Lesson 8.2 : Graphs of the Other Trigonometric Functions

New Concept

This lesson moves beyond sine and cosine to explore the graphs of tangent, cotangent, secant, and cosecant. You'll learn to analyze their unique properties, like periods and vertical asymptotes, and sketch transformations of each function.

What’s next

Next, you'll walk through interactive examples for graphing each function. Then, you'll master these skills with a series of practice cards and challenge problems.

Property

The stretching factor is $|A|$.
The period is $P = \frac{\pi}{|B|}$.
The domain is all real numbers $x$, where $x \neq \frac{\pi}{2|B|} + \frac{\pi}{|B|}k$ such that $k$ is an integer.
The range is $(-\infty, \infty)$.
The asymptotes occur at $x = \frac{\pi}{2|B|} + \frac{\pi}{|B|}k$, where $k$ is an integer.
$y = A \operatorname{tan}(Bx)$ is an odd function.

Examples

• For the function $y = 2 \operatorname{tan}(\pi x)$, the stretching factor is $2$ and the period is $\frac{\pi}{\pi} = 1$. The asymptotes are at $x = \frac{1}{2} + k$ for any integer $k$.

• For $y = \operatorname{tan}(2x)$, the stretching factor is $1$ and the period is $\frac{\pi}{2}$. The graph completes a full cycle between the asymptotes at $x = -\frac{\pi}{4}$ and $x = \frac{\pi}{4}$.

Property

The stretching factor is $|A|$.
The period is $\frac{\pi}{|B|}$.
The domain is $x \neq \frac{C}{B} + \frac{\pi}{2|B|}k$, where $k$ is an integer.
The range is $(-\infty, \infty)$.
The vertical asymptotes occur at $x = \frac{C}{B} + \frac{\pi}{2|B|}k$, where $k$ is an odd integer.
There is no amplitude.
To graph one period, identify the stretching factor $|A|$, period $P = \frac{\pi}{|B|}$, and phase shift $\frac{C}{B}$. Then, draw the graph of $y=A\operatorname{tan}(Bx)$ shifted right by $\frac{C}{B}$ and up by $D$.

Examples

• The graph of $y = \operatorname{tan}(x - \frac{\pi}{4}) + 2$ is the basic tangent graph shifted right by $\frac{\pi}{4}$ and up by $2$. The center point of a cycle is at $(\frac{\pi}{4}, 2)$.

• For $y = 3 \operatorname{tan}(2x - \pi)$, the period is $\frac{\pi}{2}$ and the phase shift is $\frac{\pi}{2}$. An asymptote that is at $x = \frac{\pi}{4}$ for $y=3\operatorname{tan}(2x)$ is now at $x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$.

Property

The stretching factor is $|A|$.
The period is $\frac{2\pi}{|B|}$.
The domain is $x \neq \frac{\pi}{2|B|}k$, where $k$ is an odd integer.
The range is $(-\infty, -|A|] \cup [|A|, \infty)$.
The vertical asymptotes occur at $x = \frac{\pi}{2|B|}k$, where $k$ is an odd integer.
There is no amplitude.
$y = A \operatorname{sec}(Bx)$ is an even function because cosine is an even function.

Examples

• For $y = 3 \operatorname{sec}(x)$, the period is $2\pi$. The graph has local minimums at $y=3$ and local maximums at $y=-3$. The range is $(-\infty, -3] \cup [3, \infty)$.

• The function $y = \operatorname{sec}(2x)$ has a period of $\frac{2\pi}{2} = \pi$. Asymptotes occur where $\operatorname{cos}(2x)=0$, such as at $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$.

Property

The stretching factor is $|A|$.
The period is $\frac{2\pi}{|B|}$.
The domain is $x \neq \frac{\pi}{|B|}k$, where $k$ is an integer.
The range is $(-\infty, -|A|] \cup [|A|, \infty)$.
The asymptotes occur at $x = \frac{\pi}{|B|}k$, where $k$ is an integer.
There is no amplitude.
$y = A \operatorname{csc}(Bx)$ is an odd function because sine is an odd function.

Examples

• In $y = 2 \operatorname{csc}(x)$, the stretching factor is $2$ and the period is $2\pi$. The range is $(-\infty, -2] \cup [2, \infty)$, with asymptotes at $x=k\pi$ for any integer $k$.

• For $y = \operatorname{csc}(\pi x)$, the period is $\frac{2\pi}{\pi} = 2$. The asymptotes occur at integer values of $x$, such as $x=0, 1, 2, ...$.

Property

For $y = A \operatorname{sec}(Bx - C) + D$ or $y = A \operatorname{csc}(Bx - C) + D$:
The stretching factor is $|A|$.
The period is $\frac{2\pi}{|B|}$.
The phase shift is $\frac{C}{B}$.
The vertical shift is $D$.
The range for both functions is $(-\infty, -|A| + D] \cup [|A| + D, \infty)$.
Vertical asymptotes for secant occur at $x = \frac{C}{B} + \frac{\pi}{2|B|}k$ for odd integers $k$. Asymptotes for cosecant occur at $x = \frac{C}{B} + \frac{\pi}{|B|}k$ for integers $k$.

Examples

• The graph of $y = 2 \operatorname{sec}(x - \frac{\pi}{2}) + 1$ is shifted right by $\frac{\pi}{2}$ and up by $1$. The range is $(-\infty, -2+1] \cup [2+1, \infty)$, which is $(-\infty, -1] \cup [3, \infty)$.

• For $y = \operatorname{csc}(\pi x + \frac{\pi}{2}) - 2$, the period is $2$, phase shift is $-\frac{1}{2}$, and vertical shift is $-2$. The asymptote at $x=0$ for the base graph is now at $x = -\frac{1}{2}$.

Property

The stretching factor is $|A|$.
The period is $P = \frac{\pi}{|B|}$.
The domain is $x \neq \frac{k\pi}{|B|}$, where $k$ is an integer.
The range is $(-\infty, \infty)$.
The asymptotes occur at $x = \frac{k\pi}{|B|}$, where $k$ is an integer.
$y = A \operatorname{cot}(Bx)$ is an odd function.

Examples

• For $y = 2 \operatorname{cot}(x)$, the graph is stretched vertically by a factor of 2. It passes through $(\frac{\pi}{4}, 2)$ and $(\frac{3\pi}{4}, -2)$, with a period of $\pi$.

• The function $y = \operatorname{cot}(2x)$ has a period of $\frac{\pi}{2}$. Its asymptotes are at $x=0, \frac{\pi}{2}, \pi, ...$. The graph decreases from positive to negative infinity between each pair of asymptotes.

Property

The stretching factor is $|A|$.
The period is $\frac{\pi}{|B|}$.
The domain is $x \neq \frac{C}{B} + \frac{k\pi}{|B|}$, where $k$ is an integer.
The range is $(-\infty, \infty)$.
The vertical asymptotes occur at $x = \frac{C}{B} + \frac{k\pi}{|B|}$, where $k$ is an integer.
There is no amplitude.
$y = A \operatorname{cot}(Bx)$ is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively).

Examples

• The function $y = \operatorname{cot}(x + \frac{\pi}{4})$ is the basic cotangent graph shifted to the left by $\frac{\pi}{4}$. The asymptote at $x=0$ is now at $x = -\frac{\pi}{4}$.

• For $y = 3 \operatorname{cot}(\pi x) - 1$, the graph is stretched by a factor of 3, has a period of 1, and is shifted down by 1. The point that was at $(\frac{1}{2}, 0)$ is now at $(\frac{1}{2}, -1)$.