Shape of Cosine Function

Theorem
The cosine function is:


 * decreasing on the interval $$\left[{0 \, . \, . \, \pi}\right]$$;
 * 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 $$\left[{\frac \pi 2 \, . \, . \, \frac {3\pi} 2}\right]$$.

Proof
From the discussion of Sine and Cosine are Periodic on Reals, we know that $$\cos x \ge 0$$ on the 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 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 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.