Question

The number of pairs $(x,y)$ satisfying the equations $sinx+siny=sin(x+y)and|x|+|y|=1$ is

• A. $2$
• B. $4$
• C. $6$
• D. infinite number of pairs

Answer 1

The correct option is C $6$

$sinx+siny=sin(x+y)$
We can write
$2sin\frac{1}{2}(x+y)cos\frac{1}{2}(x-y)=2sin\frac{1}{2}(x+y)cos\frac{1}{2}(x+y)$
or $2sin\frac{1}{2}(x+y)\left\{\begin{array}{c}cos\frac{1}{2}(x-y)-cos\frac{1}{2}(x+y)\end{array}\right\}=0$
or $2sin\frac{1}{2}(x+y).2sin\frac{1}{2}xsin\frac{1}{2}y=0$
Either $2sin\frac{x+y}{2}=0orsin\frac{x}{2}=0orsin\frac{y}{2}=0$
$\Rightarrow x+y=0,x=0,y=0and|x|+|y|=1$
$\Rightarrow x+y=1,x-y=1$
$x+y=-1,x-y=-1$
When $x+y=0$, we have to reject $x+y=1or-1$ and solve it with $x-y=1orx-y=-1$ which gives $\left(\begin{array}{c}\frac{1}{2},-\frac{1}{2}\end{array}\right)or\left(\begin{array}{c}-\frac{1}{2},\frac{1}{2}\end{array}\right)$ as the possible solution.
Again solving with $x=0$, we get $(0,\pm 1)$ and by solving with $y=0$, we get $(\pm 1,0)$ as the other solution. Thus we have $6$ pairs of solutions for $xandy$