Graphs of Secant and Cosecant Functions

Solved Problems

Solution.

1. Using the secant definition, evenness of the cosine function and reduction identity, we get
   \[\sec \left( { - \frac{{3\pi }}{4}} \right) = \frac{1}{{\cos \left( { - \frac{{3\pi }}{4}} \right)}} = \frac{1}{{\cos \frac{{3\pi }}{4}}} = \frac{1}{{\cos \left( {\pi - \frac{\pi }{4}} \right)}} = - \frac{1}{{\cos \frac{\pi }{4}}} = - \frac{1}{{\frac{{\sqrt 2 }}{2}}} = - \frac{2}{{\sqrt 2 }} = - \sqrt 2 .\]
2. The cosecant function is \(2\pi-\)periodic and odd. Therefore
   \[\csc \frac{{11\pi }}{3} = \csc \left( {\frac{{12\pi }}{3} - \frac{\pi }{3}} \right) = \csc \left( {4\pi - \frac{\pi }{3}} \right) = \csc \left( { - \frac{\pi }{3}} \right) = - \csc \frac{\pi }{3} = - \frac{1}{{\sin \frac{\pi }{3}}} = - \frac{1}{{\frac{{\sqrt 3 }}{2}}} = - \frac{2}{{\sqrt 3 }}.\]

Solution.

1. By definition,
   \[\sec \frac{\pi }{8} = \frac{1}{{\cos \frac{\pi }{8}}}.\]
   Apply the half-angle formula
   \[\cos \frac{\alpha }{2} = \pm \sqrt {\frac{{1 + \cos \alpha }}{2}} .\]
   Given that the angle \(\frac{\pi }{8}\) lies in the \(1\text{st}\) quadrant, we can write
   \[\cos \frac{\pi }{8} = \sqrt {\frac{{1 + \cos \frac{\pi }{4}}}{2}} = \sqrt {\frac{{1 + \frac{{\sqrt 2 }}{2}}}{2}} = \sqrt {\frac{{2 + \sqrt 2 }}{4}} = \frac{{\sqrt {2 + \sqrt 2 } }}{2}.\]
   By flipping the fraction, we obtain:
   \[\sec \frac{\pi }{8} = \frac{2}{{\sqrt {2 + \sqrt 2 } }}.\]
2. The cosecant function is a periodic and odd. Therefore
   \[\csc \left( { - \frac{{41\pi }}{6}} \right) = - \csc \frac{{41\pi }}{6} = - \csc \left( {\frac{{36\pi }}{6} + \frac{{5\pi }}{6}} \right) = - \csc \left( {6\pi + \frac{{5\pi }}{6}} \right) = - \csc \frac{{5\pi }}{6}.\]
   Using the definition of cosecant and reduction formula, we get
   \[\csc \left( { - \frac{{41\pi }}{6}} \right) = - \csc \frac{{5\pi }}{6} = - \frac{1}{{\sin \frac{{5\pi }}{6}}} = - \frac{1}{{\sin \left( {\pi - \frac{\pi }{6}} \right)}} = - \frac{1}{{\sin \frac{\pi }{6}}} = - \frac{1}{{\frac{1}{2}}} = - 2.\]

Solution.

The secant function is undefined at \(x = \frac{\pi }{2} + \pi n,\,n \in \mathbb{Z},\) and the cosecant function is undefined at \(x = \pi n,\,n \in \mathbb{Z}.\) These points are marked on the circle with blue circles and green squares, respectively.

We can combine both sets and represent the domain of the function in the following form:

Solution.

The function \(\sec 2x = \frac{1}{{\cos 2x}}\) is undefined at the following points

We depict these values on the circle with blue dots.

The function \(\csc 4x = \frac{1}{{\sin 4x}}\) is undefined at the points in which \(\sin 4x = 0,\) that is

These solutions are marked on the circle by green squares.

As it can be seen, the first set of points is a subset of the second set. Hence, the domain of the function is given by

Solution.

We take the basic secant function \({y_1} = \sec x\) and make the following transformations:

The function \({y_2} = \sec \frac{x}{2}\) is obtained from \({y_1} = \sec x\) by horizontal stretching by a factor of \(2.\) The function \({y_2} = \sec \frac{x}{2}\) has vertical asymptotes at the points \(\ldots - 3\pi , - \pi ,\pi ,3\pi , \ldots\) and period \(4\pi.\)

The following transformation \({y_2} \to {y_3}\) stretches the graphs of \({y_2} = \sec \frac{x}{2}\) vertically by multiplying the coordinate of each point by \(3.\)

Solution.

Consider the cosecant function \({y_1} = \csc x.\) The graph of \({y_2} = 2\csc x\) is a stretch along the \(y-\)axis by a factor of \(2.\)

To get the function \({y_3} = -2\csc x\) we reflect the graph of \({y_2}\) about the \(x-\)axis.

The final function \({y_4} = 2 - 2\csc x\) is obtained by shifting the graph of \({y_3} = -2\csc x\) \(2\) units up.