Proof:
Suppose there is a right triangle with sides $a,b,c$ and opposite angles $A,B,C$, respectively. Let $C$ be the right angle, and thus $c$ the hypotenuse. Since all interior angles of a triangle must add to $180 ^{\circ},$

(1)
\begin{align} A+B+C&=180^{\circ} \\ A+B+90^{\circ} &= 180^{\circ} \\ A+B &= 90^{\circ}. \end{align}

Find the sines and cosines of $A$ and $B$:

(2)
\begin{align} \sin A=\frac{a}{c}, \cos A = \frac{b}{c}, \sin B=\frac{b}{c}, \cos B = \frac{a}{c} \end{align}

By the sum-difference formula, $\sin(A+B) = \sin A \cos B + \cos A + \sin B$.

(3)
\begin{align} \sin A \cos B + \cos A + \sin B &= \sin 90^{\circ} \\ \frac{a}{c}\cdot \frac{a}{c} + \frac{b}{c} \cdot \frac{b}{c} = 1 \\ \frac{a^2}{c^2} + \frac{b^2}{c^2} = 1 \\ \frac{a^2 + b^2}{c^2} = 1 \\ a^2 + b^2 = c^2. \blacksquare \end{align}
page revision: 2, last edited: 24 Jan 2018 14:57