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 Trigonometric Functions, 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.