Shape of Secant Function

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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:

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



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

Secant.png


$\blacksquare$


Also see


Sources