Sum of two trig function's identity

Question

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?

Answer 1

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

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

Answer 2

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.