Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

Answers (1)

best_answer

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

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10, 12 results 2024 likely after May 20, says result portal

anam-khan

CBSE Result Date 2024 Live: CBSE Class 10, 12 board results for 39 lakh students at cbse.nic.in after May 20

anam-khan

CBSE Results 2024: Last 9 years’ CBSE Class 10, 12 result date and time; links to check score

Latest Question