Pythagoras's Theorem/Algebraic Proof

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Theorem

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

Then:

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


Proof

We start with the algebraic definitions for sine and cosine:

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


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:

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

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.