The value of is \(sin^{-1}\left(sin\frac{2\pi}{3}\right)\)A. \(\frac

1.	The value of is \(sin^{-1}\left(sin\frac{2\pi}{3}\right)\)A. \(\frac{2\pi}{3}\)B. \(\frac{5\pi}{3}\)C. \(\frac{\pi}{3}\)D. none of these
Answer» Correct Answer is \(\frac{\pi}{3}\) Now, let x = \(sin^{-1}(sin\left(\frac{2\pi}{3}\right))\) ⇒ sin x =sin ( \(\frac{2\pi}{3}\)) Here range of principle value of sine is [\(-\frac{\pi}{2},\frac{\pi}{2}\) ] ⇒ \(x=\frac{2\pi}{3}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) Hence for all values of x in range [\(-\frac{\pi}{2},\frac{\pi}{2}\) ] ,the value of \(sin^{-1}(sin\left(\frac{2\pi}{3}\right))\) is ⇒ sin x =sin ( \(\pi-\frac{\pi}{3}\)) ( \(\because\)sin ( \(\frac{2\pi}{3}\))= sin ( \(\pi-\frac{\pi}{3}\)) ) ⇒ sin x =sin ( \(\frac{\pi}{3}\)) (\(\because\) sin (π - θ) = sin θ as here θ lies in II quadrant and sine is positive) ⇒ x = \(\frac{\pi}{3}\)

The value of is \(sin^{-1}\left(sin\frac{2\pi}{3}\right)\)A. \(\frac{2\pi}{3}\)B. \(\frac{5\pi}{3}\)C. \(\frac{\pi}{3}\)D. none of these

Answer»

Correct Answer is \(\frac{\pi}{3}\)

Now, let x = \(sin^{-1}(sin\left(\frac{2\pi}{3}\right))\)

⇒ sin x =sin ( \(\frac{2\pi}{3}\))

Here range of principle value of sine is [\(-\frac{\pi}{2},\frac{\pi}{2}\) ]

⇒ \(x=\frac{2\pi}{3}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)

Hence for all values of x in range [\(-\frac{\pi}{2},\frac{\pi}{2}\) ] ,the value of

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

⇒ sin x =sin ( \(\pi-\frac{\pi}{3}\)) ( \(\because\)sin ( \(\frac{2\pi}{3}\))= sin ( \(\pi-\frac{\pi}{3}\)) )

⇒ sin x =sin ( \(\frac{\pi}{3}\))

(\(\because\) sin (π - θ) = sin θ as here θ lies in II quadrant and sine is positive)

⇒ x = \(\frac{\pi}{3}\)