How can I find $\sum_{cyc}\sin x\sin y$ $x$, $y$ & $z$ are real number such that
$$\frac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=\frac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=2$$
find the value 
$$\sum_{cyc}\sin x\sin y$$
All help would be appreciated.

I used the Blue's beautiful idea.
Let $e^{ix}=a$, $e^{iy}=b$ and $e^{iz}=c$.
Hence, $\sin{x}=\frac{a-\frac{1}{a}}{2i}=\frac{a^2-1}{2ai}$, $\cos{x}=\frac{a^2+1}{2a}$, $\sin{y}=\frac{b^2-1}{2bi}$, $\sin{x}=\frac{c^2-1}{2ci}$ and $\cos{x}=\frac{c^2+1}{2c}$.
Thus, $\sum\limits_{cyc}\sin{x}=2\sin(x+y+z)$ gives $\sum\limits_{cyc}(a^2bc-ab)=2(a^2b^2c^2-1)$ and
$\sum\limits_{cyc}\cos{x}=2\cos(x+y+z)$ gives $\sum\limits_{cyc}(a^2bc+ab)=2(a^2b^2c^2+1)$ or
$ab+ac+bc=2$ and $a+b+c=2abc$.
Thus, $$-\sum\limits_{cyc}\sin{x}\sin{y}=\sum_{cyc}\frac{(a^2-1)(b^2-1)}{4ab}=\frac{\sum\limits_{cyc}c(a^2-1)(b^2-1)}{4abc}=$$
$$=\frac{abc(ab+ac+bc)-\sum\limits_{cyc}(a^2b+a^2c)+a+b+c}{4abc}=\frac{abc(ab+ac+bc)-(a+b+c)(ab+ac+bc)+3abc+a+b+c}{4abc}=\frac{1}{2}-1+\frac{3}{4}+\frac{1}{2}=\frac{3}{4}$$
