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provide solution for rd sharma maths class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (ii)

Answers (1)

Answer: ${\frac{2\pi }{3}}$

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

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

Solution:

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

$\cos x=\frac{1}{2}=\cos \left ( {\frac{\pi }{3}}\right )$

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

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

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

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

From (1) and (2),

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

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

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

$\therefore$ Principal value of $\cos ^{-1}\left ( \frac{1}{2} \right )-2\sin ^{-1}\left ( \frac{-1}{2} \right )$ $=\frac{2\pi }{3}$

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