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Exercise 4.6 · Q14

Q.sin⁡−1(2cos⁡2x−1)+cos⁡−1(1−2sin⁡2x)=\sin^{-1}\left(2\cos^2x - 1\right) + \cos^{-1}\left(1-2\sin^2x\right) =

(1) π2\dfrac{\pi}2
(2) π3\dfrac{\pi}3
(3) π4\dfrac{\pi}4
(4) π6\dfrac{\pi}6
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Both 2cos⁡2x−12\cos^2x-1 and 1−2sin⁡2x1-2\sin^2x are standard double-angle forms of cos⁡2x\cos2x, so the whole expression collapses to sin⁡−1(cos⁡2x)+cos⁡−1(cos⁡2x)\sin^{-1}(\cos2x)+\cos^{-1}(\cos2x), which is π2\dfrac{\pi}2 for ANY valid argument.

Step 1. Recognise the double-angle identities. 2cos⁡2x−1=cos⁡2x2\cos^2x-1=\cos2x and 1−2sin⁡2x=cos⁡2x1-2\sin^2x=\cos2x — both are the same quantity.

Step 2. Rewrite the expression. sin⁡−1(2cos⁡2x−1)+cos⁡−1(1−2sin⁡2x)=sin⁡−1(cos⁡2x)+cos⁡−1(cos⁡2x)\sin^{-1}(2\cos^2x-1)+\cos^{-1}(1-2\sin^2x)=\sin^{-1}(\cos2x)+\cos^{-1}(\cos2x). …

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