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CBSE Class 12 Math 2011 Solved Paper

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Question : 2 of 29
Marks: +1, -0
Write the value of sin[π3−sin⁡−1(−12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right]
Solution:  
sin[π3−sin⁡−1(−12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right]
Let sin⁡−1(−12)\sin^{-1}\left(-\frac{1}{2}\right) = x
⇒ (−12)\left(-\frac{1}{2}\right) = sin x
⇒ sin x = - sin π6\frac{\pi}{6} = sin (−π6)\left(-\frac{\pi}{6}\right) = sin (2π−π6)\left(2\pi - \frac{\pi}{6}\right)
⇒ x = 2π - π6\frac{\pi}{6}
∴ sin [π3−sin⁡−1(−12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right] = sin [π3−(2π−π6)]\left[ \frac{\pi}{3} - \left(2\pi - \frac{\pi}{6}\right) \right]
= sin [−9π6]\left[ -\frac{9\pi}{6} \right]
= - sin [3π2]\left[ \frac{3\pi}{2} \right]
= - sin [π+π2]\left[ \pi + \frac{\pi}{2} \right]
= - [−sin⁡π2]\left[ -\sin\frac{\pi}{2} \right]
= + sin π2\frac{\pi}{2}
= 1
Thus,sin[π3−sin⁡−1(−12)]\left[ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right] = 1
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