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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
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Verified Answer
The correct answer is:
$\frac{4}{5}$
Given, $\cos x+\cos y=\frac{3}{2}$
$$
\begin{aligned}
&\Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{2} . \\
&\text { and } \\
&\Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4}
\end{aligned}
$$
On dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
\therefore \quad \sin (x+y) &=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^{2}\left(\frac{x+y}{2}\right)}=\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^{2}} \\
&=\frac{4}{4+1}=\frac{4}{5}
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{2} . \\
&\text { and } \\
&\Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4}
\end{aligned}
$$
On dividing Eq. (i) by Eq. (ii), we get
$$
\begin{aligned}
\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
\therefore \quad \sin (x+y) &=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^{2}\left(\frac{x+y}{2}\right)}=\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^{2}} \\
&=\frac{4}{4+1}=\frac{4}{5}
\end{aligned}
$$
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