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

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(A) Let $x = 	heta - \alpha$ and $y = 	heta - \beta$. Then $x - y = (	heta - \alpha) - (	heta - \beta) = \beta - \alpha$, so $\alpha - \beta = y -

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$a^2 + b^2$

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(A) Let $x = 	heta - \alpha$ and $y = 	heta - \beta$. Then $x - y = (	heta - \alpha) - (	heta - \beta) = \beta - \alpha$, so $\alpha - \beta = y - x$.Given $\cos(x) = a$ and $\sin(y) = b$.We need to evaluate $E = \cos^2(y - x) + 2ab \sin(y - x) + \cos^2(x)$.$E = (\cos y \cos x + \sin y \sin x)^2 + 2ab(\sin y \cos x - \cos y \sin x) + \cos^2 x$.Substitute $\cos x = a$ and $\sin y = b$:$E = (a \cos y + b \sin x)^2 + 2ab(b \cos x - a \sin y) + a^2$ is not the path. Let's use $\alpha - \beta = (	heta - \beta) - (	heta - \alpha) = y - x$.$E = \cos^2(y - x) + 2 \cos x \sin y \sin(y - x) + \cos^2 x$.$E = \cos^2(y - x) + 2 \cos x \sin y (\sin y \cos x - \cos y \sin x) + \cos^2 x$.$E = \cos^2(y - x) + 2 \cos^2 x \sin^2 y - 2 \cos x \sin y \cos y \sin x + \cos^2 x$.Using $\cos(y-x) = \cos y \cos x + \sin y \sin x$, the expression simplifies to $a^2 + b^2$.

If $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$, then the value of $\cos^2(\alpha - \beta) + 2ab \sin(\alpha - \beta) + \cos^2(\theta - \alpha)$ is...

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