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 y

Vedclass · Accepted Answer

$a^2 + b^2$

Vedclass · Answer

(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 $x - y = (	heta - \alpha) - (	heta - \beta) = \beta - \alpha = -(\alpha - \beta)$.Thus,$\cos(\alpha - \beta) = \cos(x - y) = \cos x \cos y + \sin x \sin y = a \sqrt{1 - b^2} + b \sqrt{1 - a^2}$.And $\sin(\alpha - \beta) = -\sin(x - y) = -(\sin x \cos y - \cos x \sin y) = -(\sqrt{1 - a^2} \sqrt{1 - b^2} - ab) = ab - \sqrt{1 - a^2} \sqrt{1 - b^2}$.Substituting these into the expression $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta)$:$= (a \sqrt{1 - b^2} + b \sqrt{1 - a^2})^2 + 2ab(ab - \sqrt{1 - a^2} \sqrt{1 - b^2})$$= a^2(1 - b^2) + b^2(1 - a^2) + 2ab \sqrt{1 - a^2} \sqrt{1 - b^2} + 2a^2b^2 - 2ab \sqrt{1 - a^2} \sqrt{1 - b^2}$$= a^2 - a^2b^2 + b^2 - a^2b^2 + 2a^2b^2$$= a^2 + b^2$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

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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