Sine and Cosine are Periodic on Reals/Sine/Proof 1

Theorem

The sine function is periodic with the same period as the cosine function.

Proof

Since Real Cosine Function is Periodic, let $K$ be its period.

Then:

$\cos K = \map \cos {0 + K} = \cos 0$

Because Cosine of Zero is One:

$\cos K = 1$

Furthermore:

\(\ds \cos^2 K + \sin^2 K\)

\(=\)

\(\ds 1\)

Sum of Squares of Sine and Cosine

\(\ds \sin^2 K\)

\(=\)

\(\ds 0\)

$\cos K = 1$

\(\ds \sin K\)

\(=\)

\(\ds 0\)

Then, the following holds:

\(\ds \map \sin {x + K}\)

\(=\)

\(\ds \sin x \cos K + \cos x \sin K\)

Sine of Sum

\(\ds \)

\(=\)

\(\ds \sin x \cdot 1 + \cos x \cdot 0\)

$\cos K = 1$; $\sin K = 0$

\(\ds \)

\(=\)

\(\ds \sin x\)

Thus $\sin$ is periodic with some period $L \leq K$.

$\Box$

The following also hold:

\(\ds \sin L\)

\(=\)

\(\ds \map \sin {0 + L}\)

\(\ds \)

\(=\)

\(\ds \sin 0\)

Period of Periodic Real Function

\(\ds \)

\(=\)

\(\ds 0\)

Sine of Zero is Zero

\(\ds \cos L\)

\(=\)

\(\ds \bigvalueat { \dfrac{\d \sin x} {\d x} } {x \mathop = L}\)

Derivative of Sine Function

\(\ds \)

\(=\)

\(\ds \lim_{h \mathop \to 0} \dfrac{\map \sin {L + h} - \map \sin L}{h}\)

Derivative of Real Function at Point

\(\ds \)

\(=\)

\(\ds \lim_{h \mathop \to 0} \dfrac{\map \sin {0 + h} - \map \sin 0}{h}\)

Period of Periodic Real Function

\(\ds \)

\(=\)

\(\ds \bigvalueat { \dfrac{\d \sin x} {\d x} } {x \mathop = 0}\)

Derivative of Real Function at Point

\(\ds \)

\(=\)

\(\ds \cos 0\)

Derivative of Sine Function

\(\ds \)

\(=\)

\(\ds 1\)

Cosine of Zero is One

So we may conclude:

\(\ds \map \cos {x + L}\)

\(=\)

\(\ds \cos x \cos L - \sin x \sin L\)

Cosine of Sum

\(\ds \)

\(=\)

\(\ds \cos x \cdot 1 - \sin x \cdot 0\)

$\cos L = 1$ and $\sin L = 0$

\(\ds \)

\(=\)

\(\ds \cos x\)

Therefore the period of cosine is at most that of sine:

$K \leq L$

But if $K \leq L \leq K$ then:

$K = L$

$\blacksquare$