Sine and Cosine are Periodic on Reals

Theorem
The sine and cosine functions are periodic on the set of real numbers $$\mathbb{R}$$.

Corollary

 * $$\cos \left({x + \pi}\right) = - \cos x$$;
 * $$\sin \left({x + \pi}\right) = - \sin x$$.


 * $$\cos x$$ is strictly positive on the interval $$\left({-\frac \pi 2 \, . \, . \, \frac \pi 2}\right)$$ and strictly negative on the interval $$\left({\frac \pi 2 \, . \, . \, \frac {3 \pi} 2}\right)$$;


 * $$\sin x$$ is strictly positive on the interval $$\left({0 \, . \, . \, \pi}\right)$$ and strictly negative on the interval $$\left({\pi \, . \, . \, 2 \pi}\right)$$;

Zeroes of Sine and Cosine

 * $$\forall n \in \mathbb{Z}: x = \left({n + \frac 1 2}\right) \pi \Longrightarrow \cos x = 0$$;
 * $$\forall n \in \mathbb{Z}: x = n \pi \Longrightarrow \sin x = 0$$.

Proof
From Basic Properties of Cosine Function we have that $$\cos 0 = 1$$ and $$\cos x = \cos \left({-x}\right)$$.

As $$\cos x$$ is continuous, it follows that $$\exists \xi > 0: \forall x \in \left({-\xi \, . \, . \, \xi}\right): \cos x > 0$$.

Now, suppose $$\cos x$$ were always positive.

From Derivative of Cosine Function, we have $$D_{xx} \left({\cos x}\right) = D_x \left({-\sin x}\right) = -\cos x$$.

Thus $$-\cos x$$ would always be negative and thus $$\cos x$$ would be concave everywhere.

But as $$\cos x$$ is bounded on $\mathbb{R}$, it can not be concave everywhere else it would be constant.

So $$\cos x$$ can not be positive all the time.

Therefore, there must exist a smallest positive $$\eta \in \mathbb{R}$$ such that $$\cos \eta = 0$$.

By definition, $$\cos \eta = \cos \left({-\eta}\right) = 0$$ and $$\cos x > 0$$ for $$-\eta < x < \eta$$.

Now we show that $$\sin \eta = 1$$.

From Sum of Squares of Sine and Cosine, we have $$\cos^2 x + \sin^2 x = 1$$.

Hence as $$\cos \eta = 0$$ it follows that $$\sin^2 \eta = 1$$, so either $$\sin \eta = 1$$ or $$\sin \eta = -1$$.

But $$D_x \left({\sin x}\right) = \cos x$$.

On the interval $$\left[{-\eta \,. \, . \, \eta}\right]$$, we have just shown that $$\cos x > 0$$.

Thus on this interval, $$\sin x$$ is increasing.

Since $$\sin 0 = 0$$ it follows that $$\sin \eta > 0$$ and so it must be that $$\sin \eta = 1$$.

Now we apply Sine and Cosine of Sum:


 * $$\sin \left({x + \eta}\right) = \sin x \cos \eta + \cos x \sin \eta = \cos x$$;
 * $$\cos \left({x + \eta}\right) = \cos x \cos \eta - \sin x \sin \eta = -\sin x$$.

Hence it follows, after some algebra, that:


 * $$\sin \left({x + 4 \eta}\right) = \sin x$$;
 * $$\cos \left({x + 4 \eta}\right) = \cos x$$.

Thus $$\sin$$ and $$\cos$$ are periodic on $$\mathbb{R}$$ with period $$4 \eta$$.

Pi
Given that the period of $$\sin$$ and $$\cos$$ as $$4 \eta$$, we define the real number $$\pi$$ (called "pi", pronounced "pie") as:

$$\pi \ \stackrel {\mathbf {def}} {=\!=} \ 2 \eta$$

See Pi.

Proof of Corollary
We have shown that:
 * $$\sin \left({x + \frac \pi 2}\right) = \cos x$$;
 * $$\cos \left({x + \frac \pi 2}\right) = -\sin x$$.

Thus:
 * $$\sin \left({x + \pi}\right) = \cos \left({x + \frac \pi 2}\right) = -\sin x$$;
 * $$\cos \left({x + \pi}\right) = -\sin \left({x + \frac \pi 2}\right) = -\cos x$$.

From the discussion and definition of $$\pi$$, it follows directly that $$\forall x \in \left[{-\frac \pi 2 \,. \, . \, \frac \pi 2}\right]: \cos x \ge 0$$.

Hence $$\forall x \in \left[{\frac \pi 2 \,. \, . \, \frac {3 \pi} 2}\right]: \cos x \le 0$$.

The result for $$\sin x$$ follows similarly, or we can use $$\sin \left({x + \frac \pi 2}\right) = \cos x$$.

The positions of the zeroes of both $$\sin x$$ and $$\cos x$$ follow directly from the above.

Note
Given that we have defined sine and cosine in terms of a power series, it is a plausible proposition to define $$\pi$$ using the same language.

$$\pi$$ is, of course, the famous irrational constant $$3.14159 \ldots$$.