cos ^2left(frac{pi}{6}+thetaright)-sin ^2left | 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 $B = \frac{\pi}{6} - 	heta$.Th

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

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

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