# Sum of Sinusoids with Same Frequency = Sinusoid (proof)

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.

• I would try and write sines and cosines as complex exponential functions and then see what happens. – busman Oct 22 '13 at 13:07
• @busman. I know you can write $e^{i\omega t+\phi}$ as $cos(\omega t+\phi) + isin(i\omega t+\phi)$ but where do I go from there. It would be great if you could point me to the right direction. I am happy to write the eq. down but I don't know where to start. thank you. – Marc Ourens Oct 22 '13 at 13:13
• I guess busman means something like this: Express $\sin$ and $\cos\;$ with complex $\exp$ via $$\cos(\omega t) = \frac{e^{i\omega t} + e^{-i\omega t}}{2}, \quad \sin(\omega t) = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}.$$ Then you can sort out sums like $a e^{i\omega t} + b e^{i\omega t}$ with more ease, and finally convert back to $\cos, \sin.$ – gammatester Oct 22 '13 at 13:35
• @Ganmaster. I am cool about that as well but how do I get to prove that the sum of these functions is another sinusoid? Shall I add these 2 identities together by adding a phase to the cosine and a phase to the sin, develop, regroup, etc? Is that the method? – Marc Ourens Oct 22 '13 at 13:47
• Okay so I think I see what you mean you start from the identities using the exponential form and then you end up with $cos(\omega t + \alpha) + i sin(\omega t + \beta)$ but that good enough as poof? but that doesn't tell me that $Acos(\omega t + \alpha) + Bi sin(\omega t + \beta) = C cos(\omega t + \gamma)$ for instance!!!? I don't understand. – Marc Ourens Oct 22 '13 at 14:09

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

1. a sinusoid can be expressed as a linear combination of a sine and a cosine,
2. conversely, a linear combination of sine and cosine can be represented as single sinusoid$^*$,
3. 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:

$$Z_0e^{i\omega t}+Z_1e^{i\omega t}+Z_2e^{i\omega t}=(Z_0+Z_1+Z_2)e^{i\omega t}.$$

(I will use the notation $\DeclareMathOperator{polar}{~\angle~}M \polar \theta$ to represent a vector in polar form, with $M$ the magnitude and $\theta$ the angle)

The general result (that I believe to be very useful) is:

If $$f(x) = A_1~\cos(\omega~x + \phi_1) + A_2~\cos(\omega~x + \phi_2)$$ then $$f(x) = A_3~\cos(\omega~x + \phi_3)$$ and $$A_3\polar \phi_3 = A_2\polar \phi_2 + A_1\polar \phi_1$$

This also holds if $\cos$ is replaced by $\sin$.

A proof of this can be done using Euler's representation of sinusoids. Given:

$$A_3 \cos(\phi_3) = A_2 \cos(\phi_2) + A_1\cos(\phi_1)$$

$$f(x) = A_1\frac {1}2 \left(e^{i(\omega x + \phi_1)} + e^{-i(\omega x + \phi_1)}\right) + A_2\frac {1}2 \left(e^{i(\omega x + \phi_2)} + e^{-i(\omega x + \phi_2)}\right)$$

Rearranging terms:

$$f(x) = \frac {1}2\left(A_1e^{ i\phi_1} + A_2e^{ i\phi_2}\right) e^{ i\omega x} + \frac {1}2\left(A_1e^{-i\phi_1} + A_2e^{-i\phi_2}\right) e^{-i\omega x}$$

Here we use the fact that since $A_3\polar \phi_3 = A_2\polar \phi_2 + A_1\polar \phi_1$, then $A_3e^{i\phi_3} = A_2e^{i\phi_2} + A_1e^{i\phi_1}$ and $A_3e^{-i\phi_3} = A_2e^{-i\phi_2} + A_1e^{-i\phi_1}$:

$$f(x) = \frac {1}2\left(A_3e^{i\phi_3}\right) e^{i\omega x} + \frac {1}2\left(A_3e^{-i\phi_3}\right)e^{-i\omega x }$$

$$f(x) = A_3\frac {1}2 \left(e^{i(\omega x + \phi_3)} + e^{-i(\omega x + \phi_3)}\right)$$

and back from Euler's representation:

$$f(x) = A_3~\cos(\omega~x + \phi_3)$$

Here is a less rigorous but (I consider) more intuitive proof:

Consider

• Point $Y$ is rotating around point $X$ at a frequency of $\frac {\omega}{2\pi}$
• Point $Z$ is rotating around point $Y$ at a frequency of $\frac {\omega}{2\pi}$
• The initial angle and magnitude of $Y$ relative to $X$ is $\varphi_1$ and $A_1$
• The initial angle and magnitude of $Z$ relative to $Y$ is $\varphi_2$ and $A_2$

As you can see, the inital vector $Z$ relative to the origin is $A_1 \polar \varphi_1 + A_2 \polar \varphi_2$. So if you can intuitively visualize that the motion $Z$ makes is a circle around $X$, then you can see that the sum of the sinusoids (either horizontal or vertical value) is itself a sinusoid.