Question

The number of pairs $(x,y)$ satisfying $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is

• A. $2$
• B. $6$
• C. $4$
• D. None of these

Answer 1

The correct option is B $6$

$\sin x(1-\cos y)+siny(1-\cos x)=0$
$4\sin \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)\{\sin \frac{y}{2}\cos \frac{x}{2}+\cos \frac{y}{2}\sin \frac{x}{2}\}=0$
$\Rightarrow \sin \left(\frac{x}{2}\right)=0$ or $\sin \left(\frac{y}{2}\right)=0$ or $\sin \left(\frac{x+y}{2}\right)=0$
Case 1 :
$\sin \left(\frac{x}{2}\right)=0$
Hence, $x=n\pi$
$|x|\le 1$
Hence, $x=0$
Hence, $y=\pm 1$
Case 2 ,
$\sin \left(\frac{y}{2}\right)=0$
Similar to the previous case, we will get $y=0$ and $x=\pm 1$
Case 3,
$\sin \left(\frac{x+y}{2}\right)=0$
$\frac{x+y}{2}=n\pi$
$x+y=2n\pi$
However, $|x|,|y|\le 1$
Hence,
$x+y=0$
Hence,
$|x|=|y|=\frac{1}{2}$
Hence, the possible solutions are $(\frac{1}{2},-\frac{1}{2})$ and $(\frac{-1}{2},\frac{1}{2})$
Hence, there are $6$ solutions.