cos^2 alpha + cos^2(alpha + 120^circ) + cos^2 | vedclass

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Question

(A) Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ) = \frac{1

Vedclass · Accepted Answer

$3/2$

Vedclass · Answer

(A) Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ) = \frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos(2\alpha + 240^\circ)}{2} + \frac{1 + \cos(2\alpha - 240^\circ)}{2}$$= \frac{1}{2} [3 + \cos 2\alpha + \cos(2\alpha + 240^\circ) + \cos(2\alpha - 240^\circ)]$Using $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:$= \frac{1}{2} [3 + \cos 2\alpha + 2\cos 2\alpha \cos 240^\circ]$Since $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -1/2$:$= \frac{1}{2} [3 + \cos 2\alpha + 2\cos 2\alpha (-1/2)]$$= \frac{1}{2} [3 + \cos 2\alpha - \cos 2\alpha]$$= \frac{3}{2}$.

$\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ)$ is equal to

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