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Copyright © 2011 Pearson Education, Inc.
Slide 9.4-1
Chapter 9: Trigonometric Identities and
Equations
9.1 Trigonometric Identities
9.2 Sum and Difference Identities
9.3 Further Identities
9.4 The Inverse Circular Functions
9.5 Trigonometric Equations and Inequalities (I)
9.6 Trigonometric Equations and Inequalities (II)
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-2
9.4 The Inverse Sine Function
Summary of Inverse Functions
1. For a one-to-one function, each x-value corresponds to
only one y-value and each y-value corresponds to only
one x-value.
2. If a function f is one-to-one, then f has an inverse
function f -1.
3. The domain of f is the range of f -1, and the range of f
is the domain f -1. That is, if (a, b) is on the graph of f,
then (b, a) is on the graph of f -1.
4. The graphs of f and f -1 are reflections of each other
about the line y = x. (continued on next slide)
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Slide 9.4-3
9.4 The Inverse Sine Function
Summary of Inverse Functions (continued)
5. To find f -1(x) from f(x), follow these steps:
Step1 Replace f(x) with y and interchange x and y.
Step 2 Solve for y.
Step 3 Replace y with f -1(x).
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-4
9.4 The Inverse Circular Functions
• The Inverse Sine Function
Apply the horizontal line test to show that y = sin x
is not one-to-one. However, by restricting the
domain over the interval  2 , 2 , a one-to-one
function can be defined.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-5
9.4 The Inverse Sine Function
The Inverse Sine Function
y = sin-1 x or y = arcsin x means that x = sin y, for
 2  y  2 .
• The domain of the inverse sine function y = sin-1 x is [–1, 1],
while the restricted domain of y = sin x, [–/2, /2], is the
range of y = sin-1 x.
• We may think of y = sin-1 x
as “y is the number in the
interval  2 , 2  whose sine
is x.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-6
9.4 Finding Inverse Sine Values
Example Find y in each equation.
(a ) y  arcsin 12 (b) y  sin 1 (1) (c) y  sin 1 (2)
Analytic Solution
1
 
(a) y is the number in  2 , 2  whose sine is . Since
2
sin /6 = ½, and /6 is in the range of the arcsine
function, y = /6.
(b) Writing the alternative equation, sin y = –1, shows
that y = –/2
(c) Because –2 is not in the domain of the inverse sine
function, y = sin-1(–2) does not exist.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-7
9.4
Finding Inverse Sine Values
Graphical Solution
To find the values with a graphing
calculator, graph y = sin-1 x and
locate the points with x-values ½
and –1.
(a) The graph shows that when
x = ½, y = /6  .52359878.
(b) The graph shows that when
x = –1, y = –/2  –1.570796.
Caution It is tempting to give
the value of sin-1 (–1) as 3/2,
however, 3/2 is not in the range
of the inverse sine function.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-8
9.4 Inverse Sine Function
y = sin-1 x or y = arcsin x
Domain: [–1, 1]
Range:  2 , 2 
• The inverse sine function is increasing and continuous on its
domain [–1, 1].
• Its x-intercept is 0, and its y-intercept is 0.
• Its graph is symmetric with respect to the origin.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-9
9.4 Inverse Cosine Function
• The function y = cos-1 x (or y = arccos x) is defined
by restricting the domain of y = cos x to the interval
[0, ], and reversing the roles of x and y.
y = cos-1 x or y = arccos x means that x = cos y, for
0  y  .
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-10
9.4
Finding Inverse Cosine Values
Example Find y in each equation.
(a ) y  arccos 1
2

(b) y  cos  

 2 
1
Solution
(a) Since the point (1, 0) lies on the graph of
y = arccos x, the value of y is 0. Alternatively,
y = arccos 1 means cos y = 1, or cos 0 = 1, so y = 0.
(b) We must find the value of y that satisfies
cos y =  2 / 2, 0 y  . The only value for y that
satisfies these conditions is 3/4.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-11
9.4
Inverse Cosine Function
y = cos-1 x or y = arccos x
Domain: [–1, 1]
Range: [0, ]
• The inverse cosine function is decreasing and continuous on
its domain [–1, 1].
• Its x-intercept is 1, and its y-intercept is /2.
• Its graph is not symmetric with respect to the y-axis nor the
origin.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-12
9.4 Inverse Tangent Function
• The function y = tan-1 x (or y = arctan x) is defined
by restricting the domain of y = tan x to the interval
 2 , 2 , and reversing the roles of x and y.
y = tan-1 x or y = arctan x means that x = tan y, for
 2  y  2 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-13
9.4
Inverse Tangent Function
y = tan-1 x or y = arctan x
Domain: (–, )
Range:  2 , 2 
• The inverse tangent function is increasing and continuous on
its domain (–, ).
• Its x-intercept is 0, and its y-intercept is 0.
• Its graph is symmetric with respect to the origin and has
horizontal asymptotes y =  2 .
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-14
9.4 The Inverse Sine Function
Inverse Cotangent, Secant, and Cosecant
Functions
y = cot-1 x or y = arccot x means that x = cot y,
for 0 < y < .
y = sec-1 x or y = arcsec x means that x = sec y,
for 0 < y < , y  /2.
y = csc-1 x or y = arccsc x means that x = csc y,
for –/2 < y < /2, y  0.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-15
9.4 Remaining Inverse Trigonometric
Functions
Function
Domain
Interval
Quadrants
y = sin-1 x
y = cos-1 x
y = tan-1 x
y = cot-1 x
[–1, 1]
[–1, 1]
(–, )
(–, )
[  2 , 2 ]
[0, ]
I and IV
I and II
I and IV
I and II
y = sec-1 x
y = csc-1 x
(  2 , 2 )
(0, )
[0, ], y 2
(–, –1]  [1, )
(–, –1]  [1, ) [  2 , 2 ], y  0
I and II
I and IV
• Inverse trigonometric functions are formally defined with real
number values.
• Sometimes we want the degree-measured angles equivalent
to these real number values.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-16
9.4 Finding Inverse Function Values
Example Find the degree measure of  in each of the
following.
(a )   arctan 1
(b)   sec 1 2
Solution
(a) Since 1 > 0 and –90° <  < 90°,  must be in
quadrant I. So tan  = 1 leads to  = 45°.
(b) Write the equation as sec  = 2. Because 2 s
positive,  must be in quadrant I and  = 60°
since sec 60° = 2.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-17
9.4 Finding Inverse Functions with a
Calculator
• Inverse trigonometric function keys on the calculator give
results for sin-1, cos-1, and tan-1.
• Finding cot-1 x, sec-1 x, and csc-1 x with a calculator is not as
straightforward.
– e.g. If y = sec-1 x, then sec y = x, must be written as follows:
1
1
If sec y  x, then
 x, or cos y  .
cos y
x
1 1
1
1 1
From this statement, y  cos
. To find sec x, we find cos
.
x
x
• Note: Since we take the inverse tangent of the reciprocal of x
to find cot-1 x, the calculator gives values of cot-1 with the same
range as tan-1, (–/2, /2), which is incorrect. The proper range
must be considered and the results adjusted accordingly.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-18
9.4 Finding Inverse Functions with a
Calculator
Example
(a) Find y in radians if y = csc-1(–3).
(b) Find  in degrees if  = arccot(–0.3541).
Solution
(a) In radian mode, enter y = csc-1(–3)
as sin-1( 13 ) to get y  –0.3398369095.
(b) In degree mode, the calculator gives
inverse tangent values of a negative
number as a quadrant IV angle. But
 must be in quadrant II for a negative
number, so we enter arccot(–0.3541) as
tan-1(1/ –0.3541) +180°,   109.4990544°.
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-19
9.4 Finding Function Values
Example Evaluate each expression without a calculator.
1
1
(a ) sin(tan 32 )
( b) tan(cos (  135 ))
Solution
(a) Let  = tan-1 32 so that tan  = 32 . Since 32
is positive,  is in quadrant I. We sketch
the figure to the right , so
3
3 13

1 3 
sin  tan   sin  

.
13
13

2
(b) Let A = cos-1(  135 ). Then cos A =  135 .
Since cos-1 x for a negative x is in
quadrant II, sketch A in quadrant II.
1
tan(cos (  135 ))  tan A   125
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-20
9.4 Writing Function Values in Terms of u
Example
Write each expression as an algebraic expression in u.
1
1
(a ) sin(tan u )
(b) cos( 2 sin u )
Solution
(a) Let  = tan-1 u, so tan  = u. Sketch  in quadrants I and IV
1



tan
 2 .
since 2
u
u u 2 1
1
sin(tan u )  sin  

2
u 2 1
u 1
(b) Let  = sin-1 u, so sin  = u.
cos 2  1  2 sin   1  2u
2
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2
Slide 9.4-21
9.4 Finding the Optimal Angle of
Elevation of a Shot Put
Example The optimal angle of elevation  a shot putter
should aim for to throw the greatest distance depends on the
velocity of the throw and the initial height of the shot. One
model for  that achieves this goal is


v2
.
  arcsin 
2
 2v  64h 
Figure 32 pg 9-73
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-22
9.4 Finding the Optimal Angle of
Elevation of a Shot Put
Suppose a shot putter can consistently throw a steel ball with
h = 7.6 feet and v = 42 ft/sec. At what angle should he throw the
ball to maximize distance?
Solution
Substitute into the model and use a calculator in
degree mode.


42 2

  arcsin 
2
 2( 42)  64(7.6) 
  41.5
Copyright © 2011 Pearson Education, Inc.
Slide 9.4-23