TITLE: tricky system of trigonometric equations
QUESTION [10 upvotes]: I am not very fresh in math, but I need to solve this system:
\begin{gather}
A\sin(x-y)+B\sin(z-y)=C\\
A\cos(x-y)+B\cos(z-y)=D
\end{gather}
where $A,B,C,D$ and $x$ are given.
I tried to expand and combine the bracket terms and I suppose that there are some tricky substitutions to get it out, but I am lost!
Thank you all!

REPLY [5 votes]: If you're not fresh at math, complex formalism may not be of much help, but for posterity:
Using
$$
e^{i\theta} = \cos \theta + i\sin\theta,
$$
your system can be written
$$
A e^{i(x - y)} + Be^{i(z - y)} = D + iC.
$$
Multiplying through by $e^{iy}$ and dividing by $D + iC$ gives
$$
\frac{(A e^{ix} + Be^{iz})(D - iC)}{D^{2} + C^{2}} = e^{iy}.
\tag{1}
$$
Conjugating,
$$
\frac{(A e^{-ix} + Be^{-iz})(D + iC)}{D^{2} + C^{2}} = e^{-iy}.
\tag{2}
$$
Multiplying (1) and (2) eliminates $y$:
\begin{align*}
1 = e^{iy}\, e^{-iy}
  &= \frac{(A e^{ix} + Be^{iz})(D - iC)}{D^{2} + C^{2}}\,
     \frac{(A e^{-ix} + Be^{-iz})(D + iC)}{D^{2} + C^{2}} \\
  &= \frac{(A e^{ix} + Be^{iz})(A e^{-ix} + Be^{-iz})}{D^{2} + C^{2}}.
\end{align*}
Expanding and rearranging,
$$
D^{2} + C^{2} = A^{2} + B^{2} + 2AB\cos(x - z),
$$
whereupon you can proceed as in John Hughes' answer.