Chapter 6, Problem 1RE

In problems $7-14,$ find the exact value of each expression. Do not use a calculator.

${\sin }^{-1}1$

To determine

The exact value of the expression ${\sin }^{-1}1$.

Answer to Problem 1RE

Solution:

The exact value of the expression ${\sin }^{-1}1$ is $\frac{\pi }{2}$.

Explanation of Solution

Given information:

${\sin }^{-1}1$.

Explanation:

Let $\theta ={\sin }^{-1}1$

Then $\theta$ is the angle, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$, whose sine equals $1$.

Since, the domain of function $y={\sin }^{-1}x$ is the interval $\left[-1,1\right]$ and its range is the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

Thus, $\theta ={\sin }^{-1}1$; $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

$\Rightarrow \sin \theta =1$; $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

Using table and figure,

$\begin{array}{cccccccccc}\theta & -\frac{\pi }{2}& -\frac{\pi }{3}& -\frac{\pi }{4}& -\frac{\pi }{6}& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}\\ \sin \theta & -1& -\frac{\sqrt{3}}{2}& -\frac{\sqrt{2}}{2}& -\frac{1}{2}& 0& \frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{\sqrt{3}}{2}& 1\end{array}$

The only angle $\theta$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ whose $\sin \theta =1$ is $\frac{\pi }{2}$.

$\Rightarrow {\sin }^{-1}1=\frac{\pi }{2}$.

Therefore, the exact value of the expression ${\sin }^{-1}1$ is $\frac{\pi }{2}$.