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I was trying to understand this answer about proving periodicity of $ \sin(x)$ and $\cos(x)$ from their series definition. I want to prove that $ \cos(x), \sin(x)$ have a smallest period, and all other periods are integer multiples of this smallest period, and this led me to this answer. However, I do not know anything about cyclic groups and I cannot understand the proof.

So, is there a simpler proof of

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous periodic function with two incommensurate periods $T_1$ and $T_2$; that is $\displaystyle \frac{T_1}{T_2}$ is irrational. Prove that $f$ is a constant function.

which an engineer can understand?

NNN
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This is a nice result worth learning how to prove, and it is possible to remove the explicit group theory language from the argument. The strategy is to argue that

  1. If $T_1$ and $T_2$ are periods then so is any integer linear combination $n_1 T_1 + n_2 T_2$, and
  2. If $T_1$ and $T_2$ are incommensurable then it is possible to make integer linear combinations $n_1 T_1 + n_2 T_2$ arbitrarily small, and
  3. having arbitrarily small periods is incompatible with being continuous unless the function is constant.

But (a classic XY problem!) you don't need this result to prove that $\cos$ and $\sin$ have smallest periods; this can also be done directly, as follows. We'll adopt the definition that $\sin x$ and $\cos x$ are the unique solutions to the differential equation

$$f''(x) = - f(x)$$

satisfying the initial conditions $\sin 0 = 0, \sin' 0 = 1$ and $\cos 0 = 1, \cos' 0 = 0$ respectively. It's not hard to show modulo analytic details that this is equivalent to the power series definitions, and personally I find this definition cleaner and more conceptual (you can tie it more closely to circles immediately). It follows from comparing initial conditions or looking at the power series that $\sin' x = \cos x$ and $\cos' x = - \sin x$, hence that

$$\frac{d}{dx} (\cos^2 x + \sin^2 x) = 0$$

(try to show this directly from the power series! It can be done but it's by no means obvious), and evaluating $\cos^2 x + \sin^2 x$ at $x = 0$ gives

The Pythagorean theorem: $\cos^2 x + \sin^2 x = 1$.

Corollary: $\cos x$ is decreasing and $\sin x$ is increasing on some interval $[0, \varepsilon)$.

Proof. Since $\cos 0 = 1$, $\cos x$ cannot increase, so it either initially decreases or stays constant. If it stayed constant then $\sin x$ would also have to stay constant, but that's incompatible with the mean value theorem. $\Box$

There are various other ways to conclude this too. We also get from the derivative calculations that the largest possible value of $\varepsilon$ is the smallest positive real such that $\cos \varepsilon = 0$ (which we later discover is, of course, $\frac{\pi}{2}$).

Corollary: Neither $\cos x$ nor $\sin x$ can have a period less than $\varepsilon$.

It's not hard to show from here that not only do $\cos x$ and $\sin x$ have a smallest period, but that that period is the arc length of the unit circle, since the Pythagorean theorem together with those derivative calculations imply that $(\cos x, \sin x)$ is not only a parameterization of the unit circle but a unit speed parameterization. This can all be rephrased in terms of either the single complex-valued function

$$e^{ix} = \cos x + i \sin x$$

($x$ is still real) or, basically equivalently, the single matrix-valued function

$$e^{\begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix}} = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix}$$

depending on taste. This is, naturally, a rotation matrix.

Qiaochu Yuan
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