Download Chapter 11: Trigonometric Identities and Equations

Chapter 11: Trigonometric Identities and Equations
11.1 Trigonometric Identities
11.2 Addition and Subtraction Formulas
11.3 Double-Angle, Half-Angle, and Product-Sum
Formulas
11.4 Inverse Trigonometric Functions
11.5 Trigonometric Equations
11.5 Trigonometric Equations
Solving a Trigonometric Equation by Linear Methods
Example
Solve 2 sin x – 1 = 0 over the interval [0, 2).
Analytic Solution Since this equation involves the first power
of sin x, it is linear in sin x.
2 sin x  1  0
2 sin x  1
1
sin x 
2
Two x values that satisfy sin x = ½ for 0  x  2
are x   and x  5
6
6
However, if we do not restrict the domain there
will be an infinite amount of answers since:

5
x   2k and x   2k
6
6
for k any integer.
11.5 Solving a Trigonometric Equation by Linear Methods
Graphing Calculator Solution
Graph y = 2 sin x – 1 over the interval [0, 2].
The x-intercepts have the same decimal
approximations as 6 and 56 .
11.5 Equations Solvable by Factoring
Example
Solve 2sin2 x – sin x – 1 = 0
Solution
This equation is of quadratic form so:
2sin 2 x  sin x 1  0
(2sin x 1)(sin x 1)  0
2 sin x 1  0
1
sin x  
2
or
sin x 1  0
sin x  1
7
The solutions for sin x = - ½ in [0, 2) are x =
6

.
The solutions for sin x = 1 in [0, 2) is x =
2
Thus the solutions are:
11
or .
6
7π
11π
π
x
 2kπ , x 
 2k , x   2k
6
6
2
11.5 Solving a Trigonometric Equation by Factoring
Example Solve sin x tan x = sin x.
Solution
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

5
x  0 or x  
x
or x 
4
4

The solutions are x  k and x   k for any integer k
4
Caution Avoid dividing both sides by sin x. The two solutions that make sin x = 0
would not appear.
11.5 Solving a Trigonometric Equation by Squaring and
Trigonometric Substitution
Example
Solve tan x  3  sec x over the interval [0, 2).
Solution
Square both sides and use the identity 1 + tan2 x = sec2 x.
2
2
tan x 3  sec x
Possible solutions are:


tan2 x  2 3 tan x  3  sec2 x
tan2 x  2 3 tan x  3 1 tan2 x
2 3 tan x  2
1
tan x  
3
5
11
and
.
6
6
Or are they? Check
answers!
11.5 Trigonometric Equations
Solving an Equation Using a Double-Number Identity
Example Solve cos 2x = cos x over the interval [0, 2).
Analytic Solution
cos 2x  cos x
2cos x 1 cos x
2 cos x  cos x 1  0
(2cos x 1)(cos x 1)  0
2
2
2 cos x  1  0
1
cos x  
2
or
or
cos x  1  0
cos x  1
Solving each equation yields the solution set
{0, 23 , 43 }.
11.5 Solving an Equation Using a Double- Number
Identity
Graphical Solution Graph y = cos 2x – cos x in an
appropriate window, and find the x-intercepts.
The x-intercept displayed is 2.0943951, an approximation
for 2/3. The other two correspond to 0 and 4/3.
11.5 Solving an Equation Using a Double-Number
Identity
Example Solve 4 sin x cos x = 3 over the interval [0, 2).
Solution
4 sin x cos x  3
2(2 sin x cos x)  3
2 sin 2 x  3
3
sin 2 x 
2
From the given domain for
x, 0  x < 2, the domain
for 2x is 0  2x < 4.
Since 2 sin x cos x = sin 2x.
3
sin 2 x 
2
2x 
x
 2 7 8
,
3 3
,
3
,
3
  7 4
, , ,
6 3 6 3
11.5 Solving an Equation that Involves Squaring Both
Sides
Example Solve tan 3x + sec 3x = 2
Solution Since the tangent and secant functions are
related by the identity 1 + tan2  = sec2  , we begin by
expressing everything in terms of secant.
tan 3x  sec 3x  2
tan 3x  2  sec 3x
2
2
tan 3x  4  4 sec 3x  sec 3x Square both sides.
Replace tan 3x
2
2
sec 3x  1  4  4 sec 3x  sec 3x with
2
sec2 3x – 1.
11.5 Solving an Equation that Involves Squaring Both
Sides
sec 2 3x  1  4  4 sec 3x  sec 2 3x
All sol. are of the form
0  5  4 sec 3x
5
sec 3 x 
1  4 
3x  cos    2k
4
5
1
5

1  4 
cos 3x 4
cos    2k
5

4
x
cos 3 x 
3
5
1  4 
1  4 
cos  
3x  cos  
2
5


5
 
x
 k
3
3