Equivalence of Definitions of Sine and Cosine

Theorem
The definitions for sine and cosine are equivalent.

That is:


 * $\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} \iff \sin x = \frac {\text{Opposite}} {\text{Hypotenuse}}$


 * $\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} \iff \cos x = \frac {\text{Adjacent}} {\text{Hypotenuse}}$

Proof
Let $\map s x: \R \to \R$, $\map c x: \R \to \R$ be two functions that satisfy:


 * $(1): \quad \map {s'} x = \map c x$
 * $(2): \quad \map {c'} x = -\map s x$
 * $(3): \quad \map s 0 = 0$
 * $(4): \quad \map c 0 = 1$
 * $(5): \quad \forall x: \map {s^2} x + \map {c^2} x = 1$

where $s'$ denotes the derivative $x$.

Let $\map f x: \R \to \R$, $\map g x: \R \to \R$ also be two functions that satisfy:
 * $(1): \quad \map {f'} x = \map g x$
 * $(2): \quad \map {g'} x = -\map f x$
 * $(3): \quad \map f 0 = 0$
 * $(4): \quad \map g 0 = 1$
 * $(5): \quad \forall x: \map {f^2} x + \map {g^2} x = 1$

It will be shown that:
 * $\map f x = \map s x$

and:
 * $\map g x = \map c x$

Define:
 * $\map h x = \paren {\map c x - \map g x}^2 + \paren {\map s x - \map f x}^2$

Notice that:
 * $\paren {\forall x: \map h x = 0} \iff \paren {\forall x: \map c x = \map g x, \map s x = \map f x}$

Then:

By taking $\map {h'} x$:

By Zero Derivative implies Constant Function, $\map h x$ is a constant function:
 * $\map h x = k$

Also:

Since $\map h x$ is constant, then:
 * $\forall x: \map h x = 0$

Then:
 * $\map c x = \map g x$

and:
 * $\map s x = \map f x$

By:
 * Derivative of Sine Function
 * Derivative of Cosine Function
 * Sine of Zero is Zero
 * Cosine of Zero is One
 * Sum of Squares of Sine and Cosine

both definitions satisfy all these properties.

Therefore they must be the same.