Question

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 $\sin (x+y)=$

• A. $\frac{2}{5}$
• B. $\frac{3}{4}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is D $\frac{4}{5}$

$\cos x+\cos y=\frac{3}{2}$
$2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{2}$
$\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{4}$ ...(i)
$\sin x+\sin y=\frac{3}{4}$
$2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{4}$
$\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{3}{8}$...(ii)
Equation $\frac{\left(ii\right)}{\left(i\right)}$ implies
$\frac{\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}=\frac{\frac{3}{8}}{\frac{3}{4}}$
$\tan \left(\frac{x+y}{2}\right)=\frac{1}{2}$
Now $\sin (x+y)=\frac{2\tan \left(\frac{x+y}{2}\right)}{1+{\tan }^2\left(\frac{x+y}{2}\right)}$
$=\frac{2\left(\frac{1}{2}\right)}{1+\frac{1}{4}}$ $=\frac{1}{\frac{5}{4}}$ $=\frac{4}{5}$