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 $\cos^2x+\sin^2x$, and notice that it is $2\cos x\cos'x+2\sin x\sin'x$. Evaluate the expressions of $\sin'x$ and $\cos'x$, and notice that they are the same as those of $\cos x$ and $-\sin x$ respectively. Then you get $\big(\cos^2x+\sin^2x\big)'=0\iff\cos^2x+\sin^2x=$ constant. How do we guess which constant exactly ? By computing $\sin^20+\cos^20$.
It is not necessary to use derivatives. It is much simpler if we connect the series for $\sin x$ and $\cos x$ with the series of $\exp(x)$. From the given series it is easy to show that $\cos x + i\sin x = \exp(ix)$ and $\cos x - i\sin x = \exp(-ix)$ where $$\exp(z) = \sum\_{n = 0}^{\infty}\frac{z^{n}}{n!}$$ and using binomial theorem it is easy to show that series of $\exp(z)$ satisfies $$\exp(z + z') = \exp(z)\cdot \exp(z')$$ and therefore $$\begin{aligned}\cos^{2}x + \sin^{2}x &= (\cos x + i\sin x)(\cos x - i\sin x)\\&= \exp(ix)\exp(-ix)\\&= \exp(0)\\&=1\end{aligned}$$