Page 195 - Calculus Demystified

Transcendental Functions
CHAPTER 6
182
on the interval (−π/2,π/2). At the endpoints of the interval, and only there, the
function Sin x takes the values −1 and +1. Therefore Sin x is increasing on its
entire domain. So it is one-to-one. Furthermore the Sine function assumes every
value in the interval [−1, 1]. Thus Sin :[−π/2,π/2]→[−1, 1] is one-to-one and
onto; therefore f(x) = Sin x is an invertible function.
We can obtain the graph of Sin −1 x by the principle of reflection in the line
y = x (Fig. 6.16). The function Sin −1 :[−1, 1]→[−π/2,π/2] is increasing,
one-to-one, and onto.

y
/2
_ 1
y = Sin x

1 x

_
/2

Fig. 6.16

The study of the inverse of cosine involves similar considerations, but we must
select a different domain for our function. We define Cos x to be the cosine function
restricted to the interval [0,π]. Then, as Fig. 6.17 shows, g(x) = Cos x is a one-to-
one function. It takes on all the values in the interval [−1, 1]. Thus Cos :[0,π]→
[−1, 1] is one-to-one and onto; therefore it possesses an inverse.
We reflect the graph of Cos x in the line y = x to obtain the graph of the function
Cos −1 . The result is shown in Fig. 6.18.

EXAMPLE 6.35
Calculate

√ √
3 2
−1 −1 −1
Sin , Sin 0, Sin − ,
2 2
√ √
3 2
Cos −1 − , Cos −1 0, Cos −1 .
2 2