Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.

If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few: 
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative. 
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer. 
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
