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Question

Mathematics Question on Trigonometric Functions

The principal value of sin⁡−1(sin⁡2π3)\sin^{-1} \left(\sin \frac{2\pi}{3}\right) is:

A

2π3\frac{2 \pi}{3}

B

−2π3- \frac{2 \pi}{3}

C

π3\frac{\pi}{3}

D

4π3\frac{4 \pi}{3}

Answer

π3\frac{\pi}{3}

Explanation

Solution

Let sin⁡−1(sin⁡2π3)=θ\sin^{-1} \left( \sin \frac{2 \pi}{3} \right) = \theta ⇒ sin⁡θ=sin⁡2π3\Rightarrow \, \sin \theta = \sin \frac{2 \pi}{3} Note: Principle value of sin x lies between −π2\frac{- \pi}{2} and π2\frac{\pi}{2} So, here sin⁡θ=sin⁡(π−π3)=sin⁡π3\sin\theta = \sin\left(\pi- \frac{\pi}{3}\right) = \sin \frac{\pi}{3} Therefore, principle value of sin⁡−1(sin⁡2π3)\sin^{-1} \left(\sin \frac{2 \pi}{3}\right) is π3\frac{\pi}{3}