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CBSE Class 12 Math 2008 Solved Paper

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Question : 2 of 29
Marks: +1, -0
Evaluate: sin [π3−sin⁡−1(12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(\frac{1}{2}\right) \right]
Solution:  
We know that the domain and range of the principal value branch of function sin⁡−1\sin^{-1} is defined as:
sin⁡−1\sin^{-1} " [- 1 , 1] → [−π2,π2]\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]
∴ sin [π3−sin⁡−1(−12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right] = sin [π3−(−π6)]\left[ \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) \right]
= sin [π3+π6]\left[ \frac{\pi}{3} + \frac{\pi}{6} \right]
= sin [π2]\left[ \frac{\pi}{2} \right]
= 1
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