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 for Derivatives

\(\ds \)

\(=\)

\(\ds 0\)

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.

$\blacksquare$