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Mathematics Question on Inverse Trigonometric Functions

Find the values of sin⁡(π3−sin⁡−1(−12))\sin(\frac{\pi}{3}-\sin^{-1}(-\frac{1}{2})) is equal to

A

12\frac{1}{2}

B

13\frac{1}{3}

C

14\frac{1}{4}

D

1

Answer

1

Explanation

Solution

Let sin⁡−1(−12)=x\sin^{-1}(-\frac{1}{2})=x
Then, sin x=−12=−sinπ6=sin⁡(−π6).-\frac{1}{2}=-sin\frac{\pi}{6}=\sin(-\frac{\pi}{6}).
We know that the range of the principal value branch of sin-1=[-π/2,π/2].
sin-1(-1/2)=-π/6
therefore sin(π/3-sin1(-1/2))=sin(π/3+π/6)=sin(3π/6)=sin(π/2)=1.
The correct answer is D.