cos ^2(x)+cos ^2left(x+frac{pi}{3}right)+cos | vedclass

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(A) We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$.Substituting this into the expression:$= \frac{1 + \cos 2x}{2} + \frac{1 + \cos(2

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) We use the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$.Substituting this into the expression:$= \frac{1 + \cos 2x}{2} + \frac{1 + \cos(2x + \frac{2\pi}{3})}{2} + \frac{1 + \cos(2x - \frac{2\pi}{3})}{2}$$= \frac{1}{2} [3 + \cos 2x + \cos(2x + \frac{2\pi}{3}) + \cos(2x - \frac{2\pi}{3})]$Using the formula $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x \cos(\frac{2\pi}{3})]$Since $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$:$= \frac{1}{2} [3 + \cos 2x + 2\cos 2x(-\frac{1}{2})]$$= \frac{1}{2} [3 + \cos 2x - \cos 2x] = \frac{3}{2}$

$\cos ^2(x)+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right) = $

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