In triangle ABC, if angle C = frac{pi}{3}, th | vedclass

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Question

(A) Given $\angle C = \frac{\pi}{3}$, so $A + B = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.We need to evaluate $X = \cos^2 A + \cos^2 B + \cos A \cos B$.

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) Given $\angle C = \frac{\pi}{3}$, so $A + B = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.We need to evaluate $X = \cos^2 A + \cos^2 B + \cos A \cos B$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$, we get:$X = \frac{1 + \cos 2A}{2} + \frac{1 + \cos 2B}{2} + \cos A \cos B$$X = 1 + \frac{1}{2}(\cos 2A + \cos 2B) + \cos A \cos B$Using $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$:$X = 1 + \cos(A+B) \cos(A-B) + \cos A \cos B$Since $A+B = \frac{2\pi}{3}$, $\cos(A+B) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$.$X = 1 - \frac{1}{2} \cos(A-B) + \cos A \cos B$Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$X = 1 - \frac{1}{2} \cos(A-B) + \frac{1}{2} [\cos(A+B) + \cos(A-B)]$$X = 1 - \frac{1}{2} \cos(A-B) + \frac{1}{2} [-\frac{1}{2} + \cos(A-B)]$$X = 1 - \frac{1}{4} = \frac{3}{4}$.

In $\triangle ABC$, if $\angle C = \frac{\pi}{3}$, then the value of $\cos^2 A + \cos^2 B + \cos A \cos B$ is...

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