1

I'd like to know if someone could derive the below wolfram alpha output for me.

I was interested in the following system of equations:

\begin{align} &\cos(\omega_1t)-\cos(\omega_2t)=0\\ &\sin(\omega_1t)-\sin(\omega_2t)=0 \end{align}

, where $\omega_1$ and $\omega_2$ are constants. The physical interpretation of this is two vectors in $\mathbb R^2$ that, at time $t=0$, are pointing to the location $(1,0)$ and rotating about the origin. I would like to know, given the different frequencies, at what times $t$ do the vectors lay directly on top of one another.

The output generated by wolfram alpha is:

$$t=\frac{2\pi n}{\omega_1-\omega_2} \text{ and $\omega_1 \neq \omega_2$ and $n \in \mathbb Z$}$$

I was able to derive this formula myself, but the method I used was not really computational, per say.

I reasoned that $t\omega_1-t\omega_2$ effectively tracks the difference in radians (starting at a phase of $0$) about the unit circle...and any time this difference equals an integer multiple of $2\pi$, we necessarily have that the vectors are laying on top of one another. Obviously, this is equivalent to wolfram alpha's output. However, my rationale for this solution is more of a visualization than a formal derivation. Could someone offer an explicit derivation?

Frightened of Sinusoids
  • 133
  • A lot depends on where you want to start. For instance, what is $\pi$? Most elementary analysis courses I have taught have more or less defined $2\pi$ as the least positive number $p$ such that $\cos (x+p)=\cos x$ for all $x$. So I find your explanation perfectly convincing. – ancient mathematician Nov 11 '22 at 07:51
  • @ancientmathematician the reason I do not like my explanation is because it is lacking symbols. In the absence of the symbols, it is difficult for me to see whether or not my solution is exhaustive. i.e. How do I know that I have covered all possible solutions? – Frightened of Sinusoids Nov 11 '22 at 07:54
  • OK: but I do not see how you can avoid using the Theorem: $\exp (2\pi \alpha i)=1$ if and only if $\alpha\in\mathbb{Z}$. And given that theorem the result is immediate using simple algebra. – ancient mathematician Nov 11 '22 at 08:05
  • @ancientmathematician I have not yet learned about complex exponentiation (so I am not sure I can fill in the implicit steps you reference). Would you be willing to demonstrate this derivation as a submitted answer? – Frightened of Sinusoids Nov 11 '22 at 08:08
  • Sorry, I can't write a sensible answer because I don't know where you are starting from. What properties of $\cos$ and $\sin$ and $2\pi$ are you allowed to use in the proof? [I still think that your proof is entirely OK. If you want to dig deeper you've got to face up to proper definitions of $\cos,\sin,\pi$.] – ancient mathematician Nov 11 '22 at 11:56
  • @ancientmathematician the definitions I am acquainted with formalize $\pi$ as twice the integral of the unit semi-circle. $\cos$ is defined as a the horizontal coordinate corresponding to a radial 'sector' of a unit circle. $\sin$ is defined similarly (where sector is a type of area)...the explicit definitions can be found here: https://math.stackexchange.com/questions/760931/different-ways-to-formally-define-trigonometric-functions – Frightened of Sinusoids Nov 12 '22 at 07:08
  • I don't find these convincing as definitions: they depend on pictures. In which case your pictorial argument is fine, except it depends on knowing that $\pi$ is not just the area of the until disk, it's also the length of the unit semi-circle. – ancient mathematician Nov 12 '22 at 07:44
  • @ancientmathematician if you're willing, I am okay using whatever definitions you want. I am just interested in seeing an analytical solution, as I do not consider mine analytical. – Frightened of Sinusoids Nov 12 '22 at 07:51

0 Answers0