sin x+sin y=sin(x+y),quad |x|+|y|=1

Solve sin x+sin y=sin(x+y) and |x|+|y|=1

Consider the system in ordered pairs (x,y) of real numbers sin x+sin y=sin(x+y) , |x|+|y|=1 . The number of ordered pairs (x,y) satisfying the system is

Consider the system in ordered pairs (x,y) of real numbers sin x+sin y=sin(x+y) , |x|+|y|=1 . The number of ordered pairs (x,y) satisfying the system is

Prove that sin x* sin y*sin(x - y) + sin y *sin z*sin(y- z) + sin z *sin x sin(z - x) + sin(x - y) *sin(y - z)*sin(z -x) = 0 .

Prove that : sin x sin y sin (x - y) + sin y sin z sin (y- z)+ sin z sin x sin (z - x) + sin (x - y) sin (y-z)sin (z- x) = 0 .

(sin x + sin y) / (sin x-sin y) = tan ((x + y) / (2)) * cot ((xy) / (2))

x sin y + y sin x = 0

Prove that : (sin x + sin y)/(sin x - sin y) = tan ((x+y)/2).cot ((x-y)/2)

Prove that: (sin x+sin y)/(sin x-sin y)=tan((x+y)/(2))*cot((x-y)/(2))