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Q. The principal value of sin$^{-1}\bigg(sin\frac{2\pi}{3}\bigg)$ is

IIT JEEIIT JEE 1986

Solution:

sin$^{-1}\bigg(sin\frac{2\pi}{3}\bigg) = sin^{-1} \bigg[ sin\bigg(\pi -\frac{\pi}{3}\bigg)\bigg]$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =sin^{-1}\bigg(sin\frac{\pi}{3}\bigg)=\frac{\pi}{3}$