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

You can also prove this identity directly from the power series
$$
\begin{align}
  \cos x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} x^{2n},\\
  \sin x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} x^{2n + 1}.
\end{align}
$$
The following is modified from the discussion on Wikipedia's article on the Pythagorean Trigonometric Identity.
Squaring each of these series using the Cauchy Product
$$\left(\sum_{i=0}^\infty a_i x^i\right) \cdot \left(\sum_{j=0}^\infty b_j x^j\right) = \sum_{k=0}^\infty \left(\sum_{l=0}^k a_l b_{k-l}\right) x^k\,,$$
and combining the factorials into a binomial coefficient we get
$$\begin{align}
\cos^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)} \\
& = \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n} \\
& = 1 +  \sum_{n = 1}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n}\,,\\
\sin^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)} \\
& = \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n} \\
& = \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}.
\end{align}
$$
Adding the squared series we can combine the odd and even terms then use the binomial theorem to simplify the internal sum to zero:
$$
\begin{align}
\cos^2 x + \sin^2 x
& = 1 + \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n}{2n \choose 2i} - \sum_{i = 0}^{n - 1}{2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \\
& = 1 + \sum_{n = 1}^\infty \left(\sum_{j = 0}^{2n}(-1)^j{2n \choose j} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}
\\
& = 1 + \sum_{n = 1}^\infty \left(1-1\right)^{2n} \frac{(-1)^{n - 1}}{(2n)!} x^{2n}
= 1\,.
\end{align}
$$
