Chapter 7, Problem 7RE

To determine

The exact value of the expression ${\sec }^{-1}\left(\sqrt{2}\right)$.

Answer to Problem 7RE

Solution:

The exact value of the expression ${\sec }^{-1}\left(\sqrt{2}\right)$ is $\frac{\pi }{4}$.

Explanation of Solution

Given information:

${\sec }^{-1}\left(\sqrt{2}\right)$.

Explanation:

Let $\theta ={\sec }^{-1}\left(\sqrt{2}\right)$.

Then, $\sec \theta =\sqrt{2}$ and $0\le \theta \le \pi ,\theta \ne \frac{\pi }{2}$.

Since, the domain of function $y={\sec }^{-1}x$ is $\left|x\right|\ge 1$ and its range is the interval $\left[0,\pi \right],y\ne \frac{\pi }{2}$.

Now find $\cos \theta$ because $y={\cos }^{-1}x$ has the same range as $y={\sec }^{-1}x$, except where undefined.

Since, $\sec \theta =\frac{1}{\cos \theta }=\sqrt{2}$, this means $\cos \theta =\frac{1}{\sqrt{2}}$.

Thus, ${\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\theta$.

$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$; $0\le \theta \le \pi ,\theta \ne \frac{\pi }{2}$.

Using table and figure,

$\begin{array}{cccccccccc}\theta & 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ \cos \theta & 1& \frac{\sqrt{3}}{2}& \frac{1}{\sqrt{2}}& \frac{1}{2}& 0& -\frac{1}{2}& -\frac{1}{\sqrt{2}}& -\frac{\sqrt{3}}{2}& -1\end{array}$

The angle $\theta$ such that $0\le \theta \le \pi$, whose cosine is $\frac{1}{\sqrt{2}}$ is $\frac{\pi }{4}$.

$\Rightarrow {\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$.

Hence, ${\sec }^{-1}\left(\sqrt{2}\right)={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$.

Therefore, the exact value of the expression ${\sec }^{-1}\left(\sqrt{2}\right)$ is $\frac{\pi }{4}$.