Sine and Cosine are Periodic on Reals

Theorem

The sine and cosine functions are periodic on the set of real numbers $\R$:

• $\cos \left({x + 2 \pi}\right) = \cos x$
• $\sin \left({x + 2 \pi}\right) = \sin x$

Corollary

• $\cos \left({x + \pi}\right) = - \cos x$
• $\sin \left({x + \pi}\right) = - \sin x$

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

• $\sin x$ 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

$(1): \quad \forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi \implies \cos x = 0$

$(2): \quad \forall n \in \Z: x = n \pi \implies \sin x = 0$

Proof

From Cosine of Zero is One we have that $\cos 0 = 1$.

From Cosine Function is Even we have that $\cos x = \cos \left({-x}\right)$.

As the Cosine Function 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 $\R$, 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 $\eta \in \R$ 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 $\R$ with period $4 \eta$.

$\blacksquare$

Pi

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

$\pi := 2 \eta$

See Pi.

Proof of Corollary

We have shown that:

• $\sin \left({x + \dfrac \pi 2}\right) = \cos x$
• $\cos \left({x + \dfrac \pi 2}\right) = -\sin x$

Thus:

• $\sin \left({x + \pi}\right) = \cos \left({x + \dfrac \pi 2}\right) = -\sin x$
• $\cos \left({x + \pi}\right) = -\sin \left({x + \dfrac \pi 2}\right) = -\cos x$

From the discussion and definition of $\pi$, it follows directly that:

$\displaystyle \forall x \in \left[{-\frac \pi 2 .. \frac \pi 2}\right]: \cos x \ge 0$

Hence:

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

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

$\blacksquare$

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 $3.14159 \ldots$.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.4$