<s>[INST] How to get the $\phi$ from $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$? $a\sin(x)+b\sin(x+\theta)=c\sin(x+\phi)$,
where $c=\sqrt{a^2+b^2+2ab\cos(\theta)}$, and $\displaystyle\tan(\phi)=\frac{b\sin(\theta)}{a+b\cos(\theta)}$.
I want to know how to get to this result.
I'm able to derive $c$ by taking the derivative of the equation, then squaring both and adding them together, and back-substituting the cosine of a double angle.
But how does one get to the expression for $\tan(\phi)$?
 [/INST] For fun, here's a trigonograph, which leads to a counterpart cosine identity:

$$\begin{align}
a \sin x + b \sin(\theta+x) &= c\sin(\phi+x) \\
a \cos x + b \cos(\theta+x) &= c\cos(\phi+x)
\end{align}$$
where
$$c^2 = a^2 + b^2 - 2 a b \cos(180^\circ-\theta) = a^2 + b^2 + 2 a b \cos \theta \qquad\text{and}\qquad \tan\phi = \frac{b\sin\theta}{a+b\cos\theta}$$
</s>