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Question

Mathematics Question on Trigonometric Functions

The principal value of sin−1(sin2π3)^{-1}\bigg(sin\frac{2\pi}{3}\bigg) is

A

−2π3-\frac{2 \pi}{3}

B

2π3\frac{2 \pi}{3}

C

π3\frac{ \pi}{3}

D

5π3\frac{5 \pi}{3}

Answer

π3\frac{ \pi}{3}

Explanation

Solution

sin−1(sin2π3)=sin−1[sin(π−π3)]^{-1}\bigg(sin\frac{2\pi}{3}\bigg) = sin^{-1} \bigg[ sin\bigg(\pi -\frac{\pi}{3}\bigg)\bigg]
                   =sin−1(sinπ3)=π3\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =sin^{-1}\bigg(sin\frac{\pi}{3}\bigg)=\frac{\pi}{3}