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CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper

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Q. Nos. 16 to bo20\text{bo} 20 are of very short answer type questions.
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Question : 16 of 36
Marks: +1, -0
Find the value of sin⁡−1[sin⁡(−17π8)]\sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right].
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sin⁡−1(sin⁡x)=x;−π2≤x≤π2\sin^{-1}(\sin x) = x ; \quad -\frac{\pi}{2} \le x \le \frac{\pi}{2}
sin⁡−1[sin⁡(−17π8)].\sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right] .
=sin⁡−1[−sin⁡(17π8)]=sin⁡−1[−sin⁡(2π+π8)].= \sin^{-1} \left[ -\sin \left( \frac{17\pi}{8} \right) \right] = \sin^{-1} \left[ -\sin \left( 2\pi + \frac{\pi}{8} \right) \right] .
=sin⁡−1[−sin⁡(17π8)]=sin⁡−1[−sin⁡(2π+π8)]= \sin^{-1} \left[ -\sin \left( \frac{17\pi}{8} \right) \right] = \sin^{-1} \left[ -\sin \left( 2\pi + \frac{\pi}{8} \right) \right].
=sin⁡−1[−sin⁡(π8)]=sin⁡−1[sin⁡(−π8)]= \sin^{-1} \left[ -\sin \left( \frac{\pi}{8} \right) \right] = \sin^{-1} \left[ \sin \left( -\frac{\pi}{8} \right) \right]
=−π8= -\frac{\pi}{8}
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