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 $\tan(\phi)$?
 A: Use the sin addition formula $\sin(\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta$
\begin{eqnarray*}
a \sin x + \underbrace{b \sin(x+\theta)}_{ b\sin x \cos \theta+b \cos x \sin \theta}= \underbrace{c \sin(x+ \phi)}_{b\sin x \cos \phi+b \cos x \sin \phi} \\
(a +  b \cos \theta) \color{red}{\sin x} + b \sin \theta \color{blue}{\cos x} = c \cos \phi \color{red}{\sin x}+c  \sin \phi \color{blue}{\cos x}
\end{eqnarray*}
Equate the coefficients of $ \sin x $ and $ \cos x $
\begin{eqnarray*}
a +  b \cos \theta = c \cos \phi \\
b \sin \theta  =c  \sin \phi
\end{eqnarray*}
Now square these equations and add to get the first equation you want. & take the ratio of these equations to get the second equation.
A: 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
$$c^2 = a^2 + b^2 - 2 a b \cos(180^\circ-\theta) = a^2 + b^2 + 2 a b \cos \theta \qquad\text{and}\qquad \tan\phi = \frac{b\sin\theta}{a+b\cos\theta}$$
A: We use $a \cdot b=|a||b|\cos(\text{angle between vectors})$.
Note,
$$\sin (x+\theta)=\sin x \cos \theta+\sin \theta \cos x$$
Hence the left hand side of your equation is,
$$a\sin x+b\sin x \cos \theta+b \cos x \sin \theta$$
$$=(a+b\cos \theta)\sin x+(b\sin \theta)\cos x$$
$$=\langle b\sin \theta,a+b\cos \theta \rangle \cdot \langle \cos x,\sin x \rangle$$
The angle between these two vectors is $\alpha-x$ or $x-\alpha$ where $\cot \alpha=\frac{b \sin \theta}{a+b\cos \theta}$, depending on the relative positioning of these vectors. In any case the cosine of the angle between these two vectors is $\cos(\alpha-x)=\cos(x-\alpha)$ do to the fact cosine is even.
Next note that $\sin (x+\frac{\pi}{2})=\cos(x)$ so that the cosines between the two angles is really:
$$\sin (x-\alpha+\frac{\pi}{2})$$
But:
$$\tan (\frac{\pi}{2}-\alpha)=\cot \alpha$$
So,
$$\tan (\phi)=\cot \alpha=\frac{b \sin \theta}{a+b\cos \theta}$$.
A: you will have $$\frac{a\sin(x)+b\sin(x+\theta)}{c}=\sin(x+\phi)$$
$$\arcsin\left(\frac{a\sin(x)+b\sin(x+\theta)}{c}\right)-x=\phi$$
