# Shape of Secant Function

Jump to navigation
Jump to search

## Theorem

The nature of the secant function on the set of real numbers $\R$ is as follows:

- $(1): \quad \sec x$ is continuous and strictly increasing on the intervals $\hointr 0 {\dfrac \pi 2}$ and $\hointl {\dfrac \pi 2} \pi$

- $(2): \quad \sec x$ is continuous and strictly decreasing on the intervals $\hointr {-\pi} {-\dfrac \pi 2}$ and $\hointl {-\dfrac \pi 2} 0$

- $(3): \quad \sec x \to + \infty$ as $x \to -\dfrac \pi 2^+$
- $(4): \quad \sec x \to + \infty$ as $x \to \dfrac \pi 2^-$
- $(5): \quad \sec x \to - \infty$ as $x \to \dfrac \pi 2^+$
- $(6): \quad \sec x \to - \infty$ as $x \to \dfrac {3 \pi} 2^-$

## Proof

From Derivative of Secant Function:

- $\map {D_x} {\sec x} = \dfrac {\sin x} {\cos^2 x}$

From Sine and Cosine are Periodic on Reals: Corollary:

- $\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos x \ne 0$

Thus, from Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:

- $\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos^2 x > 0$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Might need to find a less abstract-algebraic version of the above resultYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

From Sine and Cosine are Periodic on Reals: Corollary:

- $\sin x > 0$ on the open interval $\openint 0 \pi$

It follows that:

- $\forall x \in \openint 0 \pi \setminus \set {\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} > 0$

From Sine and Cosine are Periodic on Reals: Corollary::

- $\sin x < 0$ on the open interval $\openint {-\pi} 0$

It follows that:

- $\forall x \in \openint {-\pi} 0 \setminus \set {-\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} < 0$

Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function and Differentiable Function is Continuous.

From Zeroes of Sine and Cosine::

- $\cos - \dfrac \pi 2 = \cos \dfrac \pi 2 = \cos \dfrac {3 \pi} 2 = 0$

From Sine and Cosine are Periodic on Reals: Corollary:

- $\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$

From the same source:

- $\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$

Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem.

### Graph of Secant Function

$\blacksquare$

## Also see

- Shape of Sine Function
- Shape of Cosine Function
- Shape of Tangent Function
- Shape of Cotangent 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