If x and y are acute angles such that cos x + | vedclass

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(D) Given,$\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$.Using sum-to-product formulas:$2 \cos \left(\frac{x+y}{2}\right) \cos \l

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$\frac{4}{5}$

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(D) Given,$\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$.Using sum-to-product formulas:$2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{3}{2}$  $(1)$$2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{3}{4}$  $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)}{2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight)} = \frac{3/4}{3/2}$$	an \left(\frac{x+y}{2}ight) = \frac{1}{2}$Using the identity $\sin(x+y) = \frac{2 	an \left(\frac{x+y}{2}ight)}{1 + 	an^2 \left(\frac{x+y}{2}ight)}$:$\sin(x+y) = \frac{2 	imes \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$.

If $x$ and $y$ are acute angles such that $\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$,then $\sin(x+y)$ is equal to

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