Prove $\sin^2\theta + \cos^2\theta = 1$

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 $$f(\theta)=f(0)=1$$

Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:

$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$

Take $\theta \in\mathbb R$. The following holds: $$\begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}$$

Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,

By the Pythagorean theorem, $$ {AC}^2+{BC}^2 = {AB}^2 $$ Dividing by $AB^2$, $$ \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align} $$