Download Section 3.5 - Canton Local

Chapter 6
Trigonometric
Identities and
Equations
© 2011 Pearson Education, Inc.
All rights reserved
© 2010
2011 Pearson Education, Inc. All rights reserved
1
SECTION 6.6
Trigonometric Equations II
OBJECTIVES
1
2
3
Solve trigonometric equations involving
multiple angles.
Solve equations using sum-to-product
identities.
Solve equations containing inverse
trigonometric functions.
EQUATIONS INVOLVING MULTIPLE ANGLES
If x is the measure of an angle, then for
any real number k, the number kx is a
multiple angle of x.
© 2011 Pearson Education, Inc. All rights reserved
3
EXAMPLE 1
Solving a Trigonometric Equation
Containing Multiple Angles
1
Find all solutions of the equation cos 3x  in
2
the interval [0, 2π).
Solution

1
Recall that cos  .
3 2

5
Since cos θ > 0 in QI and QIV,   2  
.
3
3
The period of cos x is 2π. Replace θ with 3x.
5
So 3x   2n or 3x 
 2n for any
3
3
integer n.

© 2011 Pearson Education, Inc. All rights reserved
4
EXAMPLE 1
Solving a Trigonometric Equation
Containing Multiple Angles
Solution continued
2n
5 2n
x 
x

or
9
3
9
3
To find solutions in the interval [0, 2π), try the
following:
 2
5
5 2

  x

 
n = –1 x  
9
3
9
9
3
9

5
x
n=0 x 

9
9
 2 7
5 2 11



n=1 x 

 x
9
3
9
9
3
9

© 2011 Pearson Education, Inc. All rights reserved
5
Solving a Trigonometric Equation
Containing Multiple Angles
EXAMPLE 1
Solution continued

4 13
5 4 17



 x

n=2 x 
9
3
9
9
3
9

19
5
23
 x

x   2 
 2 
n=3
9
9
9
9
Values resulting from n = –1 are too small.
Values resulting from n = 3 are too large.
The solution set corresponding to n = 0, 1, and 2 is
 5 7 11 13 17 
,
,
,
,
 ,
.
9
9 
9 9 9 9
© 2011 Pearson Education, Inc. All rights reserved
6
EXAMPLE 3
The Phases of the Moon
The moon orbits the earth in 29.5 days. The
portion F of the moon visible from the earth x
days after the new moon is given by the equation
   14.75 
F  0.5 sin 
x
  0.5.
2 
14.75 
Find x when 75% of the moon is visible from the
earth.
Solution
Find x when F = 75% = 0.75.
© 2011 Pearson Education, Inc. All rights reserved
7
EXAMPLE 3
The Phases of the Moon
Solution continued
   14.75 
0.75  0.5 sin 
x
  0.5
2 
14.75 
   14.75 
0.25  0.5 sin 
x

2 
14.75 
   14.75 
1
 sin 
x

2
2 
14.75 
1
  14.75 
Thus, a solution of sin   is  
x
.
14.75 
2 
2
© 2011 Pearson Education, Inc. All rights reserved
8
EXAMPLE 3
The Phases of the Moon
Solution continued
 1
Since sin  and sin θ > 0 in QI and QII,
6 2
  14.75  5
  14.75  
or 14.75  x  2   6
x



14.75 
2  6
14.75 14.75
14.75 514.75
x

x

2
6
2
6
14.75 14.75
14.75 514.75
x

x

2
6
2
6
x  20
x  10
So, 75% of the moon is visible 10 days and
20 days after the new moon.
© 2011 Pearson Education, Inc. All rights reserved
9
EXAMPLE 6
Solving an Equation Using a
Sum-to-Product Identity
Solve cos 2x + cos 3x = 0 over the interval [0, 2π).
Solution
uv
u v
Use cos u  cos v  2 cos
cos
.
2
2
cos 2 x  cos 3x  0
5x
x
2 cos cos  0
2
2
So
5x
x
cos
 0 or cos  0.
2
2
© 2011 Pearson Education, Inc. All rights reserved
10
Solving an Equation Using a
Sum-to-Product Identity
EXAMPLE 6
Solution continued
We know if cos θ = 0, then for any integer n,


2
 2n
3
or  
 2n .
2
5x
x
Replace θ with
and and find the values of x
2
2
that lie in the interval [0, 2π).
© 2011 Pearson Education, Inc. All rights reserved
11
EXAMPLE 6
Solving an Equation Using a
Sum-to-Product Identity
Solution continued
The solution set to the equation is
© 2011 Pearson Education, Inc. All rights reserved
7 9 
 3
, .
 , , ,
5 5 12
5 5
EXAMPLE 7
Solve

6
Solution

6
Solving an Equation Containing an
Inverse Tangent Function
 2 tan
1
 2 tan
1
2 tan
tan
1
1
x  1 

2
x  1 

x  1 

x  1 

.
2
2


6


3
6
x  1  tan

6

3 3 3
x  1  tan  1 

6
3
3
© 2011 Pearson Education, Inc. All rights reserved
13