# Sine and Cosine are Periodic on Reals/Sine

## Theorem

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

## Proof 1

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:

- $K = L$

$\blacksquare$

## Proof 2

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

From Primitive of Cosine Function:

- $\ds \int \cos x \rd x = \sin x + C$

for any constant $C$.

Therefore $\sin x$ is a Primitive of $\cos x$, for the special case of $C = 0$.

From Primitive of Periodic Real Function, it follows that $\sin x$ is periodic with period $L$.

$\blacksquare$