Shape of Cosine Function

Theorem

The cosine function is:

$(1): \quad$ strictly decreasing on the interval $\closedint 0 \pi$

$(2): \quad$ strictly increasing on the interval $\closedint \pi {2 \pi}$

$(3): \quad$ concave on the interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$

$(4): \quad$ convex on the interval $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

From the discussion of Sine and Cosine are Periodic on Reals, we know that:

$\cos x \ge 0$ on the closed interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$

and:

$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$

From the same discussion, we have that:

$\map \sin {x + \dfrac \pi 2} = \cos x$

So immediately we have that $\sin x \ge 0$ on the closed interval $\closedint 0 \pi$, $\sin x > 0$ on the open interval $\openint 0 \pi$.

But $\map {D_x} {\cos x} = -\sin x$ from Derivative of Cosine Function.

$\map {D_{xx} } {\cos x} = -\cos x$

On $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ where $\cos x \ge 0$, therefore, $\map {D_{xx} } {\cos x} \le 0$.

From Second Derivative of Concave Real Function is Non-Positive it follows that $\cos x$ is concave on $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.

The rest of the result follows similarly.

$\blacksquare$