TITLE: Proof of $\sin^2(x) + \cos^2(x)=1$ using series
QUESTION [9 upvotes]: I have to prove the following identity $\sin^2 (x) + \cos^2(x)=1$. I can easily prove this, but this exercise is given in the section introducing the series expansions for $\sin(x)$ and $\cos(x)$ and I should use these in the proof but am not quite sure how to do this.

REPLY [3 votes]: I realise this thread has been well-resolved for some time, but I arrived here from another thread where I realised that there's some middle-of-the-road-like approach.
It just requires
$$
\exp(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!}
$$
for any $z\in\mathbb{C}$. Evaluating at $z=ix$ for $x\in\mathbb{R}$ and comparing with the series expansions of $\sin$ and $\cos$ clearly yields $\exp(ix)=\cos(x)+i\sin(x)$. Furthermore, by the Binomial Theorem, we see that
$$
\exp(z+w)=\sum_{n=0}^{\infty} \frac{(z+w)^n}{n!}=\sum_{n=0}^{\infty}\sum_{j=0}^n {n \choose j}\frac{z^j w^{n-j}}{n!}=\sum_{n=0}^{\infty} \sum_{j=0}^n \frac{z^j}{j!}\frac{w^{n-j}}{(n-j)!}=\sum_{j=0}^{\infty}\frac{z^j}{j!}\sum_{k=0}^{\infty} \frac{w^k}{k!}
$$
and the last expression is equal to $\exp(z)\exp(w)$.
This simplies things a lot as, clearly, $\exp(0)=1$ and so $\exp(ix)=(\exp(-ix))^{-1}$ and, by continuity of complex conjugation, $\exp(-ix)$ is the complex conjugate of $\exp(ix)$. However, this happens if and only if $|\exp(ix)|=1$. And now,
$$
1=|\exp(ix)|=(\Re\exp(ix))^2+(\Im\exp(ix))^2=\cos^2(x)+\sin^2(x)
$$