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NCERT Exemplar · Q37

Q.Differentiate w.r.t. xx: cos⁡−1(sin⁡x+cos⁡x2), −π4<x<π4\cos^{-1}\left(\dfrac{\sin x + \cos x}{\sqrt{2}}\right),\ -\dfrac{\pi}{4} < x < \dfrac{\pi}{4}.

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The key idea is to simplify the argument inside cos⁡−1\cos^{-1} using the sine addition formula, then differentiate the resulting linear function. The derivative is −1-1.

We are asked to differentiate cos⁡−1(sin⁡x+cos⁡x2)\cos^{-1}\left(\dfrac{\sin x + \cos x}{\sqrt{2}}\right) with respect to xx, for −π4<x<π4-\dfrac{\pi}{4} < x < \dfrac{\pi}{4}.

The expression inside the inverse cosine looks like it could be a single trigonometric function. Recall the identity: sin⁡Acos⁡B+cos⁡Asin⁡B=sin⁡(A+B)\sin A \cos B + \cos A \sin B = \sin(A+B). If we factor 12\frac{1}{\sqrt{2}}, we can write sin⁡x2+cos⁡x2\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}. Notice that 12=sin⁡π4=cos⁡π4\frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} = \cos \frac{\pi}{4}. So:

sin⁡x+cos⁡x2=sin⁡x⋅12+cos⁡x⋅12=sin⁡xcos⁡π4+cos⁡xsin⁡π4\frac{\sin x + \cos x}{\sqrt{2}} = \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}

This is exactly sin⁡(x+π4)\sin\left(x + \frac{\pi}{4}\right).

Tip

A quick way to spot this: any expression of the form asin⁡x+bcos⁡xa \sin x + b \cos x can be written as Rsin⁡(x+ϕ)R \sin(x + \phi) or Rcos⁡(x−ϕ)R \cos(x - \phi). Here a=b=1a = b = 1, so R=12+12=2R = \sqrt{1^2 + 1^2} = \sqrt{2}, and the angle shift is π4\frac{\pi}{4}.

So the function becomes:

y=cos⁡−1(sin⁡(x+π4))y = \cos^{-1}\left( \sin\left(x + \frac{\pi}{4}\right) \right)

Now we need to differentiate this. But cos⁡−1\cos^{-1} and sin⁡\sin are related: cos⁡−1(sin⁡θ)=π2−θ\cos^{-1}(\sin \theta) = \frac{\pi}{2} - \theta, provided θ\theta lies in the range where this holds. Let's check the domain. …

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