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CBSE Class 12 Math 2020 Delhi Set 2 Solved Paper

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Question : 5 of 11
Marks: +1, -0
Evaluate : sin⁡[  π3−sin⁡−1(−12)]\sin \left[ \;\frac{\pi}{3} - \sin^{-1} \left( \frac{-1}{2} \right) \right]
Solution:  
The value of sin⁡(  π3−sin⁡−1(−12))\sin \left( \;\frac{\pi}{3} - \sin^{-1} \left( -\frac{1}{2} \right) \right)
=sin⁡(  π3−(−π6))= \sin \left( \;\frac{\pi}{3} - \left( -\frac{\pi}{6} \right) \right)
=sin⁡π2= \sin \frac{\pi}{2}
=1=1
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