Shape of Cosine Function

Theorem
The cosine function is:


 * continuous on the whole of $\R$
 * strictly decreasing on the interval $\left[{0 \, . \, . \, \pi}\right]$
 * strictly increasing on the interval $\left[{\pi \, . \, . \, 2 \pi}\right]$
 * concave on the interval $\left[{-\dfrac \pi 2 \, . \, . \, \dfrac \pi 2}\right]$
 * convex on the interval $\left[{\dfrac \pi 2 \, . \, . \, \dfrac {3 \pi} 2}\right]$
 * $\forall n \in \Z: \cos 2 n \pi = 1$
 * $\forall n \in \Z: \cos \left({2 n + 1}\right) \pi = -1$

Proof
The fact of the continuity of $\cos x$ is established in the discussion of Derivative of Cosine Function.

From the discussion of Sine and Cosine are Periodic on Reals, we know that:
 * $\cos x \ge 0$ on the closed interval $\left[{-\dfrac \pi 2 \, . \, . \, \dfrac \pi 2}\right]$, and:
 * $\cos x > 0$ on the open interval $\left({-\dfrac \pi 2 \, . \, . \, \dfrac \pi 2}\right)$.

From the same discussion, we have that $\sin \left({x + \dfrac \pi 2}\right) = \cos x$.

So immediately we have that $\sin x \ge 0$ on the closed interval $\left[{0 \,. \, . \, \pi}\right]$, $\sin x > 0$ on the open interval $\left({0 \, . \, . \, \pi}\right)$.

But $D_x \left({\cos x}\right) = - \sin x$ from Derivative of Cosine Function.

Thus from Derivative of Monotone Function, $\cos x$ is strictly decreasing on $\left[{0 \,. \, . \, \pi}\right]$.

From Derivative of Sine Function it follows that $D_{xx} \left({\cos x}\right) = - \cos x$.

On $\left[{-\dfrac \pi 2 \,. \, . \, \dfrac \pi 2}\right]$ where $\cos x \ge 0$, therefore, $D_{xx} \left({\cos x}\right) \le 0$ and hence is concave on that interval.

The rest of the result follows similarly.


 * $\cos 2 n \pi = 1$:

From Basic Properties of Cosine Function we have that $\cos 0 = 1$, which takes care of $n = 0$.

From Sine and Cosine are Periodic on Reals, we have that $\cos \left({x + 2 \pi}\right) = \cos x$, and thus $\forall n \in \Z: \cos \left({x + 2 n \pi}\right) = \cos x$.

Hence $\cos 2 n \pi = 1$


 * $\cos \left({2 n + 1}\right) \pi = -1$:

Follows directly from the above and, from Sine and Cosine are Periodic on Reals, that $\cos \left({x + \pi}\right) = -\cos x$.

Also see

 * Nature of Sine Function