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Question 256 of 293

Q.If y=cos⁡−1(2x1−x2)y = \cos^{-1}\left(2x\sqrt{1 - x^2}\right), find dydx\dfrac{dy}{dx}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 3mImportance★★★★★
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Substitute x=sin⁡θx=\sin\theta to simplify 2x1−x22x\sqrt{1-x^2} into sin⁡2θ\sin2\theta, collapsing the inverse cosine.

y=cos⁡−1(2x1−x2)y=\cos^{-1}\left(2x\sqrt{1-x^2}\right)

Let x=sin⁡θx=\sin\theta, i.e. θ=sin⁡−1x\theta=\sin^{-1}x, with θ∈[−π2,π2]\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]. Then:

2x1−x2=2sin⁡θ1−sin⁡2θ=2sin⁡θcos⁡θ=sin⁡2θ2x\sqrt{1-x^2}=2\sin\theta\sqrt{1-\sin^2\theta}=2\sin\theta\cos\theta=\sin2\theta

So:

y=cos⁡−1(sin⁡2θ)=cos⁡−1[cos⁡(π2−2θ)]y=\cos^{-1}(\sin2\theta)=\cos^{-1}\left[\cos\left(\frac\pi2-2\theta\right)\right]

For θ∈[−π4,π4]\theta\in\left[-\dfrac\pi4,\dfrac\pi4\right] (i.e. x∈[−12,12]x\in\left[-\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right]), π2−2θ∈[0,π]\dfrac\pi2-2\theta\in[0,\pi], which is exactly the principal range of cos⁡−1\cos^{-1}, so:

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