Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $$
now using Pythagorean identity: $a^2 + b^2 = c^2$
$$\frac {c^2} {c^2} = 1, \quad 1 = 1 $$

My start, for my students is the real circunference $\mathcal C$ of center $O(0,0)$ and radius $r$, $P\equiv(x,y)\in \mathcal C$ $$\mathcal C\colon \quad x^2+y^2=r^2$$ $r>0$ and $x=r\cos\theta, y=r\sin\theta$, $\theta\in[0,2\pi[$, that it gives: $$r^2\cos^2\theta+r^2\sin^2\theta=r^2 \iff r^2(\cos^2\theta+\sin^2\theta)=r^2$$
Hence dividing by $r^2>0$ you have:
$$\cos^2\theta+\sin^2\theta=1$$
