How can I find $\sum_{cyc}\sin x\sin y$

Question

$x$, $y$ & $z$ 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.

Answer 1

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, $$-\sum\limits_{cyc}\sin{x}\sin{y}=\sum_{cyc}\frac{(a^2-1)(b^2-1)}{4ab}=\frac{\sum\limits_{cyc}c(a^2-1)(b^2-1)}{4abc}=$$ $$=\frac{abc(ab+ac+bc)-\sum\limits_{cyc}(a^2b+a^2c)+a+b+c}{4abc}=\frac{abc(ab+ac+bc)-(a+b+c)(ab+ac+bc)+3abc+a+b+c}{4abc}=\frac{1}{2}-1+\frac{3}{4}+\frac{1}{2}=\frac{3}{4}$$

Answer 2

Since \begin{align*} 2 \cos(x+y+z) &= \cos x + \cos y + \cos z \\ 2 \sin(x+y+z) &= \sin x + \sin y +\sin z \end{align*} \begin{align*} 2 e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz} \end{align*} Multiplying throughout by $e^{-ix}$, we get \begin{align*} 2e^{i(y+z)} = 1 + e^{i(y-x)} + e^{i(z-x)} \end{align*} Equating the real parts, we get \begin{align*} 2\cos(y+z) &= 1 + \cos(y-x) +\cos(z-x)\\ 2(\cos y \cos z - \sin y \sin z) & = 1 + \cos x \cos y + \sin x \sin y + \cos z \cos x + \sin z \sin x \end{align*} Similarly, we get \begin{align*} 2(\cos x \cos z - \sin x \sin z) &= 1 + \cos x \cos y + \sin x \sin y + \cos z \cos y + \sin z \sin y\\ 2(\cos x \cos y - \sin x \sin y) &= 1 + \cos x \cos z + \sin x \sin z+ \cos z \cos y + \sin z \sin y\\ \end{align*} Adding, and canceling $2 \sum \cos x \cos y$ from the sides, we get $$\sum \sin x \sin y = -\frac{3}{4}$$