Question

The number of pairs (x, y) satisfying the equations

$\sin x + \sin y = \sin ( x + y )$ and $| x | + | y | = 1$ is

• A. 2
• B. 4
• C. 6
• D. $\infty$

Answer 1

Answer: (C)

6

Answer 2

The first equation can be written as,

$2 \sin \frac { 1 } { 2 } ( x + y ) \cos \frac { 1 } { 2 } ( x - y ) = 2 \sin \frac { 1 } { 2 } ( x + y ) \cos \frac { 1 } { 2 } ( x + y )$

$\therefore$ Either $\sin \frac { 1 } { 2 } ( x + y ) = 0$ or $\sin \frac { 1 } { 2 } x = 0$ or $\sin \frac { 1 } { 2 } y = 0$

⇒. As |x| + |y| =1, therefore when $x + y = 0$we have to reject $x + y = 1$ or $x + y = - 1$ and solve it with $x - y = 1$or $x - y = - 1$ which gives $\left( \frac { 1 } { 2 } , \frac { - 1 } { 2 } \right)$ or

$\left( \frac { - 1 } { 2 } , \frac { 1 } { 2 } \right)$ as the possible solution. Again solving with we get $( 0 , \pm 1 )$ and solving with $y = 0$ we get $( \pm 1,0 )$ as the other solution. Thus we have six pairs of solution for x and y.