<s>[INST] Trigonometric elimination between two variables Eliminate $\theta$ and $\phi$ between the following equations: $$\begin{cases}\sin \theta + \sin \phi = x \\ \cos \theta + \cos \phi = y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} = z\end{cases}$$
What I've done so far
I've established that $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\sin \theta + \sin \phi}{\cos \theta + \cos \phi}$$ so that $$\tan \frac {\theta+\phi}{2} = \frac {x}{y}.$$
I then used the  trigonometric identity $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\tan \frac {\theta}{2} + \tan \frac {\phi} {2}}{1-\tan \frac {\theta}{2}\tan \frac {\phi} {2}}$$ and with a little manipulation got to $$\tan \frac {\theta}{2} + \tan \frac {\phi} {2} = \frac {x(1-z)}{y}$$
I'm stumped on the next steps...would I have to find the difference of the roots also (i.e. $\tan \frac {\theta}{2} - \tan \frac {\phi} {2}$)?  Or is there a simpler way?  (Side note: I also tried $\tan \frac {\theta}{2}$ substitution but that went nowhere.)
 [/INST] Write $u = \tan \theta/2$, $v = \tan \phi/2$. Then we get
$$
\begin{cases}
2(u+v)(1 + uv) = x(1+u^2)(1+v^2) \\
2 -2u^2v^2 = y(1+u^2)(1+v^2) \\
uv = z.
\end{cases}
$$
Now setting $s = u + v$ and using the third relation, we obtain
$$
\begin{cases}
2s(1+z) = x(1 - 2z + z^2 + s^2) \\
2(1-z^2) = y(1 - 2z + z^2 + s^2).
\end{cases}
$$
Dividing, we obtain the relation $s = x(1-z)/y$, which you've already mentioned. Substituting this relation into either of the equations, we find
$$2y(1 + z) = (x^2 + y^2)(1-z).$$
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