cos^2 left( frac{pi}{4} - beta right) - sin^2 | vedclass

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Question

(D) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B)$.Let $A = \frac{\pi}{4} - \beta$ and $B = \alpha - \frac{\pi}{4}$.

Vedclass · Accepted Answer

$\sin(\alpha + \beta)\cos(\alpha - \beta)$

Vedclass · Answer

(D) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B)$.Let $A = \frac{\pi}{4} - \beta$ and $B = \alpha - \frac{\pi}{4}$.Then,$A + B = \left( \frac{\pi}{4} - \beta ight) + \left( \alpha - \frac{\pi}{4} ight) = \alpha - \beta$.And $A - B = \left( \frac{\pi}{4} - \beta ight) - \left( \alpha - \frac{\pi}{4} ight) = \frac{\pi}{2} - (\alpha + \beta)$.Substituting these into the identity:$\cos^2 \left( \frac{\pi}{4} - \beta ight) - \sin^2 \left( \alpha - \frac{\pi}{4} ight) = \cos(\alpha - \beta)\cos\left( \frac{\pi}{2} - (\alpha + \beta) ight)$.Since $\cos\left( \frac{\pi}{2} - 	heta ight) = \sin 	heta$,we get:$= \cos(\alpha - \beta)\sin(\alpha + \beta)$.

$\cos^2 \left( \frac{\pi}{4} - \beta \right) - \sin^2 \left( \alpha - \frac{\pi}{4} \right) = $

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