Unit 6 · Chapter 6.2

6.2Graphing Tangent, Cotangent, Secant, and Cosecant

Graph tan, cot, sec, and csc functions. Identify vertical asymptotes, period (π for tan/cot, 2π for sec/csc), and apply transformations including shifts and reflections.

Tangent, secant, cosecant, and cotangent graphs have vertical asymptotes that correspond to undefined values. Recognizing these graphs and their asymptotes is essential for calculus, where these functions appear in derivatives and integrals.

Essential Question

How do the graphs of tangent, cotangent, secant, and cosecant differ from sine and cosine, and where do their asymptotes occur?

Lesson Overview

While sine and cosine are smooth, continuous waves, the other four trig functions have vertical asymptotes wherever their denominators equal zero. Understanding these graphs requires knowing the parent function, the period, and the asymptote locations.

Tangent & Cotangent

Period = π. tan(x) has asymptotes at x = π/2 + nπ; cot(x) has asymptotes at x = nπ.

Secant & Cosecant

Period = 2π. sec(x) = 1/cos(x); csc(x) = 1/sin(x). Their U-shaped branches touch the parent curve at its peaks and troughs.

Asymptotes

Occur wherever the denominator (sin or cos) equals zero. The graph approaches ±∞ on both sides of each asymptote.

Transformations

y = A·tan(Bx − C) + D: amplitude-like stretch |A|, period π/|B|, phase shift C/B, vertical shift D. Asymptotes shift accordingly.

Graph of y = tan(x)

y = tan(x)−π−π/2π/2π1−1xyasymptoteasymptote

Two branches shown over [−π, π]. Vertical asymptotes (red dashed) at x = −π/2 and x = π/2. Period = π.

Graph of y = cot(x)

y = cot(x)0ππ/23π/21−1x

Two branches over [0, 2π]. Asymptotes at x = 0, π, 2π. cot(x) decreases on each interval — opposite direction from tan(x).

Graph of y = sec(x)

y = sec(x) with parent y = cos(x) dashed−π−π/2π/2π3π/21−1xy

Pink U-shaped branches. Green dashed = parent cosine. Asymptotes at x = ±π/2, 3π/2. Branches touch cosine at its peaks (y = 1) and troughs (y = −1).

Graph of y = csc(x)

y = csc(x) with parent y = sin(x) dashed0π/2π3π/21−1x

Purple U-shaped branches. Blue dashed = parent sine. Asymptotes at x = 0, π, 2π. Branches touch sine at its peaks and troughs.

Asymptote Rules — Quick Reference

Asymptote Rules — Quick ReferenceFunctionPeriodVertical Asymptotes (n ∈ ℤ)tan(x)πx = π/2 + nπcot(x)πx = nπsec(x)x = π/2 + nπcsc(x)x = nπ

Key Vocabulary

Vertical asymptote

A vertical line x = a that the graph approaches but never crosses. Occurs where the function is undefined (denominator = 0).

Example: tan(x) has a vertical asymptote at x = π/2

Tangent function

tan(x) = sin(x)/cos(x). Period π, undefined at x = π/2 + nπ. Passes through (0, 0) and increases on each branch.

Example: tan(π/4) = 1

Cotangent function

cot(x) = cos(x)/sin(x). Period π, undefined at x = nπ. Decreases on each branch.

Example: cot(π/4) = 1

Secant function

sec(x) = 1/cos(x). Period 2π, undefined at x = π/2 + nπ. U-shaped branches above y = 1 and below y = −1.

Example: sec(0) = 1, sec(π) = −1

Cosecant function

csc(x) = 1/sin(x). Period 2π, undefined at x = nπ. U-shaped branches above y = 1 and below y = −1.

Example: csc(π/2) = 1

Reciprocal identity

sec = 1/cos, csc = 1/sin, cot = 1/tan. To graph sec or csc, first sketch the parent (cos or sin), then take reciprocals.

Worked Examples

Example 1

Find the vertical asymptotes of y = tan(x) and sketch one period.

tan(x) = sin(x)/cos(x), so asymptotes occur where cos(x) = 0.

cos(x) = 0 at x = π/2 + nπ for any integer n.

One period: from x = −π/2 to x = π/2 (width = π).

Key points in one period: (−π/4, −1), (0, 0), (π/4, 1).

Sketch: curve rises from −∞ near x = −π/2 through (0,0) to +∞ near x = π/2.

Answer:Asymptotes at x = π/2 + nπ; period = π; passes through (0, 0)
Example 2

Graph y = 2tan(x). Identify the period, asymptotes, and key points.

The coefficient 2 is a vertical stretch — it does NOT change the period.

Period = π (same as tan(x)).

Asymptotes: x = π/2 + nπ (unchanged by vertical stretch).

Key points: (−π/4, −2), (0, 0), (π/4, 2).

The graph looks like tan(x) but twice as steep.

Answer:Period = π; asymptotes at x = π/2 + nπ; key points (±π/4, ±2)
Example 3

Graph y = sec(x). Relate it to cosine, and identify asymptotes and turning points.

sec(x) = 1/cos(x), so first sketch y = cos(x) as a dashed guide.

Asymptotes occur where cos(x) = 0: x = π/2 + nπ.

Where cos(x) = 1 (at x = 0, 2π, …), sec(x) = 1 — these are the minimum turning points of the upper branches.

Where cos(x) = −1 (at x = π, 3π, …), sec(x) = −1 — these are the maximum turning points of the lower branches.

Branches open upward above y = 1 and downward below y = −1.

Answer:Asymptotes at x = π/2 + nπ; turning points at (2nπ, 1) and ((2n+1)π, −1)
Example 4

Graph y = csc(2x). Find the period, asymptotes, and key points.

csc(2x) = 1/sin(2x), so first find the period of sin(2x).

Period of sin(2x) = 2π/2 = π.

Asymptotes: sin(2x) = 0 → 2x = nπ → x = nπ/2.

Turning points: sin(2x) = 1 at 2x = π/2 → x = π/4, so csc(2x) = 1 at x = π/4.

sin(2x) = −1 at x = 3π/4, so csc(2x) = −1 at x = 3π/4.

Answer:Period = π; asymptotes at x = nπ/2; turning points at (π/4, 1) and (3π/4, −1)
Example 5

Graph y = −cot(x − π/4). Apply the phase shift and reflection.

Start with cot(x): period π, asymptotes at x = nπ, passes through (π/2, 0).

Phase shift: replace x with x − π/4, so shift every feature right by π/4.

New asymptotes: x = nπ + π/4.

New zero crossing: x = π/2 + π/4 = 3π/4.

Reflection (−): flip the graph vertically — the branch now increases instead of decreasing.

Answer:Period = π; asymptotes at x = π/4 + nπ; zero at x = 3π/4; graph increases on each branch

Guided Practice

Guided Problem 1

Find all vertical asymptotes of y = cot(x) on the interval [0, 2π].

Hint: cot(x) = cos(x)/sin(x). Set sin(x) = 0 and solve on [0, 2π].

Guided Problem 2

State the period and asymptotes of y = tan(3x).

Hint: The period of tan(Bx) is π/|B|. Asymptotes occur where 3x = π/2 + nπ.

Guided Problem 3

Describe how the graph of y = sec(x) relates to y = cos(x).

Hint: sec(x) = 1/cos(x). Where does cos(x) equal 1, −1, or 0? What happens to sec(x) at each of those points?

Guided Problem 4

Find the period and asymptotes of y = csc(x/2).

Hint: Period of csc(Bx) = 2π/|B|. Asymptotes where sin(x/2) = 0, i.e., x/2 = nπ.

Guided Problem 5

Graph y = tan(x − π/4) by applying a phase shift to y = tan(x).

Hint: Replace x with x − π/4. The asymptotes shift from x = π/2 + nπ to x = π/2 + π/4 + nπ = 3π/4 + nπ.

Quick Check

Interactive Practice — 5 Questions

1

What is the period of y = tan(x)?

2

Where are the vertical asymptotes of y = csc(x)?

3

Which function has U-shaped branches that touch y = cos(x) at its peaks and troughs?

4

What is the period of y = csc(2x)?

5

The graph of y = −tan(x) compared to y = tan(x) is:

Common Mistakes

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Common Mistakes

Thinking tan(x) has period 2π like sine and cosine.

tan(x) and cot(x) have period π — half the period of sin and cos.

Drawing sec(x) as a smooth wave without asymptotes.

sec(x) has vertical asymptotes wherever cos(x) = 0. The branches never cross those lines.

Forgetting that a vertical stretch does not change the period or asymptotes.

y = 3tan(x) has the same period (π) and same asymptotes as tan(x) — only the steepness changes.

Placing asymptotes of csc(x) at x = π/2 + nπ (confusing it with sec).

csc(x) = 1/sin(x), so asymptotes are at x = nπ. sec(x) = 1/cos(x), so asymptotes are at x = π/2 + nπ.

Math Tips

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Math Tips

📐

Always sketch the parent (sin or cos) first as a dashed guide before drawing sec or csc.

🔁

The period of tan(Bx) and cot(Bx) is π/|B|. The period of sec(Bx) and csc(Bx) is 2π/|B|.

📍

Asymptotes shift with phase shifts. If y = tan(x − C), asymptotes move from x = π/2 + nπ to x = π/2 + C + nπ.

↕️

A negative sign in front (e.g., −cot(x)) reflects the graph over the x-axis — branches flip direction.

🔗

Reciprocal identity shortcut: to evaluate sec(π/3), compute 1/cos(π/3) = 1/(1/2) = 2.