Question

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

Answer 1

Given: $sinx+siny=sin(x+y)$

$\Rightarrow 2sin\frac{x+y}{2}(cos\frac{x-y}{2}-cos\frac{x+y}{2})=0\Rightarrow 4sin\frac{x+y}{2}sin\frac{x}{2}sin\frac{y}{2}=0$

$\Rightarrow sin\frac{x+y}{2}=0$ or $sin\frac{x}{2}=0$ or $sin\frac{y}{2}=0$ ... (i)

Given $\left|x\right|+\left|y\right|=1$ .... (ii) $\left|x\right|\le 1$ and $\left|y\right|\le 1$

From (i) : $x+y=0,x=0,y=0$

Putting $x=0$ in (ii) : $y=\pm 1$ and also putting $y=0$ in (ii) : $x=\pm 1$

Again, putting $x+y=0\Rightarrow y=-x$ in (ii) : $2\left|x\right|=1\Rightarrow x=\pm \frac{1}{2}$

We get 6 pairs of $(x,y)$ such as $(0,\pm 1),(\pm 1,0),(\frac{1}{2},-\frac{1}{2}),(-\frac{1}{2},\frac{1}{2})$