If sin (theta + alpha ) = a and sin (theta + | vedclass

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(B) Given $\sin (	heta + \alpha ) = a$ and $\sin (	heta + \beta ) = b$.Let $x = 	heta + \alpha$ and $y = 	heta + \beta$. Then $\alpha - \beta = x

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$1 - 2a^2 - 2b^2$

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(B) Given $\sin (	heta + \alpha ) = a$ and $\sin (	heta + \beta ) = b$.Let $x = 	heta + \alpha$ and $y = 	heta + \beta$. Then $\alpha - \beta = x - y$.We have $\sin x = a$ and $\sin y = b$. Thus $\cos x = \pm \sqrt{1-a^2}$ and $\cos y = \pm \sqrt{1-b^2}$.$\cos (\alpha - \beta ) = \cos (x - y) = \cos x \cos y + \sin x \sin y = \pm \sqrt{1-a^2}\sqrt{1-b^2} + ab$.Let $C = \cos (\alpha - \beta )$. Then $C - ab = \pm \sqrt{1-a^2}\sqrt{1-b^2}$.Squaring both sides: $(C - ab)^2 = (1-a^2)(1-b^2) = 1 - a^2 - b^2 + a^2b^2$.$C^2 - 2abC + a^2b^2 = 1 - a^2 - b^2 + a^2b^2$.$C^2 - 2abC = 1 - a^2 - b^2$.We need to evaluate $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta ) = 2\cos^2 (\alpha - \beta ) - 1 - 4ab\cos (\alpha - \beta )$.$= 2(C^2 - 2abC) - 1$.Substituting $C^2 - 2abC = 1 - a^2 - b^2$:$= 2(1 - a^2 - b^2) - 1 = 2 - 2a^2 - 2b^2 - 1 = 1 - 2a^2 - 2b^2$.

If $\sin (\theta + \alpha ) = a$ and $\sin (\theta + \beta ) = b,$ then $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta )$ is equal to

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