Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
 A: I prefer @math110's solution, but here's a brute force method using complex exponentials, 
with 
$$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \qquad \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$
We define
$$a := e^{ix} \qquad b := e^{iy} \qquad c := e^{iz}$$
so that
$$p = \frac{\cos x + \cos y + \cos z}{\cos(x+y+z)} \implies p(a^2b^2c^2 +1) = abc (a+b+c) + bc + ca + ab \qquad (1)$$
$$p = \frac{\sin x + \sin y + \sin z}{\sin(x+y+z)} \implies p(a^2b^2c^2 - 1 ) = abc (a+b+c) - bc - ca - ab \qquad (2)$$
Thus, from $(1)-(2)$ and $(1)+(2)$, we have
$$p = bc + ca + ab \qquad\qquad p = \frac{a+b+c}{abc} = \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$$
whereupon
$$2p = bc+\frac{1}{bc}\;+\;ca+\frac{1}{ca}\;+\;ab+\frac{1}{ab} = 2\left( \cos(y+z)+\cos(z+x)+\cos(x+y)\right)$$
A: Since
\begin{align*}
p \cos(x+y+z) &= \cos x + \cos y + \cos z \\
p \sin(x+y+z) &= \sin x + \sin y +\sin z
\end{align*}
\begin{align*}
p e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz} 
\end{align*}
Multiplying throughout by $e^{-i(x+y+z)}$, we get
\begin{align*}
p = e^{-i(y+z)} + e^{-i(z+x)} + e^{-i(x+y)}
\end{align*}
Since $p$ is real, equating the real parts, we get
$$\cos(y+z) + \cos(z+x) + \cos(x+y) = p $$
A: note 
$$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$
and 
$$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$
$$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$
add this three
\begin{align*}
&\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\
&=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\
&=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\
&=p
\end{align*}
