TITLE: Sum of Sinusoids with Same Frequency = Sinusoid (proof)
QUESTION [12 upvotes]: I am studying Fourier analysis on my own, I realised that probably the first thing you want to proof in Fourier transform is that the sum of 2 sinuoids (namely a sine and cosine) with the same frequency gives another sinusoid. So I am trying to find a proof of this. In this document, I found this identity:
$$ A\cos(\omega t + \alpha) + B\sin(\omega t + \beta)
= \color{red}{\sqrt{(A\cos\alpha + \beta\sin\beta)^2 + (A \sin\alpha - B\cos\beta)^2}} \cdot \cos\left(\omega t + \color{green}{\arctan \frac{A\sin\alpha - B\cos\beta}{A\cos\alpha+B\sin\beta}}\right)
$$
EDIT: sorry I made mistake in equation.
Assuming I know how to go from the equation on the left to the equation on the right, would it be good enough as a proof since I can say that the terms that I highlighted with color are constants thus that the sum of the cosine and sine is equal to a constant multiplied by a cosine of the same frequency with some constant phase shift.
It would be great to have the confirmation from an expert.
Thank you.

REPLY [12 votes]: To avoid any confusion, let us state that 

a (pure) sine has the form $A\sin(\omega t)$,
a (pure) cosine has the form $A\cos(\omega t)$,
a sinusoid has an arbitrary phase and one of the equivalent forms $A\sin(\omega t+\phi)$ or $A\cos(\omega t+\psi)$ - where $\phi$ and $\psi$ differ by a quarter turn.

So the sine and cosine are special cases of the sinusoid.
By the well-known addition formula,
$$A\sin(\omega t+\phi)=A\sin(\omega t)\cos(\phi)+A\cos(\omega t)\sin(\phi)=A'\sin(\omega t)+A''\cos(\omega t).$$
This means that

a sinusoid can be expressed as a linear combination of a sine and a cosine,
conversely, a linear combination of sine and cosine can be represented as single sinusoid$^*$,
a linear combination of two or more sinusoids can be expressed as a linear combination of a sine and a cosine, hence can be expressed as a single sinusoid.

These properties no more hold if you mix sinusoids of different periods.

$^*$This is done by solving the system
$$A\cos(\phi)=A'\\A\sin(\phi)=A'',$$
i.e.
$$A=\sqrt{A'^2+A''^2}\\\tan(\phi)=\frac{A''}{A'}.$$

You will soon discover that complex numbers are intensively used in harmonic analysis, based on Euler's formula $e^{ix}=\cos(x)+i\sin(x)$.
Sinusoids can be represented as the imaginary part of $Ae^{i(\omega t+\phi)}=Ae^{i\phi}e^{i\omega t}=Ze^{i\omega t}$, where $Z$ is a complex number, that carries both the amplitude and the phase ($Z=A$ is real for a sine, $Z=iA$ is imaginary for a cosine).
Using this notation, adding sinusoids becomes a trivial matter:
$$Z_0e^{i\omega t}+Z_1e^{i\omega t}+Z_2e^{i\omega t}=(Z_0+Z_1+Z_2)e^{i\omega t}.$$