Question

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

• A. $2$
• B. $4$
• C. $6$
• D. infinite

Answer 1

The correct option is D infinite

$\sin x+\sin y=\sin x\cos y+\cos x\sin y$
$\Rightarrow \sin y(1-\cos x)=\sin x(\cos y-1)$
$\Rightarrow \tan \frac{x}{2}=-\tan \frac{y}{2}$
$\Rightarrow x=-y$ or $x+2\pi =y$
So, $2|x|=1$ or $x=\pm \frac{1}{2}$ which gives the pairs $(\frac{1}{2},-\frac{1}{2}),(-\frac{1}{2},\frac{1}{2}).$
Also, the second condition gives $x=\frac{1-2\pi }{2}$
Similarly there will be infinite possibilities.