Question: In an example I had to prove that $\sin^2(x)+\cos^2(x)=1$ which is fairly easy using the unit circle. My teacher then asked me to show the same thing using the following power series:$$\sin(x)=\sum\_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}$$ and $$\cos(x)=\sum\_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}$$
However, if I now take the squares of these values I get a really messy result that I can't simplify to 1. 

Could anyone give me a hint on how to deal with this question?

Answer: 
**Hint:** Differentiate the power series term by term. We get that (surprise!) the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. Let $f(x)=\sin^2 x+\cos^2 x$. Show its derivative is $0$. Thus $f(x)$ is constant.
There is an easy solution using complex numbers (but not the complex exponential):
$$
\begin{align}
\sin^2 x + \cos^2 x & = \left(\cos x + i \sin x\right)\left(\cos x - i \sin x\right) \\
&=\left(\sum\_{j=0}^{\infty} (-1)^j \frac{x^{2j}}{(2j)!} + i \sum\_{j=0}^{\infty} (-1)^j \frac{x^{2j+1}}{(2j+1)!}\right)\left(\sum\_{j=0}^{\infty} (-1)^j \frac{x^{2j}}{(2j)!} - i \sum\_{j=0}^{\infty} (-1)^j \frac{x^{2j+1}}{(2j+1)!}\right)\\
&=\left(\sum\_{k=0}^{\infty} i^k \frac{x^{k}}{k!}\right) \cdot \left(\sum\_{k=0}^{\infty} (-i)^k \frac{x^{k}}{k!}\right). \\
\end{align}
$$
(Of course these last two factors are equal to $e^{ix}$ and $e^{-ix}$, but we do not need this fact: we are just manipulating power series here.) We need to show that the above product is $1$, so that the coefficient of $x^\ell$ in the above product is $1$ if $\ell=0$ and $0$ otherwise.

Expanding the product, we get that the coefficient of $x^\ell$ equals
$$
\sum\_{k\_1+k\_2=\ell} \frac{i^{k\_1}}{k\_1!} \cdot \frac{(-i)^{k\_2}}{k\_2!}=\sum\_{k\_1+k\_2=\ell} (-1)^{k\_2} \frac{i^{\ell}}{k\_1!k\_2!} = \sum\_{k\_1=0}^\ell (-1)^{\ell-k\_1}\frac{i^{\ell}}{\ell!} \binom{\ell}{k\_1}.
$$
In view of the well-known binomial coefficient identity $\sum\_{k=0}^{\ell} (-1)^k \binom{\ell}{k}=0$ for any integer $\ell>0$, which may be obtained by expanding the identity $(1-1)^{\ell}=0$, the above sum vanishes when $\ell>0$, and when $\ell=0$ it equals $1$, as can be computed directly. Which is precisely what we had to show.