Shape of Cosine Function

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Theorem

The cosine function is:

$(1): \quad$ strictly decreasing on the interval $\left[{0 \,.\,.\, \pi}\right]$
$(2): \quad$ strictly increasing on the interval $\left[{\pi \,.\,.\, 2 \pi}\right]$
$(3): \quad$ concave on the interval $\left[{-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right]$
$(4): \quad$ convex on the interval $\left[{\dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right]$


Proof

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

$\cos x \ge 0$ on the 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, $D_{xx} \left({\cos x}\right) \le 0$.

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


The rest of the result follows similarly.


Graph of Cosine Function

Cosine.png


$\blacksquare$


Also see


Sources