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!