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?

1) With identity $a^2-b^2= (a-b)(a+b)$,
$$\cos^2x+  \sin^2x
= (\cos x-i\sin x)(\cos x+i\sin x)=e^{-i x}e^{i x}=1$$
2) With double angle identities,
$$\cos^2x+  \sin^2x
=\frac12 (1+\cos 2 x)+\frac12(1-\cos 2x)=1$$
3) with half-angle subs $t= \tan\frac x2$,
$$\cos^2x+  \sin^2x
=\left(\frac{1-t^2}{1+t^2}\right)^2 +\left(\frac{2t}{1+t^2}\right)^2
=\left(\frac{1+t^2}{1+t^2}\right)^2 =1$$
4) As an integral,
$$f(x)=\cos^2x+  \sin^2x =f(0) + \int_0^x f’(t)dt = 1+0=1$$
5) With $x=it $, apply $\sin(it)= i\sinh t$ and $\cos(it)=\cosh t$
$$\cos^2x+  \sin^2x = \cosh^2 t -  \sinh^2 t
= \left(\frac{e^t+e^{-t}}{2}\right)^2 -\left(\frac{e^t-e^{-t}}{2}\right)^2 =1$$
