248 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations
Restricted domain
[ 2 , ]
~ ~
2
Figure 4
ir
(2' 1)
IF
3Tr
1
2
2
y
27r 37r 7r1 (0, 0)7r Tr 27r
y = sin x
2
ir
( 2 , 1)
2
2
x
Reflecting the graph of y = sin x on the restricted domain, shown in Figure 5(a), across the line y = x gives the graph of the inverse function, shown in Figure 5(b). Some key points are labeled on the graph. The equation of the inverse of y = sin x is found by interchanging x and y to get
x = sin y.
This equation is solved for y by writing
y = sin' x (read inverse sine of x).
As Figure 5(b) shows, the domain of y = sin1 x is [ 1, 11, while the restricted domain of y = sin x, [  p2 , p2 1, is the range of y = sin1 x. An alternative notation for sin1 x is arcsin x.
(a) (b)
Figure 5
Inverse Sine Function
y = sin' x or y = arcsin x means that x = sin y, for  p2 ... y ... p 2.
We can think of y = sin' x or y = arcsin x as
y is the number (angle) in the interval [P2 , P2 D whose sine is x.
Thus, we can write y = sin1 x as sin y = x to evaluate it. We must pay close attention to the domain and range intervals.
ir (2' 1)
1
y = sin x
(0, 0)
1
ir
( 2' 1)
Restricted domain
[ lr
2 , 2
2
2
1
0
(2 , 6)
(1,2 ) ~
( , 3 )
^3 ~
2
2
( , 4 )
^2 ~
2
( 2, 6)
( ^2 ~r
2, 4 )
(0, 0)
x
1
y = sin1 x or y = arcsin x
( , 3 )
^3 ~ 2
(1, 2)
7T
2
y
IF
SECTION 6.1 Inverse Circular Functions
249
EXAMPLE 1
Finding Inverse Sine Values
Find y in each equation.
(a) y = arcsin 12 (b) y = sin1(1) (c) y = sin1(2)
ALGEBRAIC SOLUTION
(a) The graph of the function defined by y = arcsin x (Figure 5(b)) includes the point (12 , p6 ). Therefore, arcsin 12 = p6.
Alternatively, we can think of y = arcsin 12 as y is the number in [  p2 , p2 ] whose sine is 12.
Then we can write the given equation as sin y = 12.
Since sin p6 = 12 and p6 is in the range of the arcsine function, y = p
6.
(b) Writing the equation y = sin1(1) in the form sin y = 1 shows that y =  p2. Notice that the point ( 1,  p2 ) is on the graph of y = sin1 x.
(c) Because 2 is not in the domain of the inverse sine function, sin1(2) does not exist.
GRAPHING CALCULATOR SOLUTION
We graph the equation Y1 = sin1 X and find the points with Xvalues 12 = 0.5 and 1. For these two
Xvalues, Figure 6 indicates that Y = p6 0.52359878
and Y=  p2 = 1.570796.
1
2
2
Figure 6
Since sin1(2) does not exist, a calculator will give an error message for this input.
n^ Now Try Exercises 13, 21, and 25.
2
1 1
2
1
CAUTION In Example 1(b), it is tempting to give the value of sin1(1) as 3p2 , since sin 3p2 = 1. Notice, however, that 3p2 is not in the range of the inverse sine function. Be certain that the number given for an inverse function value is in the range of the particular inverse function being considered.
We summarize this discussion about the inverse sine function as follows.
x
y
it y = sin1 x
Y
2
p
1

2
2
N/2
2
p
4
1
1
0
0
X
0 1
1
N/2
p
4
2
7T y = sin1 x
p
2
1
2
2
Figure 7
The inverse sine function is increasing and continuous on its domain [1, 1].
Its xintercept is 0, and its yintercept is 0.
Its graph is symmetric with respect to the origin, so the function is an odd function. For all x in the domain, sin1(x) = sin1 x.
Inverse Sine Function y = sin1 x or y = arcsin x
Domain: [ 1, 1] Range: [  p2 , p2 ]
250 CHAPTER 6 Inverse Circular Functions and Trigonometric Equations
Inverse Cosine Function The function
y = cos1 x (or y = arccos x)
is defined by restricting the domain of the function y = cos x to the interval 30, p] as in Figure 8. This restricted function, which is the part of the graph in Figure 8 shown in color, is onetoone and has an inverse function. The inverse function, y = cos1 x, is found by interchanging the roles of x and y. Reflecting the graph of y = cos x across the line y = x gives the graph of the inverse function shown in Figure 9. Some key points are shown on the graph.
Inverse Cosine Function
y = cos1 x or y = arccos x means that x = cos y, for 0 < y < p.
We can think of y = cos1 x or y = arccos x as
51r y is the number (angle) in the interval [0, P] whose cosine is x.
0
1
y = cos1 x or y = arccos x
Figure 9
Figure 8
(1, 10
( , )
^3 5
2 6
7T
(2 , )
1 ~
3
( , 4 )
^2 ~
2
( , 6 )
^3 ~
2
(1, 0)
( , )
^2 3
2 4
( 2 , )
1 2 3
(~ 0, ) 2
X
1
2
(~, 1)
(0, 1)
1
(, 0) ~ 2
0
1
y = cos x
Restricted domain [0, ir]
2
4
1
1
a '\12
2 b
0.5
Find y in each equation.
(a) y = arccos 1 (b) y = cos1
EXAMPLE 2
Finding Inverse Cosine Values
51r SOLUTION
4
1
1
0.5
These screens support the results of Example 2 because
= 2.3561945.
= 0.7071068 and
^2
2
3,
4
(a) Since the point 11, 02 lies on the graph of y = arccos x in Figure 9, the value of y, or arccos 1, is 0. Alternatively, we can think of y = arccos 1 as
y is the number in 30, p] whose cosine is 1, or cos y = 1.
Thus, y = 0, since cos 0 = 1 and 0 is in the range of the arccosine function.
(b) We must find the value of y that satisfies
cos y =  '\12
2 , where y is in the interval 30, p],
which is the range of the function y = cos1 x. The only value for y that satisfies these conditions is 3p4 . Again, this can be verified from the graph in Figure 9.
n^ Now Try Exercises 15 and 23.
SECTION 6.1 Inverse Circular Functions
Our observations about the inverse cosine function lead to the following generalizations.
to the open interval 1  p2 , p2 2 yields a onetoone function. By interchanging the roles of x and y, we obtain the inverse tangent function given by y = tan1 x or y = arctan x. Figure 11 shows the graph of the restricted tangent function. Figure 12 gives the graph of y = tan1 x.
y
Inverse Tangent Function Restricting the domain of the function y = tan x
Inverse Cosine Function y = cos1x or y = arccos x
The inverse cosine function is decreasing and continuous on its domain 3 1, 14.
Its xintercept is 1, and its yintercept is p2.
Its graph is not symmetric with respect to either the yaxis or the origin.
N/2
2
N/2
2
1
x
0
1
3p
y
p
p
p
0
4
4
2
Domain: 3 1, 14 Range: 30, p4
1 0
y
y = cos
Figure 10
1
1 x
x
1 1
0
y = cos1 x
251
y = tan x
(4 , 1) ir
2
1
(0, 0)
0
2
2
(
ir
4 , 1)
1
2
Restricted domain
( 2 , 2)
y
x
2
^3
( 3 , 6 )
(1, 4 )
( , 6 )
^3 ~
3
(^3 , 3 ) ~
x
r
2
IF
4
1 (0, 0)
IT
4
IT
(^3 , 3 )
~
(1, 4)
2
y = tan1 x or y = arctan x
1 2
Figure 11 Figure 12
Inverse Tangent Function
y = tan1 x or y = arctan x means that x = tan y, for  p2 6 y 6 p2.
We can think of y = tan1 x or y = arctan x as
y is the number (angle) in the interval AP2 , P2 ~ whose tangent is x.