cos^2 left( frac{pi}{6} + theta right) - sin^ | vedclass

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Question

(A) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.Let $A = \frac{\pi}{6} + 	heta$ and $B = \frac{\pi}{6} - 	het

Vedclass · Accepted Answer

$\frac{1}{2} \cos 2	heta$

Vedclass · Answer

(A) We use the trigonometric identity: $\cos^2 A - \sin^2 B = \cos(A + B) \cos(A - B)$.Let $A = \frac{\pi}{6} + 	heta$ and $B = \frac{\pi}{6} - 	heta$.Then,$A + B = \left( \frac{\pi}{6} + 	heta ight) + \left( \frac{\pi}{6} - 	heta ight) = \frac{2\pi}{6} = \frac{\pi}{3}$.And,$A - B = \left( \frac{\pi}{6} + 	heta ight) - \left( \frac{\pi}{6} - 	heta ight) = 2	heta$.Substituting these into the identity:$\cos^2 \left( \frac{\pi}{6} + 	heta ight) - \sin^2 \left( \frac{\pi}{6} - 	heta ight) = \cos \left( \frac{\pi}{3} ight) \cos(2	heta)$.Since $\cos \left( \frac{\pi}{3} ight) = \frac{1}{2}$,the expression becomes $\frac{1}{2} \cos 2	heta$.

$\cos^2 \left( \frac{\pi}{6} + \theta \right) - \sin^2 \left( \frac{\pi}{6} - \theta \right) = $

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