Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the values of each of the expressions in Exercises 16 to 18. 16. sin inverse (sin 2pi / 3)

Find the values of each of the expressions in Exercises 16 to 18.

    16. \sin^{-1}\left (\sin\frac{2\pi}{3} \right )

Answers (1)

best_answer

Given \sin^{-1}\left (\sin\frac{2\pi}{3} \right );

We know that \sin^{-1}(\sin x) = x 

If the value of x belongs to  \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ] then we get the principal values of \sin^{-1}x.

Here, \frac{2\pi}{3} \notin \left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]

We can write \sin^{-1}\left (\sin\frac{2\pi}{3} \right )  is as:

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

 = \sin^{-1}\left [ \sin \frac{\pi}{3} \right ]  where \frac{\pi}{3} \epsilon \left [ \frac{-\pi}{2}, \frac{\pi}{2} \right ]

 \therefore \sin^{-1}\left (\sin\frac{2\pi}{3} \right )=\sin^{-1}\left [ \sin \frac{\pi}{3} \right ]=\frac{\pi}{3}

Posted by

Divya Prakash Singh

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 Board Exams 2026: Online portal for availing CWSN facilities opens from September 9
anam-khan

CBSE announces deadline for Class 10, 12 direct admissions, subject change
anam-khan

CBSE warns about unauthorised sources offering ‘quick’ ways of getting duplicate marksheets, certificates

Latest Question