Question

The number of ordered pairs $(x,y)$ satisfying the system of equations given by $\sin x+\sin y=\sin (x+y)$ and $|x|+|y|=1$ is

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

Answer 1

The correct option is B $4$

$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$

and $\sin (x+y)=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x+y}{2}\right)$

Given, $\sin (x+y)=\sin x+$ siny

$\Rightarrow \cos \left(\frac{x+y}{2}\right)=\cos \left(\frac{x-y}{2}\right)$

$\Rightarrow \cos \left(\frac{x+y}{2}\right)-\cos (x-y)=0$

$\Rightarrow 2\sin \left(\frac{x}{2}\right)\sin \left(\frac{y}{2}\right)=0$

$\Rightarrow x=0$ or $y=0.$

$|x|+|y|=1$

$\to$ when $x=0$,

$\begin{array}{c}|y|=1\Rightarrow y=1,-1\\ (0,1),(0,-1)\end{array}$

$\Rightarrow$ when $y=0,$

$|x|=1\Rightarrow x=1,-1$

(1,0),(-1,0)

4 pairs

option $B$ is correct.