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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 $\sin (x+y)$ equals to
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Verified Answer
The correct answer is:
$\frac{4}{5}$
Given, $\cos x+\cos y=\frac{3}{2}$
$$
\Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{2}
$$
and
$$
\begin{aligned}
& \sin x+\sin y=\frac{3}{4} \\
& \Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4} \\
& \therefore \quad \frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \\
& \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
& \therefore \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\
& =\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}=\frac{4}{4+1}=\frac{4}{5} \\
&
\end{aligned}
$$
$$
\Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{2}
$$
and
$$
\begin{aligned}
& \sin x+\sin y=\frac{3}{4} \\
& \Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4} \\
& \therefore \quad \frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \\
& \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
& \therefore \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\
& =\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}=\frac{4}{4+1}=\frac{4}{5} \\
&
\end{aligned}
$$
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