Precalculus by Richard Wright

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4-07 Graphs of Other Trigonometric Functions

Mr. Wright teaches the lesson.

Summary: In this section, you will:

Graph tangent, secant, cosecant, and cotangent
Graph a damped trigonometric function

SDA NAD Content Standards (2018): PC.4.1, PC.5.3

Sandhill cranes are large birds native to North America that can be almost 4 feet high when they stand. They migrate between the southern United States and southern Canada, although they have occasionally been spotted in Great Britain and China. Pretend you are standing in your yard as a sandhill crane flies over. Trigonometric functions can be used to calculate the distance between you and the crane. This lesson is about sketching graphs of the other trigonometric functions.

Graph the Tangent and Cotangent Functions

The process for graphing the other trigonometric functions is almost the same as sketching the graph of sine or cosine like in lesson 4-06. There are two main differences: the shape of the graph and the period of tangent and cotangent.

There is no amplitude for tangent and cotangent, but there is still a vertical stretch that takes the place of amplitude.

A graph of y = tan x is given in figure 2. Notice there are asymptotes for angles where tangent is undefined, such as $x = –\frac{π}{2}$ and $x = \frac{π}{2}$. Key points are $\left(–\frac{π}{4}, –1\right)$, $\left(0, 0\right)$, and $\left(\frac{π}{4}, 1\right)$. The shape of the tangent graph repeats every π instead of the 2π of sine and cosine, so the period of tangent is π.

A graph of y = cot x is given in figure 3. Notice there are asymptotes for angles where cotangent is undefined, such as x = 0 and x = π. Key points are $\left(\frac{π}{4}, 1\right)$, $\left(\frac{π}{2}, 0\right)$, and $\left(\frac{3π}{4}, –1\right)$. The shape of the cotangent graph repeats every π like tangent, so the period of cotangent is π.

Transformations of Tangent and Cotangent: y = a tan(bx – c) + d and y = a cot(bx – c) + d

Vertical Stretch = | a | (if a < 0, the graph is inverted.)
Period: $T = \frac{π}{b}$
Phase Shift: $PS = \frac{c}{b}$ (if c > 0, shifts right)
Midline: y = d

How to Sketch a Tangent or Cotangent Graph

Identify the vertical stretch, period, phase shift, and midline.
Draw the midline, y = d.
Use the vertical stretch to label the inflection point heights on the y-axis.
Use the period and phase shift to find and plot the 5 key points (two asymptotes, a zero, and two inflection points).
Draw the graph

Example 1: Sketch a Tangent Graph

Sketch the graph of $y = tan (π x - \frac{π}{2})$ .

Solution

Compare $y = tan (π x - \frac{π}{2})$ to y = a tan(bx – c) + d.

a = 1
b = π
c = $\frac{π}{2}$
d = 0

The midline is y = d, so y = 0. There is no vertical shift.

The vertical stretch is 1, so label the y-axis so the inflection point of the curve is 1 above the midline, 1, and the other inflection point is 1 below the midline, −1.

The period is

$$ T=\frac{π}{b} $$

$$ T=\frac{π}{π} $$

$$ T=1 $$

The phase shift is

$$ PS = \frac{c}{b} $$

$$ PS = \frac{\frac{π}{2}}{π} $$

$$ PS = \frac{1}{2} $$

The middle key point for tangent is at (0, 0), but it has shifted right $\frac{1}{2}$, so it is $\left(\frac{1}{2}, 0\right)$. The asymptotes are half a period on either side of the middle key point, x = 0 and x = 1. The left inflection point is halfway between the 1^st asymptote and the middle key point with height of the vertical stretch below the midline, $\left(\frac{1}{4}, -1\right)$. The other inflection point is halfway between the middle key point and right asymptote and the height of the vertical shift above the midline, $\left(\frac{3}{4}, 1\right)$.

Sketch the tangent curve through the key points and continue the shape so it repeats at either end.

Example 2: Sketch a Cotangent Graph

Sketch the graph of y = 2 cot(x) + 1.

Solution

Compare y = 2 cot(x) + 1 to y = a cot(bx – c) + d.

a = 2
b = 1
c = 0
d = 1

The midline is y = d, so y = 1. There is no vertical shift.

The vertical stretch is 2, so label the y-axis so the inflection point of the curve is 2 above the midline, 3, and the other inflection point is 2 below the midline, −1.

The period is

$$ T=\frac{π}{b} $$

$$ T=\frac{π}{1} $$

$$ T=π $$

The phase shift is

$$ PS = \frac{c}{b} $$

$$ PS = \frac{0}{1} $$

$$ PS = 0 $$

The left asymptote is at the origin, but shifted the phase shift, x = 0. The other asymptote is the period to the right, x = π. The middle key point for cotangent is half way between the asymptotes on the midline, $\left(\frac{π}{2}, 0\right)$. The left inflection point is halfway between the 1^st asymptote and the middle key point with height of the vertical stretch above the midline, $\left(\frac{π}{2}, 3\right)$. The other inflection point is halfway between the middle key point and right asymptote and the height of the vertical shift below the midline, $\left(\frac{3π}{2}, -1\right)$.

Sketch the cotangent curve through the key points and continue the shape so it repeats at either end.

Try It 1

Sketch a graph of $y = 2 \tan{\left(x - \frac{π}{2}\right)}$.

Answer

Graph of Secant and Cosecant

Secant and cosecant are reciprocals of cosine and sine respectively, so their graphs should be related. Secant is graphed in figure 6a and cosecant is in figure 7a. Notice the graphs are identical except for a phase shift of $\frac{π}{2}$. Figure 6b graphs both secant and cosine on the same graph. Notice that x-intercepts on the cosine graph correspond to asymptotes on the secant graph. This is because the functions are reciprocals. The reciprocal of 0 is $\frac{1}{0}$ which is undefined. Also notice, that when cosine is 1, secant is also 1. Again this is because of the reciprocal relationship; the reciprocal of 1 is 1. The same pattern exists between cosecant and sine as in figure 7b.

The period of both secant and cosecant is 2π like sine and cosine. There is no amplitude for secant and cosecant, but there is a vertical stretch that is used instead.

Figure 6a: y = sec x Figure 6b: y = sec x and y = cos x

Figure 7a: y = csc x Figure 7b: y = csc x and y = sin x

Transformations of Secant and Cosecant: y = a sec(bx – c) + d and y = a csc(bx – c) + d

Vertical stretch = | a | (if a < 0, the graph is inverted.)
Period: $T = \frac{2 π}{b}$
Phase Shift: $PS = \frac{c}{b}$ (if c > 0, shifts right)
Midline: y = d

How to Sketch a Secant or Cosecant Graph

Start by sketching basic shape of the reciprocal (sine or cosine) function.
Draw asymptotes at all the places the sine or cosine function crosses the midline.
Sketch the basic shape of the secant or cosecant graph.

Example 3: Sketch a Graph of Secant

Sketch a graph of $y = sec (π x + \frac{π}{2})$ .

Solution

Compare $y = sec (π x + \frac{π}{2})$ to y = a sec(bx – c) + d.

a = 1
b = π
c = $-\frac{π}{2}$
d = 0

Since, the desired function is secant, start by sketching the reciprocal function, cosine. Then, sketch the basic secant graph, the asymptotes are where the cosine graph crosses the x-axis.

The midline is y = d, so y = 0 which is the x-axis.

The amplitude is 1, so label the y-axis so the maximum of the curve is 1 above the midline, 1, and the minimum is 1 below the midline, −1.

The period is

$$ T = \frac{2π}{b} $$

$$ T = \frac{2π}{π} $$

$$ T = 2 $$

The phase shift is

$$ PS = \frac{c}{b} $$

$$ PS = \frac{-\frac{π}{2}}{π} $$

$$ PS = -\frac{1}{2} $$

The first key points for cosine is at (0, a), but it has shifted left $\frac{1}{2}$, so it is $\left(-\frac{1}{2}, 1\right)$. The fifth key point is one period to the right or at $\left(-\frac{1}{2} + 2, 1) = (\frac{3}{2}, 1). These are the maximums. The minimum is the middle key point and is halfway between the 1^st and 5^th key points but an amplitude below the midline, \(\left(\frac{1}{2}, -1\right)$. One zero is halfway between the 1^st and 3^rd key points on the midline, (0, 0). The other zero is halfway between the 3^rd and 5^th key points on the midline, (1, 0).

Sketch the sine curve through the key points and continue the shape so it repeats at either end.

Draw the vertical asymptotes everywhere the cosine graph crosses the midline, x-axis. The draw the secant shaped graph so that it touches the minimums and maximums of the cosine graph.

Figure 8: y = sec(πx + π/2) and y = cos(πx + π/2)

Example 4: Sketch a Graph of Cosecant

Sketch a graph of $y = \frac{1}{2} csc (x) - 1$ .

Solution

Compare $y = \frac{1}{2} csc (x) - 1$ to y = a sec(bx – c) + d.

a = ½
b = 1
c = 0
d = –1

Since, the desired function is cosecant, start by sketching the reciprocal function, sine. Then, sketch the basic cosecant graph, the asymptotes are where the sine graph crosses the x-axis.

The midline is y = d, so y = −1. Draw this line on the graph as a light dotted line.

The amplitude is ½, so label the y-axis so the maximum of the curve is ½ above the midline, −½, and the minimum is ½ below the midline, −3/2.

The period is

$$ T = \frac{2π}{b} $$

$$ T = \frac{2π}{1} $$

$$ T = 2π $$

The phase shift is

$$ PS = \frac{c}{b} $$

$$ PS = \frac{0}{1} $$

$$ PS = 0 $$

The first key points for sine is at (0, 0), but it has shifted down 1, so it is (0, −1). The fifth key point is one period to the right or at (2π, −1). The middle key point is halfway between the 1^st and 5^th key points, (π, −1). The maximum is halfway between the 1^st and 3^rd key points and the amplitude higher, $\left(\frac{π}{2}, -\frac{1}{2}\right)$. The minimum is halfway between the 3^rd and 5^th key points and the amplitude lower, $\left(\frac{3π}{2}, -\frac{3}{2}\right)$.

Sketch the sine curve through the key points and continue the shape so it repeats at either end.

Draw the vertical asymptotes everywhere the cosine graph crosses the midline, x-axis. The draw the secant shaped graph so that it touches the minimums and maximums of the cosine graph.

Figure 9: y = ½ csc(x) – 1 and y = ½ sin(x) – 1

Try It 2

Sketch a graph of $y = csc (\frac{π}{2} x)$ .

Answer

Damped Trigonometric Functions

Trigonometric functions can be modified, or damped, by multiplying it by another function. The graph of sine or cosine is then constrained between the damping function and its x-axis reflection.

Graph of a Damped Sine or Cosine Function

Identify the damping function and the trigonometric function.
Graph the damping function and its reflection over the x-axis.
Graph the sine or cosine function using the damping function as the amplitude.

Example 5: Graph a Damped Cosine Function

Sketch a graph of y = x cos x.

Solution

The damping function is y = x and the trigonometric function is y = cos x. Start by graphing y = x. Then reflect the graph over the x-axis. Then graph y = cos x, but the minimums and maximums are at the y = x and y = –x lines.

Example 6: Graph a Damped Sine Function

Sketch a graph of y = x² sin x.

Solution

The damping function is y = x². Start by graphing y = x² and its reflection over the x-axis, y = –x². Then graph y = sin x, but the minimums and maximums are at the y = x² and y = –x² curves.

Try It 3

Sketch a graph of y = e^x sin x.

Answer

Lesson Summary

y = a tan(bx – c) + d and y = a cot(bx – c) + d

Vertical Stretch = | a | (if a < 0, the graph is inverted.)
Period: $T = \frac{π}{b}$
Phase Shift: $PS = \frac{c}{b}$ (if c > 0, shifts right)
Midline: y = d

How to Sketch a Tangent or Cotangent Graph

Identify the vertical stretch, period, phase shift, and midline.
Draw the midline, y = d.
Use the vertical stretch to label the inflection point heights on the y-axis.
Use the period and phase shift to find and plot the 5 key points (two asymptotes, a zero, and two inflection points).
Draw the graph

y = a sec(bx – c) + d and y = a csc(bx – c) + d

Vertical stretch = | a | (if a < 0, the graph is inverted.)
Period: $T = \frac{2 π}{b}$
Phase Shift: $PS = \frac{c}{b}$ (if c > 0, shifts right)
Midline: y = d

How to Sketch a Secant or Cosecant Graph

Start by sketching basic shape of the reciprocal (sine or cosine) function.
Draw asymptotes at all the places the sine or cosine function crosses the midline.
Sketch the basic shape of the secant or cosecant graph.

Graph of a Damped Sine or Cosine Function

Identify the damping function and the trigonometric function.
Graph the damping function and its reflection over the x-axis.
Graph the sine or cosine function using the damping function as the amplitude.

Helpful videos about this lesson.

Practice Exercises

Explain how the graph of cosine can be used to graph secant.

Sketch two periods of the graph of the function. Identify the stretching factor, period, and horizontal shift.

n(x) = 2 sec(πx)
$p(x) = \tan\left(\frac{π}{2}x-\frac{π}{4}\right)$
$r(x) = \frac{1}{2} \cot(x)$
t(x) = −3 csc (2πx)
$f(x) = 2 \tan\left(x-\frac{π}{8}\right)$
y = sec(4x) + 2

Write an equation for the graph.

Use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to input csc x as $\frac{1}{sin x}$ .

f(x) = |sec(x)|
f(x) = cot(x)tan(x)

Graph the damped trigonometric function.

$y = \frac{1}{2} x \sin(x)$
f(x) = |x| cos (x)

Mixed Review

(4-06) Graph two full periods of f(x) = 2 sin(πx).
(4-06) Graph two full periods of $g (x) = cos (2 x - \frac{1}{2})$ .
(4-05) Evaluate $cot \frac{17 π}{6}$ using reference angles.
(4-03) Evaluate all six trigonometric functions of $\frac{π}{3}$ using special right triangles.
(4-02) Evaluate all six trigonometric functions of $\frac{5 π}{4}$ using the unit circle.

Answers

Since y = sec x is the reciprocal function of y = cos x, plot the reciprocal of the coordinates on the graph of y = cos x to obtain the y-coordinates of y = sec x. The x-intercepts of the graph y = cos x are the vertical asymptotes for the graph of y = sec x.
; Stretching factor = 2; Period = 2; Horizontal shift = 0
; Stretching factor = 1; Period = 2; Horizontal shift = \frac{1}{2}\) to the right.
; Stretching factor = $\frac{1}{2}$; Period = π; Horizontal shift = 0.
; Stretching factor = 3; Period = 1; Horizontal shift = 0.
; Stretching factor = 2; Period = π; Horizontal shift = $\frac{π}{8}$ to the right.
; Stretching factor = 1; Period = $\frac{π}{2}$; Horizontal shift = 0.
y = cot(πx)
$y = \csc\left(\frac{1}{2}x\right)$
y = sec(4x)
y = 2 csc(x)
$- \sqrt{3}$
$\begin{array}{l} sin \frac{π}{3} = \frac{\sqrt{3}}{2} & cos \frac{π}{3} = \frac{1}{2} & tan \frac{π}{3} = \sqrt{3} \\ csc \frac{π}{3} = \frac{2 \sqrt{3}}{3} & sec \frac{π}{3} = 2 & cot \frac{π}{3} = \frac{\sqrt{3}}{3} \end{array}$
$\begin{array}{l} sin \frac{5 π}{4} = - \frac{\sqrt{2}}{2} & cos \frac{5 π}{4} = - \frac{\sqrt{2}}{2} & tan \frac{5 π}{4} = 1 \\ csc \frac{5 π}{4} = - \sqrt{2} & sec \frac{5 π}{4} = - \sqrt{2} & cot \frac{5 π}{4} = 1 \end{array}$