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CBSE Class 12 Math 2009 Solved Paper

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Question : 1 of 29
Marks: +1, -0
Using principal value, evaluate the following: sin⁡−1(sin⁡3π5)\sin^{-1}\left(\sin\frac{3\pi}{5}\right)
Solution:  
As sin⁡−1\sin^{-1} (sin θ) = θ so sin⁡−1(sin⁡(3π5))\sin^{-1}\left(\sin\left(\frac{3\pi}{5}\right)\right) = 3π5\frac{3\pi}{5}
But 3π5\frac{3\pi}{5} ∉ [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
So,
sin⁡−1(sin⁡(3π5))\sin^{-1}\left(\sin\left(\frac{3\pi}{5}\right)\right) = sin⁡−1(sin⁡(π−2π5))\sin^{-1}\left(\sin\left(\pi - \frac{2\pi}{5}\right)\right)
= sin⁡−1(sin⁡2π5)\sin^{-1}\left(\sin\frac{2\pi}{5}\right)
= 2π5\frac{2\pi}{5} ∊ [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]
∴ Principal value is 2π5\frac{2\pi}{5}
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