AP Precalculus 3.11 Secant, Cosecant & Cotangent — Notes

🔄

The Reciprocal Functions

sec, csc, cot are each defined as 1 divided by their corresponding trig function

Secant

sec x = 1 / cos x

Reciprocal of cosine.
Undefined when cos x = 0.

VA: x = π/2 + πk

Cosecant

csc x = 1 / sin x

Reciprocal of sine.
Undefined when sin x = 0.

VA: x = πk

Cotangent

cot x = 1 / tan x

= cos x / sin x

Reciprocal of tangent.
Undefined when sin x = 0.

VA: x = πk

Function	Definition	Undefined when	Vertical Asymptotes	Range
sec x	`1/cos x`	cos x = 0	`x = π/2 + πk`	(−∞,−1] ∪ [1,∞)
csc x	`1/sin x`	sin x = 0	`x = πk`	(−∞,−1] ∪ [1,∞)
cot x	`cos x / sin x`	sin x = 0	`x = πk`	(−∞, ∞)

⚠️

Range of sec and csc — never between −1 and 1

Since sec x = 1/cos x and |cos x| ≤ 1, dividing 1 by a number between −1 and 1 gives a result ≤ −1 or ≥ 1. The values between −1 and 1 are simply impossible for sec and csc.
cot x has range all reals (like tan x) because it is cos/sin with no such restriction.

📈

Graphs of sec, csc, and cot

Each graph "hugs" its corresponding sin/cos/tan curve — the reciprocal flips and creates U-shapes

y = sec x (blue) with cos x (light dashed) — asymptotes at x = π/2 + πk

y = csc x (teal) with sin x (light dashed) — asymptotes at x = πk

y = cot x (orange) — asymptotes at x = πk (same as csc), decreasing on each interval

🔑

How to find asymptotes for transformed functions

For sec(bx) or csc(bx), set the inner function's denominator = 0:
— sec: set cos(bx) = 0 → solve for x
— csc or cot: set sin(bx) = 0 → solve for x
Then add ±period to generate all asymptotes.

✏️

Examples 1–4 — Asymptotes and Range

Set denominator = 0, solve for x · scale range by the multiplier outside

📌 Example 1 — f(x) = 3sec(2x). Which is a vertical asymptote?

1

sec(2x) = 1/cos(2x). Set cos(2x) = 0.

2

cos(2x) = 0 → 2x = π/2 + πk → x = π/4 + πk/2. Simplest positive: x = π/4.

(A)

x = π

cos(2·π)=cos(2π)=1 ≠ 0 ✗

(B)

x = π/2

cos(2·π/2)=cos(π)=−1 ≠ 0 ✗

✓ (C)

x = π/4

cos(2·π/4)=cos(π/2)=0 ✓ → asymptote

(D)

x = 0

cos(0)=1 ≠ 0 ✗

📌 Example 2 — g(x) = 4 − 2csc(πx). Which is a vertical asymptote?

1

csc(πx) = 1/sin(πx). Set sin(πx) = 0.

2

sin(πx) = 0 → πx = πk → x = k (all integers). So x = 0, 1, 2, −1, … are all asymptotes.

(A)

x = π/2

sin(π·π/2)=sin(π²/2) ≠ 0 ✗

(B)

x = π

sin(π·π)=sin(π²) ≠ 0 ✗

(C)

x = 1/2

sin(π/2)=1 ≠ 0 ✗

✓ (D)

x = 1

sin(π·1)=sin(π)=0 ✓ → asymptote

📌 Example 3 — h(θ) = 3csc(θ/2). Which gives the range of h?

1

Range of csc(x) is (−∞, −1] ∪ [1, ∞).

2

Multiply by 3: values ≤ −1 become ≤ −3, values ≥ 1 become ≥ 3.

3

Range of 3csc(x): (−∞, −3] ∪ [3, ∞).

(A)

(−∞,−1] ∪ [1,∞)

That's the range of csc(x) before multiplying by 3 ✗

(B)

(−∞,−2] ∪ [2,∞)

Would need a multiplier of 2, not 3 ✗

✓ (C)

(−∞,−3] ∪ [3,∞)

3 × (−∞,−1] = (−∞,−3] and 3 × [1,∞) = [3,∞) ✓

(D)

[−3, 3]

Completely wrong — csc/sec NEVER produce values between −1 and 1 ✗

📌 Example 4 — k(x) = −5cot(2πx). Which is a vertical asymptote?

1

cot(2πx) = cos(2πx)/sin(2πx). Set sin(2πx) = 0.

2

sin(2πx) = 0 → 2πx = πk → x = k/2. So x = 0, ±1/2, ±1, ±3/2, … are all asymptotes.

(A)

x = 1/4

sin(2π·1/4)=sin(π/2)=1 ≠ 0 ✗

✓ (B)

x = 1/2

sin(2π·1/2)=sin(π)=0 ✓ → asymptote

(C)

x = π/2

sin(2π·π/2)=sin(π²)≠0 ✗

(D)

x = 2π

Not the simplest asymptote ✗

🧮

Solving Equations with sec, csc, cot — Examples 5–7

Isolate → convert to sin/cos/tan → solve as usual

🔑 Strategy: Reciprocal → Back to sin/cos/tan

1

Isolate the reciprocal trig function (sec, csc, or cot) on one side.

2

Convert: replace sec = 1/cos, csc = 1/sin, or cot = cos/sin. Flip both sides if needed.

3

Solve the resulting sin/cos/tan equation using the unit circle and domain restrictions.

📌 Example 5 — f(x)=4csc(x)+3 and g(x)=11. Find x-coordinates of intersections on [0, 2π).

1

Set f = g: 4csc(x) + 3 = 11 → 4csc(x) = 8 → csc(x) = 2.

2

csc(x) = 2 → 1/sin(x) = 2 → sin(x) = 1/2.

3

sin = 1/2 → x = π/6 (Q1) and x = 5π/6 (Q2).

x = π/6 and x = 5π/6

📌 Example 6 — h(x)=3−(1/2)sec(x) and k(x)=4. Find x-coordinates of intersections on [0, 2π).

1

Set h = k: 3 − (1/2)sec(x) = 4 → −(1/2)sec(x) = 1 → sec(x) = −2.

2

sec(x) = −2 → 1/cos(x) = −2 → cos(x) = −1/2.

3

cos = −1/2 → x = 2π/3 (Q2) and x = 4π/3 (Q3).

x = 2π/3 and x = 4π/3

📌 Example 7 — p(x) = 3 − 1.7cot(0.5x − 1) = 2. Find x in [0, 2π). 🖩 Calculator required.

1

Isolate: 3 − 1.7cot(0.5x − 1) = 2 → −1.7cot(0.5x − 1) = −1 → cot(0.5x − 1) = 1/1.7.

2

cot = 1/tan → tan(0.5x − 1) = 1.7.

3

0.5x − 1 = arctan(1.7) ≈ 1.0385 → 0.5x = 2.0385 → x ≈ 4.0781.

4

Check for second solution: add π to the arctan result (tan has period π): 0.5x − 1 = 1.0385 + π ≈ 4.1801 → x ≈ 10.36 — outside [0,2π). Only one solution.

x ≈ 4.0781

⚡

Quick Reference

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🔑 Key Facts

sec = 1/cos · VA: cos=0 → x=π/2+πk

csc = 1/sin · VA: sin=0 → x=πk

cot = cos/sin · VA: sin=0 → x=πk

Range sec/csc: (−∞,−1]∪[1,∞)  cot: all reals

To solve: isolate → flip → solve sin/cos/tan

⚠️ Common Mistakes

❌ csc = 1/cos

csc = 1/sin. sec = 1/cos. Memory trick: co-secant pairs with sine (the non-co function), and secant pairs with cosine.

❌ VA of csc at x=π/2+πk

csc VA at x=πk (where sin=0). It's sec that has VA at π/2+πk (where cos=0).

❌ Range of 3csc(x) is [−3,3]

csc and sec NEVER produce values between −1 and 1. Multiplying by 3 gives (−∞,−3]∪[3,∞), not [−3,3].

❌ Solving csc(x)=2 directly

Flip first: csc=2 → sin=1/2. Then use unit circle. Never try to invert csc directly without converting to sin.

Secant, Cosecant,& Cotangent

Range of sec and csc — never between −1 and 1

How to find asymptotes for transformed functions

Secant, Cosecant,
& Cotangent