Get Answers to all your Questions

header-bg qa

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)


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