Pythagoras's Theorem/Algebraic Proof

Theorem

Given any right triangle $\triangle ABC$ with $c$ as the hypotenuse, we have $a^2 + b^2 = c^2$.

Proof

We start with the algebraic definitions for sine and cosine:

• $\displaystyle \sin x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$

• $\displaystyle \cos x = \sum_{n=0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$

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

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

• $\displaystyle \sin \theta = \frac {\text{Opposite}} {\text{Hypotenuse}}$

• $\displaystyle \cos \theta = \frac {\text{Adjacent}} {\text{Hypotenuse}}$

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

Thus:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \cos^2 x + \sin^2 x\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({\frac a c}\right)^2 + \left({\frac b c}\right)^2\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a^2 + b^2\)	\(=\)	\(\displaystyle c^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	multiplying both sides by $c^2$

$\blacksquare$