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:


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


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

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:


 * SineCosine.png


 * $$\sin \theta = \frac {\text{Opposite}} {\text{Hypotenuse}}$$;
 * $$\cos \theta = \frac {\text{Adjacent}} {\text{Hypotenuse}}$$.

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

Thus:

$$ $$ $$