For the principal values, evaluate each of the following:
Question:

For the principal values, evaluate each of the following:

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

(ii)

(iii) $\sin ^{-1}\left(-\frac{1}{2}\right)+2 \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

(iv) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Solution:

(i) $\cos ^{-1}(\cos x)=x$

$\sin ^{-1}(\sin x)=x$

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

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

$=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)$

$=\frac{2 \pi}{3}$

(ii) 

(iii) $\sin ^{-1}\left(-\frac{1}{2}\right)+2 \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

$=\sin ^{-1}\left\{\sin \left(-\frac{\pi}{6}\right)\right\}+2 \cos ^{-1}\left(\cos \frac{5 \pi}{6}\right) \quad\left[\because\right.$ Range of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ;-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and range of cosine is $\left.[0, \pi] ; \frac{5 \pi}{6} \in[0, \pi]\right]$

$=-\frac{\pi}{6}+2\left(\frac{5 \pi}{6}\right)$

$=-\frac{\pi}{6}+\frac{5 \pi}{3}$

$=\frac{9 \pi}{6}$

$=\frac{3 \pi}{2}$

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

(iv) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

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

$=-\frac{\pi}{3}+\frac{\pi}{6}$                $\left[\because\right.$ Range of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ;-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

and range of cosine is $\left.[0, \pi] ; \frac{\pi}{6} \in[0, \pi]\right]$

$=-\frac{\pi}{6}$

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

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