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Question

Mathematics Question on Inverse Trigonometric Functions

Find the principal value sin⁡−1(−12)\sin^{-1}\left(\frac{-1}{2}\right)

A

π3\frac{\pi}{3}

B

−π3-\frac{\pi}{3}

C

π6\frac{\pi}{6}

D

−π6-\frac{\pi}{6}

Answer

−π6-\frac{\pi}{6}

Explanation

Solution

Let sin⁡−1(−12)=θ\sin^{-1}\left(\frac{-1}{2}\right) = \theta ⇒sin⁡ θ=−12\Rightarrow \sin\,\theta = \frac{-1}{2} =−sin⁡π6=- \sin \frac{\pi}{6} =sin⁡(−π6)= \sin \left(\frac{-\pi }{6}\right) ⇒ θ=−π6∈[−π2,π2]\Rightarrow \, \theta = \frac{-\pi }{6}\in\left[\frac{-\pi }{2}, \frac{\pi }{2}\right] ∴ \therefore \, Principal value of sin⁡−1(−12)\sin^{-1} \left(\frac{-1}{2}\right) is (−π6)\left(\frac{-\pi }{6}\right)