# Graphs of Tangent and Cotangent Functions

## Solved Problems

### Example 1.

Find the domain of the function $y = \tan \left( {\frac{x}{2} - \frac{\pi }{6}} \right).$

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 $$x:$$

$\frac{x}{2} - \frac{\pi }{6} = \frac{\pi }{2} + \pi n, \Rightarrow \frac{x}{2} = \frac{\pi }{2} + \frac{\pi }{6} + \pi n, \Rightarrow \frac{x}{2} = \frac{{2\pi }}{3} + \pi n, \Rightarrow x = \frac{{4\pi }}{3} + 2\pi n,\,n \in \mathbb{Z}.$

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

$\text{dom}\left( y \right) = \left\{ {x \in \mathbb{R} \left|\, x \ne \frac{{4\pi }}{3} + 2\pi n\right., \,n \in \mathbb{Z}} \right\}.$

### Example 2.

Find the domain of the function $y = \sqrt {\cot x}.$

Solution.

The domain of the given function is determined by the inequalities

$\left\{ {\begin{array}{*{20}{l}} {\cot x \ge 0}\\ {x \ne \pi n,n \in \mathbb{Z}} \end{array}} \right..$

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

$x \in \left( {\pi n,\frac{\pi }{2} + \pi n} \right],\,n \in \mathbb{Z}.$

### Example 3.

Find the points of discontinuity of the function $y = \tan x + \cot 2x.$

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 $$x:$$

$2x = \pi n, \Rightarrow x = \frac{{\pi n}}{2},\,n \in \mathbb{Z}.$

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:

$x = \frac{{\pi n}}{2},\,n \in \mathbb{Z}.$

### Example 4.

Sketch a graph of the function $y = \tan 2x \cot 2x.$

Solution.

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

$2x = \frac{\pi }{2} + \pi n, \Rightarrow x = \frac{\pi }{4} + \frac{{\pi n}}{2},\,n \in \mathbb{Z}.$

We mark these points with blue circles.

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

$2x = \pi n, \Rightarrow x = \frac{{\pi n}}{2},\,n \in \mathbb{Z}.$

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

$\text{dom}\left( y \right) = \left\{ {x \in \mathbb{R} \left| \,x \ne \frac{{\pi n}}{4},\right. n \in \mathbb{Z}} \right\}.$

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

$y = \tan 2x\cot 2x = \frac{{\sin 2x}}{{\cos 2x}} \cdot \frac{{\cos 2x}}{{\sin 2x}} = \frac{{\cancel{{\sin 2x}}\cancel{{\cos 2x}}}}{{\cancel{{\cos 2x}}\cancel{{\sin 2x}}}} = 1.$

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

### Example 5.

Sketch a graph of the function $y = 2\tan 3x.$

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

$3x = \frac{\pi }{2} + \pi n, \Rightarrow x = \frac{\pi }{6} + \frac{{\pi n}}{3},\,n \in \mathbb{Z}.$

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 $$6.$$

### Example 6.

Sketch a graph of the function $y = \left| {\cot x} \right|.$

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 $$x = \frac{\pi }{2} + \pi n,$$ $$n \in \mathbb{Z}.$$
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