<s>[INST] Prove $\sin(\pi/2)=1$ using Taylor series 
Prove $\sin(\pi/2)=1$ using the Taylor series definition of $\sin x$,
  $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

It seems rather messy to substitute in $\pi/2$ for $x$. So we have 
$$\sin(\pi/2)=\sum_{n=0}^{\infty} \frac{(-1)^n(\pi/2)^{2n+1}}{(2n+1)!}.$$
I'm not too sure where to go from here.  Any help would be appreciated! Thanks!
 [/INST] [ A sketch: ]
Let $f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$. You can easily show that
$$\begin{eqnarray}
f(x + c) & = & f(c) + f'(c)x + f''(c)\frac{x^2}{2!} + \cdots \quad(1) \\
f''(x) & = & -f(x) \quad(2)\\
f'(x) & = & 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \quad(3)
\end{eqnarray}$$
Then it is straight forward to prove that for any $c$: $$f(x+c) = f'(c)f(x) + f(c)f'(x) \quad(4)$$
We can also prove that there is a $b$ such that $f'(b) = 0$. Then $(1),(4) \Rightarrow f(b) = 1$ and so $$f(x+b) = f'(x) \quad(5)$$
Going a little further, $(2),(5) \Rightarrow f(x+4b) = f(x)$ so $b = \frac{1}{4}\mathrm{period}$.
But $f^2(x) + f'^2(x) = 1$ (all terms of expansion except the first of $f'(x)$ cancel out) so the period of $f$ is $2\pi$, $$b = \frac{\pi}{2}$$
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