# Equivalence of Definitions of Sine and Cosine

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## 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 with respect to $x$.

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:

\(\ds \map h x\) | \(=\) | \(\ds \map {c^2} x - 2 \map c x \map g x + \map {g^2} x + \map {s^2} x - 2 \map s x \map f x + \map {f^2} x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 - 2 \paren {\map c x \map g x + \map s x \map f x}\) | Property $(5)$ |

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

\(\ds \map {h'} x\) | \(=\) | \(\ds -2 \paren {\map c x \paren {-\map f x} + \map g x \paren {-\map s x} + \map s x \map g x + \map c x \map f x}\) | Properties $(1)$ and $(2)$ and Product Rule | |||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

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

- $\map h x = k$

Also:

\(\ds \map h 0\) | \(=\) | \(\ds \paren {1 - 1}^2 + \paren {0 - 0}^2\) | Properties $(3)$ and $(4)$ | |||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

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.

$\blacksquare$