Shape of Cosine Function

Theorem
The cosine function is:


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

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 $$\left[{-\frac \pi 2 \,. \, . \, \frac \pi 2}\right]$$, and $$\cos x > 0$$ on the open interval $$\left({-\frac \pi 2 \, . \, . \, \frac \pi 2}\right)$$.

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

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 $$\left[{-\frac \pi 2 \,. \, . \, \frac \pi 2}\right]$$ 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 $$\forall n \in \mathbb{Z}: \cos \left({x + 2 n \pi}\right) = \cos x$$.

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