Proof. $$

1.

In Figure 5a, by Pythagorean theorem, $a^2+b^2=c^2$.

2.

In Figure 5b, by definition of sin and cosine,

$\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃={\displaystyle \frac{a}{c}},\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃={\displaystyle \frac{b}{c}}$

3.

Combining step 1 and 2:

${\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃+{\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃={\displaystyle \frac{a^2}{c^2}}+{\displaystyle \frac{b^2}{c^2}}={\displaystyle \frac{a^2+b^2}{c^2}}=1$

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