Use cos 210  
Pencil Problem #3
3.
2
and a half-angle formula to
2
find the exact value of cos157.5.
Use cos 315 
Since an angle that measures 105° lies in quadrant II
and cosine is negative in quadrant II, we choose the
negative sign in the half-angle formula for cosine.
210
2
1  cos 210

2
cos105  cos

3
1  

 2 

2

2 3
4

2 3
2

3
1  

 2 
we
Note that to simplify the fraction
2
multiplied the numerator and denominator both by
 

3 
3
1   
  2 1 2   

2
 2 
  2  
2 3


2:
.
22
4
4
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231
Algebra and Trigonometry 5e
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1a. sin 2θ  
240
161
240
; cos 2θ  
; tan 2θ 
289
289
161
(6.3 #7)
1
(6.3 #15)
2
1c. sin 4t  sin(2  2t )  2sin 2t cos 2t  2(2 sin t cos t )(cos2 t  sin 2 t )  4 sin t cos3 t  4 sin 3 t cos t
1b.
2.
1 1
 cos 4 x (6.3 #37)
8 8
3. 
232
2 2
2
(6.3 #41)
Copyright © 2014 Pearson Education Inc.
(6.3 #33)
Section 6.4
Product-to-Sum and Sum-to-Product Formulas
Music to Your Ears
Each time you push a button on a touch-tone phone a sound is produced. This
sound can be modeled by the sum of two sine functions. In the Exercise Set, you
will learn how to play Mary Had a Little Lamb on your phone and you will write
the sum of sines producing a note as a product.
Objective #1: Use the product-to-sum formulas.
Solved Problem #1
1a. Express sin 5 x sin 2 x as a sum or difference.
Pencil Problem #1
1a. Express sin 6 x sin 2 x as a sum or difference.
We use the formula for the product of two sines:
1
sin α sin β  [cos(α  β )  cos(α  β )].
2
1
[cos(5 x  2 x )  cos(5 x  2 x )]
2
1
 [cos 3 x  cos 7 x ]
2
sin 5 x sin 2 x 
1b. Express cos 7 x cos x as a sum or difference.
1b. Express sin x cos 2 x as a sum or difference.
We use the formula for the product of two cosines:
1
cos α cos β  [cos(α  β )  cos(α  β )].
2
1
[cos(7 x  x )  cos(7 x  x )]
2
1
 [cos 6 x  cos8 x ]
2
cos 7 x cos x 
Copyright © 2014 Pearson Education Inc.
233
Algebra and Trigonometry 5e
Objective #2: Use the sum-to-product formulas.
Solved Problem #2
Pencil Problem #2
2a. Express sin 7 x  sin 3x as a product.
2a. Express sin 6 x  sin 2 x as a product.
We use the formula for the sum of two sines:
α β
α β
sin α  sin β  2 sin
cos
.
2
2
7 x  3x
7 x  3x
cos
2
2
10 x
4x
 2 sin
cos
2
2
 2 sin 5 x cos 2 x
sin 7 x  sin 3x  2 sin
2b. Express cos 3x  cos 2 x as a product.
2b. Express cos 4 x  cos 2 x as a product.
We use the formula for the sum of two cosines:
α β
αβ
cos α  cos β  2 cos
cos
.
2
2
3x  2 x
3x  2 x
cos
2
2
x
5x
 2 cos cos
2
2
cos 3 x  cos 2 x  2 cos
Answers for Pencil Problems (Textbook Exercise references in parentheses):
1a.
1
[cos 4 x  cos8 x ] (6.4 #1)
2
2a. 2 sin 4 x cos 2 x
234
(6.4 #9)
1b.
1
[sin 3x  sin x ] (6.4 #5)
2
2b. 2 cos 3x cos x (6.4 #13)
Copyright © 2014 Pearson Education Inc.
Section 6.5
Trigonometric Equations
How Many Solutions are There?
We have seen that trigonometric functions can be used to model cyclic
phenomena such as tides, temperatures, and hours of daylight. For example, we
have worked with models where we could estimate the number of hours of
daylight at a location on a particular day of the year by evaluating the function
for a value of the independent variable representing the day.
In this section’s Exercise Set, you will solve trigonometric equations to find the
dates on which a certain city has 10.5 hours of daylight. In general, the equation
has infinitely many solutions, but when we restrict ourselves to one cycle, one
year, there are exactly two solutions. In this section, pay attention to whether
you are looking for all solutions or just the solutions that are in a given interval.
Objective #1: Find all solutions of a trigonometric equation.
Solved Problem #1
1.
Find all solutions of 5sin x  3sin x  3.
Pencil Problem #1
1.
Find all solutions of 4 sin θ  1  2 sin θ .
First isolate sin x on the left side by first
subtracting 3sin x from both sides and then
dividing both sides by the resulting coefficient of
sin x.
5sin x  3sin x  3
5sin x  3sin x  3sin x  3sin x  3
2 sin x  3
3
2 sin x

2
2
sin x 
3
2
The sine function is positive in quadrants I and II
π
3
at
in quadrant I. In quadrant
3
2
π 2π
3
. In one period of
II, sine equals
at π  
3
3
2
the sine function, there are two solutions of the
π
2π
and x 
given equation, x 
.
3
3
and sine equals
(continued on next page)
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235
Algebra and Trigonometry 5e
We are asked to find all solutions. Because of the
periodic property of the sine function, if we add or
subtract any integer multiple of 2π to either of the
solutions on the previous page, we will get another
solution. Thus, the solutions are
x
π
3
2π
 2nπ ,
3
 2nπ and x 
where n is any integer.
Objective #2: Solve equations with multiple angles.
Solved Problem #2
2a. Solve the equation: tan 2 x  3, 0  x  2π .
Pencil Problem #2
2a. Solve the equation: tan 3 x 
The equation already has the tangent expression
isolated on the left. Notice that the argument, 2x, is
a double-angle. We solve for 2x first and then for x.
We also note that we are only looking for solutions
x that satisfy 0  x  2π . However, we first identify
all solutions and then select those in the given
interval.
In Solved Problem #1, we worked with sine which
has period 2π. In this problem, we are working with
tangent which has period π. In one period, tangent
equals
2x 
π
3
3 only once, at
π
3
. Thus,
 nπ , where n is any integer.
Dividing by 2, we find all solutions of the given
equation:
nπ

, where n is any integer.
6
2
(continued on next page)
236
Copyright © 2014 Pearson Education Inc.
x
π
3
, 0  x  2π .
3
Section 6.5
We want to find the solutions x that satisfy
0  x  2π . For any negative integer n, x is
negative and does not satisfy 0  x  2π . For any
integer n ≥ 4, x ≥ 2π and does not satisfy
0  x  2π . Here are the values of x for n = 0, 1, 2,
and 3:
π 0π π
x 

6
2
6
π 1  π π 3π 4π 2π
x 
 


6
2
6 6
6
3
π 2  π π 6π 7π
x 
 

6
2
6
6
6
π 3  π π 9π 10π 5π
x 
 


6
2
6
6
6
3
The solutions that satisfy 0  x  2π are
π 2π 7π
5π
,
,
, and
.
6 3
6
3
2b. Solve the equation: sin
x 1
 , 0  x  2π .
3 2
2b. Solve the equation: sin
2θ
 1, 0  θ  2π .
3
The expression involving sine is already isolated on
1
π
at
the left. In one period, the value of sine is
2
6
π 5π
in quadrant I and at π  
in quadrant II.
6
6
Thus,
x π
x 5π
  2nπ or 
 2nπ ,
3 6
3
6
where n is any integer. Multiplying by 3 in each, we
obtain
π
5π
x   6nπ or x 
 6nπ ,
2
2
where n is any integer. Letting n = 0 in
x
π
2
 6nπ , we obtain x 
results in a value of x 
π
2
π
2
. No other integer n
 6nπ that satisfies
5π
 6nπ , there is no integer
2
n that results in a value of x that satisfies
0  x  2π .
0  x  2π . For x 
The only solution is
π
2
.
Copyright © 2014 Pearson Education Inc.
237
Algebra and Trigonometry 5e
Objective #3: Solve trigonometric equations quadratic in form.
Solved Problem #3
3a. Solve the equation:
2sin 2 x  3sin x  1  0, 0  x  2π .
Pencil Problem #3
3a. Solve the equation:
2sin 2 x  sin x  1  0, 0  x  2π .
Notice that if we replace sin x with u on the left side
of the equation we obtain 2u 2  3u  1, which
factors as (2u  1)(u  1). Thus, the left side of the
equation factors as (2 sin x  1)(sin x  1). Thus, we
can solve this equation by factoring and setting each
factor equal to 0.
2 sin 2 x  3sin x  1  0
(2 sin x  1)(sin x  1)  0
2 sin x  1  0 or sin x  1  0
sin x  1
2sin x  1
sin x 
1
2
The equation sin x 
0 ≤ x < 2π:
π
6
and
1
has two solutions satisfying
2
5π
. The equation sin x  1 has
6
one solution satisfying 0 ≤ x < 2π:
are
π
6
,
π
2
, and
π
2
. The solutions
5π
.
6
3b. Solve the equation: 4 cos2 x  3  0, 0  x  2π .
3b. Solve the equation: 4 cos2 x  1  0, 0  x  2π .
We solve the equation by isolating the squared
expression and then using the square root property.
4 cos2 x  3  0
4 cos 2 x  3
3
cos 2 x 
4
cos x 
238
3
3
3
3

or cos x  

4
2
4
2
Copyright © 2014 Pearson Education Inc.
Section 6.5
3
has two solutions
2
π
11π
. The equation
and
satisfying 0 ≤ x < 2π:
6
6
3
has two solutions satisfying
cos x  
2
5π
7π
0 ≤ x < 2π:
. The solutions are
and
6
6
π 5π 7π
11π
,
, and
.
,
6
6
6 6
The equation cos x 
Objective #4: Use factoring to separate different functions in trigonometric equations.
Solved Problem #4
4.
Solve the equation: sin x tan x  sin x, 0  x  2π .
Pencil Problem #4
4.
Solve the equation:
sin x  2sin x cos x  0, 0  x  2π .
We begin by subtracting sin x from both sides,
obtaining 0 on the right. Then we factor out the
common factor of sin x from the terms on the left
and set each factor equal to 0.
sin x tan x  sin x
sin x tan x  sin x  0
sin x (tan x  1)  0
sin x  0 or tan x  1  0
tan x  1
The equation sin x  0 has solutions 0 and π
satisfying 0 ≤ x < 2π. The equation tan x  1 has
π
5π
and
satisfying 0 ≤ x < 2π. The
solutions
4
4
π
5π
solutions are 0, , π, and
.
4
4
Copyright © 2014 Pearson Education Inc.
239
Algebra and Trigonometry 5e
Objective #5: Use identities to solve trigonometric equations.
Solved Problem #5
5a. Solve the equation: cos 2 x  sin x  0, 0  x  2π .
Pencil Problem #5
5a. Solve the equation: sin 2 x  cos x, 0  x  2π .
We use the double-angle formula
cos 2 x  1  2 sin 2 x to rewrite the left side in terms
of sine only so the equation becomes quadratic in
form.
cos 2 x  sin x  0
1  2 sin 2 x  sin x  0
2 sin 2 x  sin x  1  0
2 sin 2 x  sin x  1  0
(2 sin x  1)(sin x  1)  0
2 sin x  1  0 or sin x  1  0
sin x  1
2 sin x  1
sin x  
1
2
The equation sin x  
1
7π
has solutions
and
2
6
11π
satisfying 0 ≤ x < 2π. The equation sin x  1
6
has only the solution
solutions are
240
π
2
,
π
2
satisfying 0 ≤ x < 2π. The
7π
11π
, and
.
6
6
Copyright © 2014 Pearson Education Inc.
Section 6.5
5b. Solve the equation: cos x  sin x  1, 0  x  2π .
5b. Solve the equation: sin x  cos x  1, 0  x  2π .
We begin by squaring each side of the equation and
then applying an identity. Remember that squaring
may introduce extraneous solutions, so it is
important to check all proposed solutions.
cos x  sin x  1
(cos x  sin x ) 2  ( 1) 2
cos 2 x  2 cos x sin x  sin 2 x  1
(cos2 x  sin 2 x )  2 cos x sin x  1
Next we apply the identity sin 2 x  cos 2 x  1.
1  2 cos x sin x  1
2 cos x sin x  0
cos x  0 or sin x  0
3π
2
2
satisfying 0 ≤ x < 2π. The second equation has
solutions 0 and π satisfying 0 ≤ x < 2π.
The first equation has solutions
Check
π
2
: cos
π
2
 sin
π
and
π
 1?
2
0  1  1?
1  1, true
Check
3π
3π
3π
: cos
 sin
 1?
2
2
2
0  ( 1)  1?
1  1, false
Check 0: cos 0  sin 0  1?
1  0  1?
1  1, false
Check π: cos π  sin π  1?
1  0  1?
1  1, true
The solutions are
π
2
and π.
Copyright © 2014 Pearson Education Inc.
241
Algebra and Trigonometry 5e
Objective #6: Use a calculator to solve trigonometric equations.
Solved Problem #6
6a. Solve tan x  3.1044, correct to four decimal
places, for 0 ≤ x < 2π.
Pencil Problem #6
6a. Solve sin x  0.8246, correct to four decimal
places, for 0 ≤ x < 2π.
Note that tangent is positive in quadrants I and III
and that tan 1 3.1044 will be a value in quadrant I.
The solution in quadrant I is
x  tan 1 3.1044  1.2592.
The solution in quadrant III is
x  π  1.2592  4.4008.
The solutions are 1.2592 and 4.4008.
6b. Solve sin x  0.2315, correct to four decimal
places, for 0 ≤ x < 2π.
6b. Solve tan x  3, correct to four decimal places,
for 0 ≤ x < 2π.
Sine is negative in quadrants III and IV and
sin 1 0.2315  0.2336 is in quadrant I.
The solution in quadrant III is
x  π  0.2336  3.3752.
The solution in quadrant IV is
x  2π  0.2336  6.0496.
The solutions are 3.3752 and 6.0496.
Answers for Pencil Problems (Textbook Exercise references in parentheses):
6a.
5π
 2nπ , where n is any integer (6.5 #21)
6
6
π 7π 13π 19π 25π 31π
,
,
,
,
,
(6.5 #29) 2b. no solution (6.5 #33)
18 18 18
18
18
18
π 7π 11π
π 2π 4π 5π
,
,
,
,
,
(6.5 #39) 3b.
(6.5 #47)
2 6
6
3 3
3
3
2π
4π
0,
, π,
(6.5 #59)
3
3
π π 5π 3π
π
, ,
,
(6.5 #69) 5b. 0,
(6.5 #77)
6 2 6
2
2
0.9695, 2.1721 (6.5 #85) 6b. 1.8926, 5.0342 (6.5 #89)
242
Copyright © 2014 Pearson Education Inc.
1. θ 
2a.
3a.
4.
5a.
π
 2nπ , θ 