#### Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 2 sub question (iii) maths textbook solution

Answer:   $\frac{2\pi }{3}$

Hint: The principal value branch of function $\cos ^{-1}$ is $\left [ 0,\pi \right ]$
Given: $\cos ^{-1}\left ( \cos \frac{4\pi }{3} \right )$

Explanation:

First we solve $\cos \left ( \frac{4\pi }{3} \right )$

\begin{aligned} &\cos \frac{4 \pi}{3}=\cos \left(\pi+\frac{\pi}{3}\right) \\ &\therefore \cos [\pi+\theta]=-\cos \theta \end{aligned}

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

By substituting these value in $\cos ^{-1}\left\{\cos \left(\frac{4 \pi}{3}\right)\right\}$

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

Now,

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

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

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

$\cos^{-1}(\cos\frac{2\pi}{3})=\frac{2\pi}{3}$