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.

$\blacksquare$