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 $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

Proof
From the discussion of Cosine is 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.

Thus from Derivative of Monotone Function, $\cos x$ is strictly decreasing on $\closedint 0 \pi$.

From Derivative of Sine Function it follows that:
 * $\map {D_{xx} } {\cos x} = -\cos x$

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.

Also see

 * Properties of Real Cosine Function


 * Shape of Sine Function
 * Shape of Tangent Function
 * Shape of Cotangent Function
 * Shape of Secant Function
 * Shape of Cosecant Function