If cos(alpha + beta) = 0,then sin(alpha + 2be | vedclass

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(A) Given $\cos(\alpha + \beta) = 0$.Since $\cos(	heta) = 0$ implies $	heta = \frac{\pi}{2} + n\pi$,we take the principal value $\alpha + \beta = \f

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$\sin \alpha$

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(A) Given $\cos(\alpha + \beta) = 0$.Since $\cos(	heta) = 0$ implies $	heta = \frac{\pi}{2} + n\pi$,we take the principal value $\alpha + \beta = \frac{\pi}{2}$.Thus,$\beta = \frac{\pi}{2} - \alpha$.Now,substitute $\beta$ into the expression $\sin(\alpha + 2\beta)$:$\sin(\alpha + 2(\frac{\pi}{2} - \alpha)) = \sin(\alpha + \pi - 2\alpha) = \sin(\pi - \alpha)$.Using the identity $\sin(\pi - 	heta) = \sin 	heta$,we get $\sin(\pi - \alpha) = \sin \alpha$.

If $\cos(\alpha + \beta) = 0$,then $\sin(\alpha + 2\beta) =$

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