$\sin^2(x)+\cos^2(x) = 1$ using power series 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?

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.
