In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on both sides:
$$1 = \frac{p^2}{h^2}+\frac{b^2}{h^2}$$
since $\sin x = \frac ph$ and $\cos x = \frac bh$, 
$$1 = \sin^2x+\cos^2x$$
Are there any more innovative ways of proving this common identity?

Here's a couple of different methods:
Proof using the cosine angel sum formula:$$
1=\cos(0) =\cos(x + (-x)) =\cos(x)\cos(-x) - \sin(x)\sin(-x) = \cos^2(x) + \sin^2(x)
$$
Proof using differential equations:
Recall the differential equation definitions of $\sin$ and $\cos$: They are the solutions to $f'' = - f$ with the appropriate initial conditions $f(0) = 0$, $f'(0) = 1$ for $\sin(x)$ and $f(0) = 1$, $f'(0) = 0$ for $\cos(x)$. Since solutions to differential equations are unique given the initial conditions, we immediately get $\sin'(x) = \cos(x)$ under this definition. Then:\begin{eqnarray}
\frac{d}{dx} \left(\sin^2(x) + \cos^2(x)\right) &=& \frac{d}{dx} \left(\sin^2(x) + \left(\sin'(x)\right)^2\right)\\
&=& 2\sin(x)\sin'(x) + 2\sin'(x)\sin''(x)\\
&=& 2\sin(x)\sin'(x) +2\sin'(x)(-\sin(x)) = 0
\end{eqnarray}
hence $\sin^2(x) + \cos^2(x)$ is constant, since its derivative is 0. Plugging in $x=0$ we see that it must equal 1.
Proof using Euler's formula:
$$
\sin^2(x) + \cos^2(x) = \left(\cos x + i \sin x\right)\left(\cos x - i \sin x\right) =\left(\cos x + i \sin x\right)\left(\cos (-x) + i \sin (-x)\right)= e^{i x} e^{-i x} = 1
$$
Proof using Taylor series:
Using the Taylor series:\begin{eqnarray}
\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\\
\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}
\end{eqnarray}
we have $$
\sin^2(x) = \left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m + 2}}{(2n+1)!(2m+1)!} = \sum_{k=0}^\infty x^{2k+2}(-1)^k \sum_{j=0}^k \frac{1}{(2j+1)!(2(k-j)+1)!}= \sum_{k=0}^\infty \frac{x^{2k+2}(-1)^k}{(2k+2)!}\sum_{j=0}^k \binom{2k+2}{2j+1} = -\sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^{k-1} \binom{2k}{2j+1}
$$
and $$
\cos^2(x) =\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m}}{(2n)!(2m)!} = \sum_{k=0}^\infty x^{2k}(-1)^k \sum_{j=0}^k \frac{1}{(2j)!(2(k-j))!} = \sum_{k=0}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j} = 1 + \sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j}
$$
Adding these together:$$
\sin^2(x) + \cos^2(x) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^k \binom{2k}{2j}-\sum_{j=0}^{k-1} \binom{2k}{2j+1}\right)=1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^{2k} (-1)^j\binom{2k}{j}\right) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^l}{(2k)!} (1 + (-1))^{2k} = 1
$$
We use the binomial theorem on the second last step.
Carlson's Theorem: Major overkill to use this one, but it still works. Notice $\cos(x)$, and $\sin(x)$ have exponential type 1. Thus $$
f(x) = \cos^2(\frac{\pi}{4}x) + \sin^2(\frac{\pi}{4}x) - 1
$$
has exponential type at most $\frac{\pi}{2} < \pi$. By inspection, you can see that $f(x) = 0$ for $x\in \mathbb{N}$. Hence Carlson's theorem implies $f(x) = 0$ identically.
