If cos(theta - alpha) = a and sin(theta - bet | vedclass

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(C) Given: $\cos(	heta - \alpha) = a$ and $\sin(	heta - \beta) = b$.Let $x = 	heta - \alpha$ and $y = 	heta - \beta$. Then $\cos x = a$ and $\sin

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$a^2 + b^2$

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(C) Given: $\cos(	heta - \alpha) = a$ and $\sin(	heta - \beta) = b$.Let $x = 	heta - \alpha$ and $y = 	heta - \beta$. Then $\cos x = a$ and $\sin y = b$.Note that $x - y = (	heta - \alpha) - (	heta - \beta) = \beta - \alpha$,so $\alpha - \beta = y - x$.We need to find $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta) = \cos^2(y - x) + 2ab\sin(y - x)$.Using $\cos(y - x) = \cos y \cos x + \sin y \sin x = \cos y \cdot a + b \cdot \sin x$.Since $\sin x = \sqrt{1 - a^2}$ and $\cos y = \sqrt{1 - b^2}$,we have $\cos(y - x) = a\sqrt{1 - b^2} + b\sqrt{1 - a^2}$.Using $\sin(y - x) = \sin y \cos x - \cos y \sin x = ba - \sqrt{1 - b^2}\sqrt{1 - a^2}$.Substituting these into the expression:$\cos^2(y - x) + 2ab\sin(y - x) = (a\sqrt{1 - b^2} + b\sqrt{1 - a^2})^2 + 2ab(ab - \sqrt{1 - b^2}\sqrt{1 - a^2})$$= a^2(1 - b^2) + b^2(1 - a^2) + 2ab\sqrt{1 - b^2}\sqrt{1 - a^2} + 2a^2b^2 - 2ab\sqrt{1 - b^2}\sqrt{1 - a^2}$$= a^2 - a^2b^2 + b^2 - a^2b^2 + 2a^2b^2 = a^2 + b^2$.

If $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$,then $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta)$ is equal to

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