# Pythagoras's Theorem/Algebraic Proof

## Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.

Then:

$a^2 + b^2 = c^2$

## Proof

$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

From these, we derive the proof that $\cos^2 x + \sin^2 x = 1$.

Then from the Equivalence of Definitions of Sine and Cosine, we can use the geometric interpretation of sine and cosine:

$\sin \theta = \dfrac {\text{Opposite}} {\text{Hypotenuse}}$
$\cos \theta = \dfrac {\text{Adjacent}} {\text{Hypotenuse}}$

Let $\text{Adjacent} = a, \text{Opposite} = b, \text{Hypotenuse} = c$, as in the diagram at the top of the page.

Thus:

 $\ds \cos^2 x + \sin^2 x$ $=$ $\ds 1$ Sum of Squares of Sine and Cosine $\ds \leadsto \ \$ $\ds \paren {\frac a c}^2 + \paren {\frac b c}^2$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds a^2 + b^2$ $=$ $\ds c^2$ multiplying both sides by $c^2$

$\blacksquare$

## Source of Name

This entry was named for Pythagoras of Samos.