Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Please solve rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 5 sub question (i) maths textbook solution

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

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 Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers

anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question