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.
Let me contribute by this so let
$$f(\theta)=\cos^2\theta+\sin^2\theta$$
then it's simple to see that
$$f'(\theta)=0$$
then
$$f(\theta)=f(0)=1$$
Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.

Define the matrix

$$ A(\theta) = \left( \begin{matrix} \cos(\theta) &amp; \sin(\theta) \\ -\sin(\theta) &amp; \cos(\theta) \end{matrix} \right) $$

By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.

There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives

$$ A(\theta)^n = A(n \theta) \approx I $$

where $I$ is the identity matrix. By taking determinants, we get

$$ (\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$

for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that

$$ \cos(\theta)^2 + \sin(\theta)^2 = 1$$



To get the details right, we select sequences of integers $a_n, b_n$ with

$$ \lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0 $$
$$ \lim_{n \to \infty} a_n = +\infty $$

e.g. this can be done by continued fractions. Then,

$$ \lim_{n \to \infty} A(a_n \theta) =
\lim_{n \to \infty} A(a_n \theta - 2 \pi b_n)
= A(0) = I$$

and consequently

$$ \lim_{n \to \infty}  (\det A(\theta))^{a_n} = 1 $$



Another method in the same vein is to also define vectors

$$ v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right) $$

so that $A(\theta) v(\varphi) = v(\theta + \varphi)$

However, we know that for every angle $\theta$:

$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1 $$

In particular, we have

$$ \frac{1}{2} \leq \| v(\theta) \| \leq 2 $$

and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.

By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that  all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.

However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.