Finding $\pi$ through sine and cosine series expansions I am working on a problem in Partha Mitra's book Observed Brain Dynamics (the problem was originally from Rudin's textbook Real and Complex Analysis, and appears on page 54 of Mitra's book). Unfortunately, the book I have does not contain any solutions... Here is the question:

By considering the first few terms of the series expansion of sin(x) and cos(x), show that there is a real number $x_0$ between 0 and 2 for which $\cos(x_0)=0$  and $\sin(x_0)=1$. Then, define $\pi=2x_0$, and show that $e^{i\pi/2}=i$ (and therefore that $e^{i\pi}=-1$ and $e^{2\pi i}=1$.

Attempt at a solution: In a previous problem I derived the series expansions for sine and cosine as
$$
\sin(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n+1}}{(2n+1)!} \right) \right]
$$
$$
\cos(x) = \sum_{n=0}^{\infty} \left[ (-1)^n \left( \frac{x^{2n}}{(2n)!} \right) \right]
$$
My thought is that you can show that $\cos(0)=1$ (trivially), and that $\cos(2)<0$ (less trivially). This then implies that there is a point $x_0$ between 0 and 2 where $\cos(x_0)=0$, since the cosine function is continuous. However, I do not understand how you could then show that $\sin(x_0)=1$ at this same point. My approach may be completely off here.
I believe that the second part of this problem ("Then, define $\pi=2x_0$...") will be easier once I get past this first part.
Thanks so much for the help. Also - I swear this is not a homework assignment. I am reading through this book on my own to improve my math.

How to show that $\sin (x_0)=1$ if $\cos (x_0)=0$? Quite simply:
$$\sin^2 x+\cos^2 x=1$$
(you may also want to specify that $\sin x$ is positive in the given range)
