TITLE: Prove $\cos^2(\theta)+\sin^2(\theta) = 1$
QUESTION [8 upvotes]: $$\cos^2(\theta) + \sin^2(\theta) = 1$$
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: $a^2 + b^2 = c^2$
$$\frac {c^2} {c^2} = 1, \quad 1 = 1 $$

REPLY [5 votes]: Well, the fundamental theorem of trigonometry is an extra useful identity for us, we all know that it's important.
Here's an approach:
It's basic that:
$$0=0$$
Wich is the same if we wrote:
\begin{align}
0&=0.t\\
&=0.x\bigg\vert_0^t\\
&=0\times\int_0^t\mathrm{d}x\\
&=\int_0^t2\cos(x)\sin(x)-2\cos(x)\sin(x)\mathrm{d}x\\
&=\underbrace{\int_0^t 2\cos(x)\sin(x)\mathrm{d}x}_{I}-\underbrace{\int_0^t2\cos(x)\sin(x)\mathrm{d}x}_{J}\\
\end{align}
Now using some substituions in the integral $I$:
$$u=\sin(x)\ \ \ \ \ \ \ \ \ \mathrm{d}u=\cos(x)$$
We'll have:
$$I=2\int_0^{\sin(t)}u\ \mathrm{d}u$$
For the integral $J$:
$$v=\cos(x)\ \ \ \ \ \ \ \ \ \ \ \mathrm{d}v=-\sin(x)$$
So:
$$J=2\int_1^{\cos(t)}v\ \mathrm{d}v$$
Note that the minus sign will dissapear due to the substitution:
\begin{align}
0&=2\int_0^{\sin(t)}u\ \mathrm{d}u+2\int_1^{\cos(t)}v\ \mathrm{d}v\\
&=u^2\bigg\vert_0^{\sin(t)}+v^2\bigg\vert_1^{\cos(t)}\\
\sin^2(t)+\cos^2(t)-1&=0\end{align}
And here's our identity:
$${\cos^2(\varphi)+\sin^2(\varphi)=1}$$
I Hope that you've liked it!