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In calculus, $\cos x$ is defined as $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots$. With this definition, how can we find the period of $\cos x$

jimjim
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Jonathan Cui
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    maybe look at this link: https://math.stackexchange.com/questions/1285810/periodicity-of-trigonometric-functions-directly-from-their-power-series – atul ganju Jun 11 '20 at 07:37
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    well, you would need a definition for the number $\pi$. How do you propose that? Anyway, one reference I can suggest (because it's the only one I know that treats trigonometric functions this way) is Henri Cartan's Complex analysis textbook, Chapter 1.3 (page 30). – peek-a-boo Jun 11 '20 at 07:37
  • I thought it is defined via unit circle, this makes things more simple. – Alexey Burdin Jun 11 '20 at 07:40
  • that is one way to define $\pi$ – jimjim Jun 11 '20 at 08:07
  • What do you call "find" exactly ? –  Jun 11 '20 at 08:53

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Define $\sin(x) = x - x^3/3! + ...$. Note that $d/dx \sin(x) = \cos(x)$ and $d/dx \cos(x) = -\sin(x)$.

Taylor expand $f(x) = \cos(x + \alpha)$ about $x = 0$ to get $\cos(\alpha + x) = f(x) = \cos(\alpha) - \sin(\alpha) - \cos(\alpha) x^2/2 + \sin(\alpha) x^3 / 3! + ... = \cos(\alpha) (1 - x^2 / 2! + ...) - \sin(\alpha) (x - x^3 / 3! + ...) = \cos(\alpha) \cos(x) - \sin(\alpha) \sin(x)$. That is, for all $x, \alpha$, $\cos(x + \alpha) = \cos(x) \cos(\alpha) - \sin(x) \sin(\alpha)$.

Furthermore, note that $\cos^2(x) + \sin^2(x)$ has a derivative of 0 and thus is constant; therefore, $\cos^2(x) + \sin^2(x) = \cos^2(0) + \sin^2(0) = 1$ for all $x$.

We now seek some $\eta > 0$ such that $\cos(\eta) = 0$. We know such an $\eta$ exists since if one didn't, we would have $d^2/dx^2 \cos(x) = -\cos(x)$ always negative for $x > 0$ and therefore $\cos(x)$ would have a negative, decreasing derivative for all $x > 0$ and yet still remain positive, which is clearly impossible. Take the least such $\eta$; then clearly we must have $\cos'(\eta) = -\sin(\eta) \leq 0$; then $\sin(\eta) = 1$. Define $\tau = 4 \eta$. We have $\cos(\tau / 2) = \cos(\eta + \eta) = \cos^2(\eta) - \sin^2(\eta) = -1$ and therefore $\sin(\tau / 2) = 0$. Then $\cos(\tau) = \cos(\tau/2 + \tau/2) = \cos^2(\tau / 2) - \sin^2(\tau / 2) = 1$, and therefore $\sin(\tau) = 0$. Then for all $x$, we have $\cos(x + \tau) = \cos(x) \cos(\tau) - \sin(x) \sin(\tau) = \cos(x)$. A bit more analysis shows that no $\tau'$ such that $0 < \tau' < \tau$ satisfies $\cos(0 + \tau') = \cos(\tau') = 1$; then $\tau$ is the period of the cosine function.

Mark Saving
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