Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let f(α)=sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)f(\alpha)=\sin^{-1}(\sin\alpha)+\cos^{-1}(\cos\alpha) and g(β)=sin⁡−1(sin⁡β)+tan⁡−1(tan⁡β)g(\beta)=\sin^{-1}(\sin\beta)+\tan^{-1}(\tan\beta). Find the value of f(100)+g(8)f(100)+g(8).
Solution
Answer: 0
Step 1: Compute f(100)f(100). Note 32π≈100.5332\pi\approx 100.53, so 100=32π−c100=32\pi-c where c≈0.53∈(0,π2)c\approx 0.53\in\left(0,\dfrac{\pi}{2}\right). sin⁡100=sin⁡(32π−c)=−sin⁡c\sin 100=\sin(32\pi-c)=-\sin c, so sin⁡−1(sin⁡100)=sin⁡−1(−sin⁡c)=−c=100−32π\sin^{-1}(\sin 100)=\sin^{-1}(-\sin c)=-c=100-32\pi. cos⁡100=cos⁡(32π−c)=cos⁡c\cos 100=\cos(32\pi-c)=\cos c, so cos⁡−1(cos⁡100)=c=32π−100\cos^{-1}(\cos 100)=c=32\pi-100. Step 2: Sum:
f(100)=(100−32π)+(32π−100)=0f(100)=(100-32\pi)+(32\pi-100)=0
 Then sin⁡−1(sin⁡8)=d=3π−8\sin^{-1}(\sin 8)=d=3\pi-8 (since this lies in [−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]). For tan⁡8\tan 8: tan⁡(3π−d)=−tan⁡d=tan⁡(8−3π)\tan(3\pi-d)=-\tan d=\tan(8-3\pi) since tan⁡\tan has period π\pi. Actually tan⁡(8)=tan⁡(8−3π)\tan(8)=\tan(8-3\pi) where 8−3π≈−1.42∈(−π2,π2)8-3\pi\approx -1.42\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right). So tan⁡−1(tan⁡8)=8−3π\tan^{-1}(\tan 8)=8-3\pi. Step 3: Sum:
g(8)=(3π−8)+(8−3π)=0g(8)=(3\pi-8)+(8-3\pi)=0
 Step 4: Therefore:
f(100)+g(8)=0+0=0f(100)+g(8)=0+0=0
 Answer: 00
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