Chapter 7.2, Problem 21AYU

To determine

To calculate: The exact value of expression ${\cos }^{-1}\left(\sin \frac{5\pi }{4}\right)$ .

Answer to Problem 21AYU

The exact value is $\frac{3\pi }{4}$ .

Explanation of Solution

Given information:

  ${\cos }^{-1}\left(\sin \frac{5\pi }{4}\right)$ .

Formula used:

Trigonometric Ratios :

  $\sin \left(\pi +\theta \right)=-\sin \theta$ .

Calculation:

Consider ,

  ${\cos }^{-1}\left(\sin \frac{5\pi }{4}\right)$

Let

  $\theta =\sin \frac{5\pi }{4}$

where $-\pi <\theta <\pi$ .

It follows $\pi <\theta \le -\frac{\pi }{2}$ .

This implies $\theta$ lies in quadrant III.

Now,

  $\begin{array}{l}\sin \frac{5\pi }{4}=\sin \left(\pi +\frac{\pi }{4}\right)=-\sin \frac{\pi }{4}\\ =-\sin \frac{\pi }{4}=-\left(\frac{1}{\sqrt{2}}\right)\\ =-\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{array}$

Where the range of ${\cos }^{-1}\theta$ is $0\le \theta \le \pi$

and the negative value of ${\cos }^{-1}\theta$ lies between zero and $\frac{\pi }{2}$

  $\begin{array}{l}={\cos }^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\theta \\ =\cos \theta =-\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}\cos \theta =-\frac{1}{\sqrt{2}}\\ \cos \theta =-\frac{\pi }{4}\end{array}$

  $=\left(\pi -\frac{\pi }{4}\right)=\frac{3\pi }{4}$

Hence , the exact value of ${\cos }^{-1}\left(\sin \frac{5\pi }{4}\right)$ is $\frac{3\pi }{4}$ .