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, $D_{xx} \left({\cos x}\right) \le 0$.
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
$\blacksquare$
Also see
- Shape of Sine Function
- Shape of Tangent Function
- Shape of Cotangent Function
- Shape of Secant Function
- Shape of Cosecant Function
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.5 \ (1)$
