Question

The number of pairs ordered (x, $y)$ satisfying the equation $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is equal to

Answer 1

Answer: 6.00

From the given equation we have,
$2\sin \frac{x+y}{2}\cos \frac{x-y}{2}=2\sin \frac{x+y}{2}\cos \frac{x+y}{2}$
$\Rightarrow \sin \frac{x+y}{2}\left[\cos \frac{x-y}{2}-\cos \frac{x+y}{2}\right]=0$
$\Rightarrow \sin \frac{x+y}{2}\times 2\sin \frac{x}{2}\sin \frac{y}{2}=0$
Which holds if either $\sin \frac{x+y}{2}=0$ or $\sin \frac{x}{2}=0$
Also since $|x|+|y|=1\Rightarrow |x|\le 1,|y|\le 1$. So, the required solution is either $x+y=0$ or $x=0$ or $y=0$
If $x=0,y=\pm 1$, and if $y=0,x=\pm 1$ and if $x+y=0$
$\Rightarrow \mathrm{y}=-\mathrm{x}\Rightarrow |\mathrm{y}|=|-\mathrm{x}|=|\mathrm{x}|=\frac{1}{2}$
So that the required pairs $(\mathrm{x},\mathrm{y})$ are $(0,\pm 1),(\pm 1,0),\left(\frac{1}{2},-\frac{1}{2}\right),\left(-\frac{1}{2},\frac{1}{2}\right)$ which are 6 in number.