Q99 / 160 · TS EAMCET 2014 · Mathematics
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
- A$\frac{2}{5}$
- B$\frac{3}{4}$
- C$\frac{3}{5}$
- Correct answerD$\frac{4}{5}$
Solution
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
$$
\begin{aligned}
& \sin x+\sin y=\frac{3}{4} \\
& \Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4} \\
& \therefore \quad \frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \\
& \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
& \therefore \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\
& =\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}=\frac{4}{4+1}=\frac{4}{5} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'D'.