# 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

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From the discussion of Real Cosine Function is Periodic, 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 $\map {\dfrac \d {\d x} } {\cos x} = -\sin x$ 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:

- $\map {\dfrac {\d^2} {\d x^2} } {\cos x} = -\cos x$

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

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.

$\blacksquare$

### Graph of Cosine Function

## 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): $\S 16.5 \ (1)$