What is the period of $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\cdots$? 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 $\cos x$
 A: 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.
