My question is very simple yet I've gotten nowhere with it. Is there any way one can, without directly or indirectly referencing any differential equations satisfied by the circular trigonometric functions, demonstrate that their power series expansions are periodic? Just looking at the power series, I never would have guessed in a million year that they were periodic and with period $2\pi$, of all things. Any and all insights are welcome!
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Not referring to the differential equations is pretty crippling. They're arguably equivalent to the power series expansions, but in a way that actually makes it possible to answer this question. – Qiaochu Yuan May 17 '15 at 05:04
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I have a feeling that there's some sneaky trick we're missing but you may be right. PS- next year , a gentleman and a scholar named Christopher Eur who is interested in homotopy theory will be studying at Berkley. I recommend you get to know him, as he is a mathemagician of epic proportions. – Archaick May 17 '15 at 05:16
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Does this answer your question? How to prove periodicity of $\sin(x)$ or $\cos(x)$ starting from the Taylor series expansion? – Arnaud D. Jun 11 '20 at 08:12
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Sine and cosine are immediately seen to be odd and even, respectively. Also, it's not too hard to show that sine and cosine are continuous.
It's not hard to prove the sum rules for $\sin(A+B)$ and $\cos(A+B)$, with the series definition. (Complex numbers are recommended but not necessary.) Note that, using the sum rules, we have: $$\cos(x-x)=\cos(x)\cos(-x)-\sin(x)\sin(-x)$$ and thus $1=\cos^2(x)+\sin^2(x)$. In addition, the sum rules give us the double angle formulae.
Then, notice that $\cos0=1$, and it can be shown that $-1<\cos2<-\frac13$. (We get these bounds by using, respectively, two and three terms of the series.) Thus, by the IVT, there is a zero of the cosine function, between $x=0$ and $x=2$. Call this number $\frac\pi2$.
We thus have $\sin(\pi)=2\sin(\frac\pi2)\cos(\frac\pi2)=0$, and $\cos(\pi)=2\cos^2(\frac\pi2)-1=-1$. Using the double angle formulae again, we have $\sin(2\pi)=0$ and $\cos(2\pi)=1$.
Thus: \begin{align} \sin(x+2\pi)&=\sin(x)\cos(2\pi)+\cos(x)\sin(2\pi)\\ &=\sin(x)1+\cos(x)0\\ &=\sin(x) \end{align} and \begin{align} \cos(x+2\pi)&=\cos(x)\cos(2\pi)-\sin(x)\sin(2\pi)\\ &=\cos(x)1-\sin(x)0\\ &=\cos(x) \end{align} which means that $\sin$ and $\cos$ are periodic with period $2\pi$.
Akiva Weinberger
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Here's an approach that avoids differential equations.
Euler's identity $e^{ix}=\cos x+i\sin x$ is proved by simple inspection of the power series of the functions involved.
The sum-product formula $e^{a+b}=e^ae^b$ can be proved by series manipulation: $$ e^{a+b}=\sum_{n=0}^\infty \frac{(a+b)^n}{n!} = \sum_{n=0}^\infty\sum_{k=0}^n \frac{a^kb^{n-k}}{k!(n-k)!} = \sum_{k=0}^\infty \frac{a^k}{k!}\sum_{m=0}^\infty \frac{b^m}{m!}=e^ae^b $$ (where $m=n-k$ substitution was made; rearrangement is legal because the double series converges absolutely)
Since the coefficients of the power series of $e^z$ are real, $\overline{e^z}=e^{\bar z}$.
Using 3, we get for $x\in \mathbb{R}$ $$ |e^{ix}|^2 = e^{ix}\overline{e^{ix}} = e^{ix}e^{\overline{ix}}=e^{ix}e^{-ix}=1 $$
The image of $\mathbb{R}$ under the function $f(x)=e^{ix}$ is a nonempty connected subset of the unit circle $\mathbb{T}$ in the complex plane, and is closed under multiplication. The only such subsets are $\{1\}$ and $\mathbb{T}$. It can't be $\{1\}$ since $f$ is nonconstant (e.g., because $f(x)=1+ix+o(|x|)$ for small $x$.)
Thus, there exists $a\in \mathbb{R}$ such that $e^{ia}=1$. It follows that $e^{i(x+a)}=e^{ix}$ for all real $x$, and subsequently $\cos(x+a)=\cos x$ and $\sin(x+a)=\sin x$.