# 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

## 3 Answers

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.

This is a small addition to the Answer 1.

The original question is about calculating A3 and ϕ3.

This can be done easily from the drawing in Answer 1. If Zx, Zy, Yx, and Yy are coordinates of points Z and Y than A3 and tan(ϕ3) can be easily expressed as a functions of Zx, Zy, Yx, and Yy. Than Zx, Zy, Yx, and Yy could be expressed as a functions of A1, ϕ1, A2, and ϕ2.

Another approach: To find A3 start from multiplying formulas for A3eiϕ3 and A3e−iϕ3 (see the Answer 1). (Euler's representation of expression [cos(ϕ1)cos(ϕ2)+sin(ϕ1)sin(ϕ2)] is required for conversions.) To find tan(ϕ3) start from dividing the same formulas.

• continuation of the last sentence:To find tan(ϕ3) start from dividing the same formulas by A3 than substitute eiϕ3 and e−iϕ3 in Euler's representation of tan(ϕ3). Following conversions are similar as used for finding A3 – Jan Bejnar Mar 15 '15 at 14:57