Question

If $\sin x+\sin y=\frac{7}{5}$ and $\cos x+\cos y=\frac{1}{5}$, then $\sin (x+y)$ equals

• A. $\frac{7}{25}$
• B. $\frac{24}{25}$
• C. $\frac{-7}{25}$
• D. $\frac{-24}{25}$

Answer 1

Answer: (A)

We have,
$\sin x+\sin y=\frac{7}{5}$
and $\cos x+\cos y=\frac{1}{5}$
$\Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{7}{5}\dots \left(i\right)$
and $2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{1}{5}\dots \left(ii\right)$
On dividing Eq. $\left(ii\right)$ by Eq. $\left(i\right)$, we get
$\tan \left(\frac{x+y}{2}\right)=7$
$\Rightarrow \sin \left(x+y\right)=\frac{2\tan \left(\frac{x+y}{2}\right)}{1+{\tan }^2\left(\frac{x+y}{2}\right)}$
$\Rightarrow \sin \left(x+y\right)=\frac{14}{1+49}=\frac{14}{50}$
$\Rightarrow \sin \left(x+y\right)=\frac{7}{25}$