Trigonometric Functions — Maths STD 12 Science — Question

$\therefore \cos \alpha=\frac{1}{2}=\cos \frac{\pi}{3}$

$\therefore \alpha=\frac{\pi}{3} \quad \cdots\left[\because 0<\frac{\pi}{3}<\pi\right]$

$\therefore \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$$\ldots(1)$

Let $\sin ^{-1}\left(\frac{1}{2}\right)=\beta$, where $\frac{-\pi}{2} \leq \beta \leq \frac{\pi}{2}$

$\therefore \sin \beta=\frac{1}{2}=\sin \frac{\pi}{6}$

$\therefore \beta=\frac{\pi}{6} \quad \cdots\left[\because \frac{-\pi}{2} \leq \frac{\pi}{6} \leq \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

...(2)

$\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ and $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

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

$\begin{aligned} & =\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \\ & =\frac{\pi}{3}+\frac{\pi}{3} \\ & =\frac{2 \pi}{3}\end{aligned}$