Free preview

5.4 More Trigonometric Graphs

Focus on tangent period and tangent asymptote.

Before you solve

Ideas in this lesson

Explore the lesson brief and the worked example for Idea $1$ . Create a free account to continue with the remaining ideas and practice.

01Construct reciprocal base graphsWorked steps · five questions 02Transform tangent and cotangent branchesWorked steps · five questions 03Transform secant and cosecant branchesWorked steps · five questions 04Match formulas to actual reciprocal graphsWorked steps · five questions 05Analyze physical tangent and shadow modelsWorked steps · five questions 06Prove reciprocal and quotient periodsWorked steps · five questions 07Explain shift identities with domainsWorked steps · five questions

Lesson brief and rules

5.4 More Trigonometric Graphs

The tangent function has period $\pi$ ; vertical asymptotes occur where cosine is zero.

Worked example

1

01 / Use the sine/cosine quotient or reciprocal definition to construct

y=\tan x

(- \dfrac{\pi }{2},\dfrac{\pi }{2})

. State its real domain, least period, parity, full pole set, range, zeros or vertices, and both one-sided behaviors at the first nonnegative pole. Evaluate it at

\dfrac{\pi }{3}

where defined. Explain why no finite sinusoidal amplitude applies. Curve

1

\tan{\left(x \right)}

Show the requested calculations and reasoning. Written explanations and sketches require teacher review; choosing a summary does not verify that work. Complete the written work or drawing first. The selected answer checks the result only; your reasoning or drawing needs teacher/self-review.

02 / Use

\tan x=\dfrac{\sin x}{\cos x}

. Denominator

\ne0

, so exclude

x=\dfrac{\pi }{2}+n\pi

03 / The parity is odd from the sine/cosine parity rules. The least period is

\pi

. The zero set of tangent/cotangent or the alternating sign of consecutive reciprocal branches rules out half this period.

04 / The range is

\mathbb{R}

and zeros are

x=n\pi

05 / At the first nonnegative pole

\dfrac{\pi }{2}

, the left limit is positive infinity and the right limit is negative infinity. Check the denominator sign on each side; do not connect the branches.

06 / The value at

\dfrac{\pi }{3}

\sqrt{3}

07 / These functions are unbounded. A vertical scale factor is not a finite amplitude.

Worked example 2 · R01

Use the sine/cosine quotient or reciprocal definition to construct $y=\cot x$ on $(0,\pi )$ . State its real domain, least period, parity, full pole set, range, zeros or vertices, and both one-sided behaviors at the first nonnegative pole. Evaluate it at $\dfrac{\pi }{4}$ where defined. Explain why no finite sinusoidal amplitude applies. Curve $1$ : $\cot{\left(x \right)}$ .

Use $\cot x=\dfrac{\cos x}{\sin x}$ . Denominator $\ne0$ , so exclude $x=n\pi$ .
The parity is odd from the sine/cosine parity rules. The least period is $\pi$ . The zero set of tangent/cotangent or the alternating sign of consecutive reciprocal branches rules out half this period.
The range is $\mathbb{R}$ and zeros are $x=\dfrac\pi2+n\pi$ .
At the first nonnegative pole $0$ , the left limit is negative infinity and the right limit is positive infinity. Check the denominator sign on each side; do not connect the branches.
The value at $\dfrac{\pi }{4}$ is $1$ .
These functions are unbounded. A vertical scale factor is not a finite amplitude.
Written-work review:

Use the defining quotient or reciprocal to derive the entire domain and pole set.
Give least period, parity, range and every zero or vertex family.
Draw disconnected branches, justify both one-sided pole signs and evaluate the specified input.

Read the graphs

Construct reciprocal base graphs

Use the sine/cosine quotient or reciprocal definition to construct

y=\tan x

(- \dfrac{\pi }{2},\dfrac{\pi }{2})

. State its real domain, least period, parity, full pole set, range, zeros or vertices, and both one-sided behaviors at the first nonnegative pole. Evaluate it at

\dfrac{\pi }{3}

where defined. Explain why no finite sinusoidal amplitude applies. Curve

1

\tan{\left(x \right)}

Use $\tan x=\dfrac{\sin x}{\cos x}$ . Denominator $\ne0$ , so exclude $x=\dfrac{\pi }{2}+n\pi$ .
The parity is odd from the sine/cosine parity rules. The least period is $\pi$ . The zero set of tangent/cotangent or the alternating sign of consecutive reciprocal branches rules out half this period.
The range is $\mathbb{R}$ and zeros are $x=n\pi$ .
At the first nonnegative pole $\dfrac{\pi }{2}$ , the left limit is positive infinity and the right limit is negative infinity. Check the denominator sign on each side; do not connect the branches.
The value at $\dfrac{\pi }{3}$ is $\sqrt{3}$ .
These functions are unbounded. A vertical scale factor is not a finite amplitude.
Written-work review:

Use the defining quotient or reciprocal to derive the entire domain and pole set.
Give least period, parity, range and every zero or vertex family.
Draw disconnected branches, justify both one-sided pole signs and evaluate the specified input.

Answer: Least period $\pi$ ; parity odd; poles $x=\dfrac{\pi }{2}+n\pi$ , $n\in\mathbb{Z}$ .

Transform tangent and cotangent branches

For

y=2 \tan{\left(3 x - \dfrac{3 \pi }{7} \right)} + 1

, factor the input and find the least period and horizontal shift. Give the entire excluded pole set, a convenient cycle interval with every pole removed, signed branch direction or vertices, and range. Draw all branches in that interval with the marked key coordinates. Curve

1

1 - 2 \cot{\left(3 x + \dfrac{\pi }{14} \right)}

Normalize the frequency to $3$ using oddness for tangent, cotangent and cosecant, and evenness for secant. The equivalent formula is $y=1+2\tan{\left(3\left(x-\dfrac{\pi }{7}\right)\right)}$ .
The shift is $\dfrac{\pi }{7}$ . Dividing the base period $\pi$ by $3$ gives $T=\dfrac{\pi }{3}$ . Nonzero scale is essential.
Denominator $\cos(3 x - \dfrac{3 \pi }{7})\ne0$ , so every excluded input is $x=\dfrac{13 \pi }{42}+n(\dfrac{\pi }{3}),\ n\in\mathbb{Z}$ .
One cycle spans $[- \dfrac{\pi }{42},\dfrac{13 \pi }{42}]$ , with all listed poles removed. The poles in this window are $- \dfrac{\pi }{42}$ , $\dfrac{13 \pi }{42}$ .
Key coordinates: $\left( \dfrac{5 \pi }{84}, \ -1\right)$ , $\left( \dfrac{\pi }{7}, \ 1\right)$ , $\left( \dfrac{19 \pi }{84}, \ 3\right)$ . Apply the signed outside scale before the vertical translation; a negative scale reverses branch orientation.
The range is $\mathbb{R}$ . Each tangent branch is increasing when the normalized outside scale is positive and decreasing when it is negative; cotangent reverses those directions.
Here the normalized outside scale is $2$ . Therefore every branch is increasing; this conclusion includes the original input reflection.
Written-work review:

Factor and normalize the input, preserving signed tangent/cotangent orientation.
Compute the full pole lattice, least period and complete between-pole interval.
Calculate key points, range and branch direction and draw every branch without crossing a pole.

Answer: Least period $\dfrac{\pi }{3}$ ; pole set $x=\dfrac{13 \pi }{42}+n(\dfrac{\pi }{3})$ , $n\in\mathbb{Z}$ .

Transform secant and cosecant branches

For

y=2 \sec{\left(3 x - \dfrac{3 \pi }{7} \right)} + 1

1

\sin{\left(3 x + \dfrac{\pi }{14} \right)}

. Curve

2

2 \csc{\left(3 x + \dfrac{\pi }{14} \right)} + 1

Normalize the frequency to $3$ using oddness for tangent, cotangent and cosecant, and evenness for secant. The equivalent formula is $y=1+2\sec{\left(3\left(x-\dfrac{\pi }{7}\right)\right)}$ .
The shift is $\dfrac{\pi }{7}$ . Dividing the base period $2 \pi$ by $3$ gives $T=\dfrac{2 \pi }{3}$ . Nonzero scale is essential.
Denominator $\cos(3 x - \dfrac{3 \pi }{7})\ne0$ , so every excluded input is $x=\dfrac{13 \pi }{42}+n(\dfrac{\pi }{3}),\ n\in\mathbb{Z}$ .
One cycle spans $[- \dfrac{4 \pi }{21},\dfrac{10 \pi }{21}]$ , with all listed poles removed. The poles in this window are $- \dfrac{\pi }{42}$ , $\dfrac{13 \pi }{42}$ .
Key coordinates: $\left( - \dfrac{4 \pi }{21}, \ -1\right)$ , $\left( \dfrac{\pi }{7}, \ 3\right)$ , $\left( \dfrac{10 \pi }{21}, \ -1\right)$ . Apply the signed outside scale before the vertical translation; a negative scale reverses branch orientation.
The range is $(-\infty,-1]\cup[3,\infty)$ . Reciprocal guide curves locate poles at zero-denominator inputs and branch vertices where the guide has values $1$ or $-1$ . These are unbounded branches, not finite-amplitude waves.
The continuous first curve is the guide $g(x)=\sin{\left(3 x + \dfrac{\pi }{14} \right)}$ . The second is $y=1+\dfrac{2}{g(x)}$ . Guide zeros give the full pole set. Guide values $1$ and $-1$ give the displayed vertices $3$ and $-1$ . At a guide value $\dfrac{1}{2}$ , the reciprocal contribution is doubled and the output is $5$ . Apply reciprocation first, then the signed scale, then the vertical translation. The positive scale preserves the reciprocal branch signs relative to the midline.
Written-work review:

Factor the input and use a sine/cosine reciprocal guide.
Compute the full pole lattice, least period and all internal exclusions in the cycle.
Calculate exact vertices and range, showing both branch signs and any reflection or vertical translation.

Answer: Least period $\dfrac{2 \pi }{3}$ ; pole set $x=\dfrac{13 \pi }{42}+n(\dfrac{\pi }{3})$ , $n\in\mathbb{Z}$ .

Match formulas to actual reciprocal graphs

Match each of the six formulas to exactly one of the actual panels. For every match, justify the pole spacing, horizontal shift, branch direction and vertical scale or offset. Formulas:

1

y=- 2 \cot{\left(x + \dfrac{3 \pi }{10} \right)}

;

2

y=- 2 \tan{\left(x \right)}

;

3

y=2 \cot{\left(2 x \right)}

;

4

y=2 \sec{\left(x \right)}

;

5

y=\sec{\left(2 x \right)}

;

6

y=2 \csc{\left(x \right)} + 1

. Panel

A

. Curve

1

: unidentified curve.

Panel

B

. Curve

1

: unidentified curve.

Panel

C

. Curve

1

: unidentified curve.

Panel

D

. Curve

1

: unidentified curve.

Panel

E

. Curve

1

: unidentified curve.

Panel

F

. Curve

1

: unidentified curve.

Formula $1$ matches panel $A$ . In the explicit $\tan$ form $y=2\tan(x - \dfrac{\pi }{5})$ , the horizontal shift is $\dfrac{\pi }{5}$ , outside scale $2$ and vertical offset $0$ . Its complete poles are $x=\dfrac{7 \pi }{10}+n(\pi ),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\pi$ . Its branch direction is increasing. The sign of the outside scale and the tangent/cotangent base direction give this actual orientation.
Formula $2$ matches panel $B$ . In the explicit $\tan$ form $y=-2\tan(x)$ , the horizontal shift is $0$ , outside scale $-2$ and vertical offset $0$ . Its complete poles are $x=\dfrac{\pi }{2}+n(\pi ),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\pi$ . Its branch direction is decreasing. The sign of the outside scale and the tangent/cotangent base direction give this actual orientation.
Formula $3$ matches panel $C$ . In the explicit $\cot$ form $y=2\cot(2 x)$ , the horizontal shift is $0$ , outside scale $2$ and vertical offset $0$ . Its complete poles are $x=n(\dfrac{\pi }{2}),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\dfrac{\pi }{2}$ . Its branch direction is decreasing. The sign of the outside scale and the tangent/cotangent base direction give this actual orientation.
Formula $4$ matches panel $D$ . In the explicit $\sec$ form $y=2\sec(x)$ , the horizontal shift is $0$ , outside scale $2$ and vertical offset $0$ . Its complete poles are $x=\dfrac{\pi }{2}+n(\pi ),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\pi$ . Its branch direction is not one monotone direction across a U-shaped reciprocal branch; the two sides meet at a vertex. The reciprocal vertices have heights $-2$ and $2$ .
Formula $5$ matches panel $E$ . In the explicit $\sec$ form $y=\sec(2 x)$ , the horizontal shift is $0$ , outside scale $1$ and vertical offset $0$ . Its complete poles are $x=\dfrac{\pi }{4}+n(\dfrac{\pi }{2}),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\dfrac{\pi }{2}$ . Its branch direction is not one monotone direction across a U-shaped reciprocal branch; the two sides meet at a vertex. The reciprocal vertices have heights $-1$ and $1$ .
Formula $6$ matches panel $F$ . In the explicit $\csc$ form $y=1+2\csc(x)$ , the horizontal shift is $0$ , outside scale $2$ and vertical offset $1$ . Its complete poles are $x=n(\pi ),\ n\in\mathbb{Z}$ , so adjacent pole spacing is $\pi$ . Its branch direction is not one monotone direction across a U-shaped reciprocal branch; the two sides meet at a vertex. The reciprocal vertices have heights $-1$ and $3$ .
Thus Panels for formulas $1$ – $6$ : $A, B, C, D, E, F$ . All four feature types agree for every match; a shared intercept alone would not identify a graph.
Written-work review:

Match all six actual panels, not just one selected graph.
Justify every match with pole spacing, horizontal shift, branch shape/direction and scale or offset.
Explain why matching one intercept is insufficient.

Answer: Panels for formulas $1$ – $6$ : $A, B, C, D, E, F$ .

Analyze physical tangent and shadow models

A lighthouse is

2

kilometers from a straight shore. Its beam turns once every

4

minutes, and the shoreline distance from the nearest point is

D(t)=2 \tan{\left(\dfrac{\pi t}{2} \right)}

0\le t<1

. Evaluate the distance at

\dfrac{1}{4}

\dfrac{1}{2}

\dfrac{9}{10}

minutes to three decimals. Graph the complete physical interval, find when the distance first equals the offshore distance, and explain the one-sided boundary blow-up. If the sweep begins at

9

:00 a.m., convert the threshold time to a clock time, rounded to the nearest second, with an exact half-second rounded upward. Curve

1

D(t)=2 \tan{\left(\dfrac{\pi t}{2} \right)}

The angular speed is $\omega=\dfrac{2\pi }{4}$ radians per minute. The initial beam is perpendicular to the shore. The boundary beam becomes parallel to the shore, at $\omega t=\dfrac{\pi }{2}$ , so the physical boundary is $1$ minutes.

Time $($ minutes $)$	Distance $($ kilometers $)$
$\dfrac{1}{4}$	$0.828$
$\dfrac{1}{2}$	$2.000$
$\dfrac{9}{10}$	$12.628$

D(t)=2

becomes

\tan(\omega t)=1

. On the physical branch

\omega t=\dfrac{\pi }{4}

, giving

t=\dfrac{1}{2}

minutes, or 9:00:30 a.m..
As time approaches

1

from below, the tangent denominator approaches zero positively while the numerator approaches one. The modeled distance tends to positive infinity. The endpoint is excluded; this is a geometric idealization, not a finite shoreline endpoint.
Written-work review:

Convert every supplied time and compute the full value table with units.
Draw the full physical interval with the correct open boundaries and any absolute value.
Recover all threshold times, convert clock times and justify the appropriate one-sided positive blow-up.

Answer: Distances $0.828$ , $2.000$ , $12.628$ km; threshold $0.5000$ $\min$ ; blow-up boundary $1$ $\min$ .

Prove reciprocal and quotient periods

Let

f

be defined on all real inputs, with least positive period

p

. Define

r(x)=\dfrac{1}{f(x)}

exactly where

f(x)\ne0

. Prove that

r

has exactly the same periods as

f

, including invariance of its domain and recovery of the zero set. Then prove that a quotient of two

p

-periodic functions inherits

p

on its denominator-nonzero domain, but may have a smaller least period. Use

\dfrac{\sin(x)}{\cos(x)}

as a counterexample to equal least periods. Finally compare

\dfrac{2+\cos(x)}{2+\cos(x)}

with

\dfrac{\sin(x)}{\sin(x)}

: both have constant values, but do they have the same domain and least-period behavior? Also explain the unchanged pole exclusions in

0\tan(x)

. Curve

1

\dfrac{1}{\cos{\left(x \right)} + 2}

. Curve

2

\tan{\left(x \right)}

Let $D=\{x:f(x)\ne0\}$ . If $P$ is a period of $f$ , then $x\in D\iff x+P\in D$ and $r(x+P)=\dfrac{1}{f(x+P)}=\dfrac{1}{f(x)}=r(x)$ .
Conversely, a period of $r$ must preserve its domain $D$ . On $D$ , taking reciprocals gives $f(x+P)=f(x)$ . On its complement both values are zero, since domain invariance also preserves the complement. Thus every period of $r$ is a period of $f$ . Their sets of periods, and hence least periods, coincide. The all-real original domain matters.
Apply this to sine and cosine. Their reciprocals cosecant and secant have least period $2\pi$ , not merely a known period. The nonzero restricted domain has not erased the zero-set information because it is the complement in the full real line.
If $f(x+p)=f(x)$ and $g(x+p)=g(x)$ , then $g(x)\ne0\iff g(x+p)\ne0$ and $\dfrac{f(x+p)}{g(x+p)}=\dfrac{f(x)}{g(x)}$ . This proves a period, not necessarily the least one.
The sine and cosine numerator and denominator each have least period $2 \pi$ , but their tangent quotient has least period $\pi$ . Its zeros are spaced by that amount, proving minimality.
The positive-denominator quotient equals $1$ on all real inputs, so every positive number is a period and there is no least one. The sine-over-sine quotient equals $1$ only off $x=\dfrac{n\pi }{1}$ . Its excluded lattice forces least period $\pi$ . Constant values alone do not remove domain restrictions.
Likewise $0\tan(x)$ is zero only on the original tangent domain; the poles remain excluded unless an extension is explicitly defined. Its domain has least positive translation $\pi$ , unlike a constant defined on all real numbers.
Written-work review:

Prove reciprocal period inheritance in both directions using the nonzero domain and its zero complement.
Prove quotient period inheritance and exhibit a strictly smaller fundamental-period example.
Distinguish all-real constants, punctured constants and zero-scaled functions that retain poles.

Answer: Reciprocal base period $2 \pi$ ; tangent quotient least period $\pi$ ; all-real constant has no least positive period; punctured constant least period $\pi$ .

Explain shift identities with domains

Use actual shifted graphs and then algebra to establish both identities

\tan(x-\dfrac{\pi }{2})=-\cot(x)

and

\sec(x-\dfrac{\pi }{2})=\csc(x)

. Explain the horizontal translation before scaling back to the input axis, list the complete common domain, and justify the branch signs. Curve

1

- \cot{\left(x \right)}

. Curve

2

- \cot{\left(x \right)}

Curve

1

\csc{\left(x \right)}

. Curve

2

\csc{\left(x \right)}

Let $u=x$ . The shift by $\dfrac{\pi }{2}$ in the argument corresponds to $\dfrac{\pi }{2}$ in the horizontal coordinate, not $\dfrac{\pi }{2}$ unless the frequency is $1$ .
$\sin(u-\dfrac{\pi }{2})=-\cos u$ and $\cos(u-\dfrac{\pi }{2})=\sin u$ . Therefore $\tan(u-\dfrac{\pi }{2})=\dfrac{-\cos u}{\sin u}=-\cot u$ .
$\sec(u-\dfrac{\pi }{2})=\dfrac1{\cos(u-\dfrac{\pi }{2})}=\dfrac1{\sin u}=\csc u$ .
Both sides of both identities require $\sin u\ne0$ , giving $x\ne 0+\dfrac{n\pi }{1},\ n\in\mathbb{Z}$ . The coincident drawn branches have exactly these poles and signs. Coincidence on a finite plot suggests the identities; the algebra proves them everywhere on the common domain.
Written-work review:

Convert the argument shift into the correctly frequency-scaled horizontal shift.
Give both quotient/reciprocal algebraic identity proofs and the complete common domain.
Use actual coincident graph branches as illustrations, not substitutes for the global proof.

Answer: First multiplier $-1$ ; second multiplier $1$ ; common domain $x\ne 0+\dfrac{n\pi }{1},\ n\in\mathbb{Z}$ .

15 questions with a worked answer for each.Start practice ↗

Textbook reference: Stewart, Redlin & Watson, Precalculus: Mathematics for Calculus, 7th ed. (Cengage, 2016). The explanations and practice here are original, organized by textbook section.

Read this lesson preview

MathLab with Mr. Ahmed

5.4 More Trigonometric Graphs

Ideas in this lesson

5.4 More Trigonometric Graphs

Read the graphs

Construct reciprocal base graphs

Transform tangent and cotangent branches

Transform secant and cosecant branches

Match formulas to actual reciprocal graphs

Analyze physical tangent and shadow models

Prove reciprocal and quotient periods

Explain shift identities with domains