frac{d}{dx} left( cos^{-1} sqrt{frac{1 + cos | vedclass

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Question

(B) We know that $1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right)$.Substituting this into the expression,we get:$\cos^{-1} \sqrt{\frac{2 \cos^2 (x/2)

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) We know that $1 + \cos x = 2 \cos^2 \left( \frac{x}{2} ight)$.Substituting this into the expression,we get:$\cos^{-1} \sqrt{\frac{2 \cos^2 (x/2)}{2}} = \cos^{-1} \sqrt{\cos^2 (x/2)} = \cos^{-1} \left( \cos \frac{x}{2} ight)$.Assuming the principal value branch,$\cos^{-1} (\cos 	heta) = 	heta$.Thus,the expression simplifies to $\frac{x}{2}$.Now,differentiating with respect to $x$:$\frac{d}{dx} \left( \frac{x}{2} ight) = \frac{1}{2}$.

$\frac{d}{dx} \left( \cos^{-1} \sqrt{\frac{1 + \cos x}{2}} \right) = $

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