Sum of two trig function's identity We all know that
$\sin(x) + \sin(y) = 2\sin((x+y)/2)\cos((x-y)/2)$
But is there an identity for
$\sin(x) + z\sin(y) = ?$
Or do I need to figure it out using Euler's formula
$\sin(x) = (e^{ix} - e^{-ix})/2$ and put it back into trigonometric form?
 A: According to 

Alan Jeffrey, Hui-Hui Dai: Handbook of Mathematical Formulas and
  Integrals. 4th Edition, Academic Press, 2008.
at page 131, 2.4.1.9 Sum of Multiples of $\sin x$ and $\cos x$

There is a formula for that


*

*$A \sin x + B \cos x = R \sin(x + \theta)$, where $R = (A^2+B^2)^{1/2}$ with $\theta = \arctan B/A$ when $A>0$ and $\theta = \pi + \arctan B/A$ when $A<0$.

*$A \cos x + B \sin x = R \cos(x - \theta)$, where $R = (A^2+B^2)^{1/2}$ with $\theta = \arctan B/A$ when $A>0$ and $\theta = \pi + \arctan B/A$ when $A<0$.


It is not exactly what you are looking for, but maybe it could help.
A: As far as I know, there is no simple relation turning a sum in a product, except in the case $|z|=1$.
To convince yourself, consider the zero set of the LHS expression:
$$\sin(x)+\sin(y)=0,$$
giving the solutions
$$\sin(y)=-\sin(x),$$
$$y=-x+2k\pi\lor y=x+\pi+2k\pi,$$
$$y+x=2k\pi\lor y-x=(2k+1)\pi.$$
This zero set is formed of two pencils of parallel straight lines, which correspond to the zero sets of two independent sine waves, having for arguments a linear combination of $x$ and $y$.
With $|z|\ne1$, the relation is more complicated:
$$\sin(x)+k\sin(y)=0$$
does not define straight lines but transcendent curves and no linear combination is possible.
