Shape of Cosine Function

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, $D_{xx} \left({\cos x}\right) \le 0$ and hence is concave on that interval.

The rest of the result follows similarly.

$\blacksquare$

Also see

• Basic Properties of Cosine Function
• Shape of Sine Function
• Shape of Tangent Function
• Shape of Cotangent Function

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.5 \ (1)$