#### explain solution RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (ii) maths

Answer:   $\frac{3\pi }{4}$

Hint: The principal value branch of function  $\cos ^{-1}$  is  $[0,\pi ]$

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

Explanation:

First we solve $\cos \frac{5\pi }{4}$

\begin{aligned} &\cos \frac{5 \pi}{4}=\cos \left(\pi+\frac{\pi}{4}\right) \\ &\cos \left(\pi+\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \quad \because[\cos (-\theta)=\cos \theta] \end{aligned}

As we know that $\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

$\cos \left(\frac{5 \pi}{4}\right)=-\frac{1}{\sqrt{2}}$

By substituting these value in $\cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}$

we get,

$\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Since   $[\cos (-\theta)=\cos \theta]$

Let            $y=\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

\begin{aligned} &\cos y=-\frac{1}{\sqrt{2}} \\ &-\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \\ &\cos \left(\pi-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}} \\ &\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}} \end{aligned}

Hence, range of principal value of  $\cos ^{-1} \text { is }[0, \pi] \text { and } \cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}$

$\cos ^{-1}\left\{\cos \left(\frac{5 \pi}{4}\right)\right\}=\frac{3 \pi}{4} \therefore \cos ^{-1}(\cos x)=x, x \in[0, \pi]$