Characterisation of Sine and Cosine
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Theorem
Let $\map s x: \R \to \R$, $\map c x: \R \to \R$ be differentiable real functions that satisfy:
- $(1): \quad \map {s'} x = \map c x$
- $(2): \quad \map {c'} x = -\map s x$
- $(3): \quad \map s 0 = 0$
- $(4): \quad \map c 0 = 1$
- $(5): \quad \forall x: \map {s^2} x + \map {c^2} x = 1$
Then, for every $x \in \R$:
- $\map s x = \map \sin x$
- $\map c x = \map \cos x$
Proof
Define:
- $\map h x = \paren {\map c x - \map \cos x}^2 + \paren {\map s x - \map \sin x}^2$
Then:
\(\ds \map h x\) | \(=\) | \(\ds \map {c^2} x - 2 \map c x \map \cos x + \map {\cos^2} x + \map {s^2} x - 2 \map s x \map \sin x + \map {\sin^2} x\) | Square of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 - 2 \paren {\map c x \map \cos x + \map s x \map \sin x}\) | Property $(5)$, Sum of Squares of Sine and Cosine |
By taking $\map {h'} x$:
\(\ds \map {h'} x\) | \(=\) | \(\ds -2 \paren {\map c x \map \cos x + \map s x \map \sin x}'\) | Linear Combination of Derivatives, Derivative of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \paren {\map c x \map {\cos'} x + \map {c'} x \map \cos x + \map s x \map {\sin'} x + \map {s'} x \map \sin x}\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \paren {-\map c x \map \sin x - \map s x \map \cos x + \map s x \map \cos x + \map c x \map \sin x}\) | Properties $(1)$ and $(2)$, Derivative of Cosine Function, Derivative of Sine Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \paren {\map c x \map \sin x - \map c x \map \sin x} - 2 \paren {\map s x \map \cos x - \map s x \map \cos x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
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 $(4)$, Cosine of Zero is One, Sine of Zero is Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Since $\map h x$ is constant:
- $\map h x = 0$
Therefore:
- $\paren {\map c x - \map \cos x}^2 + \paren {\map s x - \map \sin x}^2 = 0$
for every $x \in \R$.
But since Square of Real Number is Non-Negative and Square of Non-Zero Real Number is Strictly Positive:
- $\map c x - \map \cos x = 0$
- $\map s x - \map \sin x = 0$
The result follows.
$\blacksquare$