Question

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

• A. $2$
• B. $4$
• C. $6$
• D. $\infty$

Answer 1

The correct option is C $6$

$\sin x+\sin y=\sin (x+y)$
$\Rightarrow 2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right)$
$\Rightarrow \sin \left(\frac{x+y}{2}\right)=0$ and $\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)=0$
$\Rightarrow x+y=2n\pi$ and $\sin \frac{x}{2}\cos \frac{y}{2}=0$
$\Rightarrow x+y=2n\pi$ and $x=2k\pi$ or $y=2m\pi$
$\therefore x+y=0$ and $x=0,y=0$
$\therefore (0,1),(1,0),(0,-1),(-1,0)(-\frac{1}{2},\frac{1}{2}),(\frac{1}{2},-\frac{1}{2})$
Hence 6 solutions.