Pythagoras's Theorem/Algebraic Proof

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:


 * 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: