Graphs of Tangent and Cotangent Functions

Solved Problems

Solution.

We know that the domain of the tangent function \(y = \tan t\) is all real numbers except the points \(t = \frac{\pi }{2} + \pi n,\) \(n \in \mathbb{Z}\) in which the tangent function has a discontinuity. To determine these points we denote \(t = \frac{x}{2} - \frac{\pi }{6}\) and solve the equation for \(x:\)

These points should be excluded from the domain of the function. Hence,

Solution.

The domain of the given function is determined by the inequalities

Consider one period from \(0\) to \(\pi.\) Cotangent is decreasing over this interval and changes its sign at \(x = \frac{\pi }{2}.\) It is non-negative in the interval \(x \in \left( {0,\frac{\pi }{2}} \right].\) Given that the period of the cotangent function is \(\pi\) and using interval notation, we can write the domain of \(y = \sqrt {\cot x}\) in the form

Solution.

The tangent function \(\tan x\) is discontinuous at \(x = \frac{\pi }{2} + \pi n,\,n \in \mathbb{Z}.\) There are two such points in the interval \(\left[ {0,2\pi } \right):\) \(x = \frac{\pi }{2}\) and \(x = \frac{{3\pi }}{2}.\) These points are depicted on the unit circle by blue circles.

The cotangent function \(\cot t\) is discontinuous at \(t = \pi n,\,n \in \mathbb{Z}.\) By denoting \(t = 2x,\) solve the equation for \(x:\)

Hence, there are \(4\) points on the unit circle in which the function \(\cot 2x\) has a discontinuity: \(x = 0,\) \(x = \frac{\pi }{2},\) \(x = \pi ,\) and \(x = \frac{{3\pi }}{2}.\) These points are shown in the figure by green squares.

Both sets of points can be represented by one formula:

Solution.

First we determine the domain of the function. The function \(y = \tan 2x\) is not defined at the points

We mark these points with blue circles.

Similarly, the function \(y = \cot 2x\) is undefined at

These points are shown on the unit circle by green squares. Both families of points are described by the formula \(x = \frac{{\pi n}}{4},\,n \in \mathbb{Z}.\) Hence, the domain of the given function is

Using the definition of tangent and cotangent, we can simplify the function:

The graph of the function is the horizontal line \(y = 1\) with the punctured points at \(x = \frac{{\pi n}}{4},\,n \in \mathbb{Z}.\)

Solution.

We take the basic tangent function \({y_1} = \tan x\) and shrink its graph by a factor of \(3\) along the \(x-\)axis. This gives us the function \({y_2} = \tan 3x,\) which has the vertical asymptotes at the points

To get the final function \({y_3} = 2\tan 3x,\) we stretch the graph of \({y_2}\) vertically by a factor of \(2.\)

These transformations are shown in Figure \(6.\)

Solution.

Absolute values are never negative. Therefore, to graph the absolute value of cotangent, we need to take the portion of the graph with negative \(y-\)values and reflect it about the \(x-\)axis to the upper half-plane. The resulting graph has vertices at the points \(x = \frac{\pi }{2} + \pi n,\) \(n \in \mathbb{Z}.\)