If cos x+cos y=p and sin x+sin y=q,then cos l | vedclass

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Question

(A) Given,$\cos x + \cos y = p$ and $\sin x + \sin y = q$. Squaring and adding both equations,we get: $(\cos x + \cos y)^2 + (\sin x + \sin y)^2 = p^2

Vedclass · Accepted Answer

$\pm \frac{\sqrt{p^2+q^2}}{2}$

Vedclass · Answer

(A) Given,$\cos x + \cos y = p$ and $\sin x + \sin y = q$. Squaring and adding both equations,we get: $(\cos x + \cos y)^2 + (\sin x + \sin y)^2 = p^2 + q^2$ $(\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y + \sin x \sin y) = p^2 + q^2$ $1 + 1 + 2\cos(x - y) = p^2 + q^2$ $2 + 2\cos(x - y) = p^2 + q^2$ $2(1 + \cos(x - y)) = p^2 + q^2$ Using the identity $1 + \cos 	heta = 2\cos^2(	heta/2)$,we have: $2(2\cos^2(\frac{x-y}{2})) = p^2 + q^2$ $4\cos^2(\frac{x-y}{2}) = p^2 + q^2$ $\cos^2(\frac{x-y}{2}) = \frac{p^2 + q^2}{4}$ Taking the square root on both sides: $\cos(\frac{x-y}{2}) = \pm \frac{\sqrt{p^2 + q^2}}{2}$

If $\cos x+\cos y=p$ and $\sin x+\sin y=q$,then $\cos \left(\frac{x-y}{2}\right) = $

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