If cos x + cos y = frac{2}{3} and sin x - sin | vedclass

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(B) Given: $\cos x + \cos y = \frac{2}{3}$  $(i)$  $\sin x - \sin y = \frac{3}{4}$  $(ii)$  Using sum-to-product formulas:  $2 \cos \left(\frac{x+y}{2

Vedclass · Accepted Answer

$\frac{127}{145}$

Vedclass · Answer

(B) Given: $\cos x + \cos y = \frac{2}{3}$  $(i)$  $\sin x - \sin y = \frac{3}{4}$  $(ii)$  Using sum-to-product formulas:  $2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{2}{3}$  $(iii)$  $2 \cos \left(\frac{x+y}{2}ight) \sin \left(\frac{x-y}{2}ight) = \frac{3}{4}$  $(iv)$  Dividing $(iv)$ by $(iii)$:  $	an \left(\frac{x-y}{2}ight) = \frac{3/4}{2/3} = \frac{9}{8}$  Let $	heta = \frac{x-y}{2}$,so $	an 	heta = \frac{9}{8}$.  Then $\sin(x-y) = \sin(2	heta) = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \frac{2(9/8)}{1 + (81/64)} = \frac{9/4}{145/64} = \frac{9}{4} 	imes \frac{64}{145} = \frac{144}{145}$.  $\cos(x-y) = \cos(2	heta) = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \frac{1 - (81/64)}{1 + (81/64)} = \frac{-17/64}{145/64} = -\frac{17}{145}$.  Therefore,$\sin(x-y) + \cos(x-y) = \frac{144}{145} - \frac{17}{145} = \frac{127}{145}$.

If $\cos x + \cos y = \frac{2}{3}$ and $\sin x - \sin y = \frac{3}{4}$,then find the value of $\sin(x - y) + \cos(x - y)$.

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