Recall: What is tan θ?
Three equivalent definitions — all from the unit circle
Ratio Definition
tan θ = sin θ / cos θ
Slope Definition
tan θ = m (slope of terminal ray)
Coordinate Definition
tan θ = y/x (for point (x,y) on circle)
Why does tan have vertical asymptotes?
tan θ = sin θ / cos θ. When cos θ = 0, the denominator is zero → undefined.
This happens at θ = π/2, 3π/2, −π/2, … i.e. at θ = π/2 + kπ for any integer k.
At these angles, the terminal ray is vertical, and its slope is undefined (infinite). Each half-turn around the circle creates another vertical asymptote.
The Graph of y = tan θ
Passes through origin · period = π · vertical asymptotes at π/2 + kπ · always increasing
y = tan θ — period = π, vertical asymptotes (orange dashed) at π/2 and 3π/2
📊 tan θ Properties
Period:
π (half of sin/cos)Amplitude: None — no max or min
Range: (−∞, +∞) — all real numbers
Asymptotes:
θ = π/2 + kπZeros:
θ = kπAlways: increasing on each interval
🌊 sin θ / cos θ (for contrast)
Period:
2πAmplitude: 1 (max and min exist)
Range: [−1, 1]
Asymptotes: None
Zeros: at multiples of π (for sin)
Shape: wave (concave up & down)
Standard Form
f(θ) = a·tan(b(θ+c)) + d — note: period formula is DIFFERENT from sin/cos
📌 Standard Form of the Tangent Function
f(θ) = a · tan(b(θ + c)) + d
a
Vertical dilation of tan θ by factor |a|
a = vert. scale
b
Horizontal dilation — determines period
P = π / |b|
c
Phase shift (horizontal translation by −c)
shift = −c
d
Vertical translation
midline: y = d
⚠️ Period Formula is DIFFERENT for tan vs sin/cos
Tangent
P = π / |b|
Sine / Cosine
P = 2π / |b|
The base period of tan is π (not 2π). So the formula uses π instead of 2π. Same b, same idea — just different base period.
Finding vertical asymptotes of a·tan(b(θ+c)) + d
Asymptotes occur when the argument of tan = π/2 + kπ.
For tan(bx): set bx = π/2 + kπ → solve for x → x = π/(2b) + kπ/b.
For tan(x/3): set x/3 = π/2 + kπ → x = 3π/2 + 3kπ.
Examples 1–4 — Working with Tangent
Reading a, b, d from graphs · period from b · asymptotes from argument
📌 Example 1 — Graph of f (asymptotes at ±π/2, passes through (π/8, ≈1)). f(x) = a·tan(bx) + d. Find a, b, d.
1
Period: Asymptotes at x = −π/2 and x = π/2 → period = π/2. Set π/|b| = π/2 → b = 2.2
d: No vertical shift visible (midline is y = 0) → d = 0.3
a: At x = π/8: f(π/8) = a·tan(2·π/8) = a·tan(π/4) = a·1 = a ≈ 1 (from graph). So a = 1.a
1
b
2
d
0
f(x) = tan(2x)
📌 Example 2 — h(x) = 4tan(2x) + 5. What is the period of h?
1
b = 2 → Period = π/|b| = π/2.2
Note: a = 4 changes the vertical scale, d = 5 shifts up — neither affects the period.Period = π / |2| = π/2
📌 Example 3 — g(x) = 2tan(x/3) − 1. Which gives the vertical asymptotes?
1
b = 1/3. Asymptotes when argument = π/2 + kπ: x/3 = π/2 + kπ → x = 3π/2 + 3kπ.2
Period check: π/(1/3) = 3π ✓ — asymptotes are 3π apart, confirming period = 3π.3
Note: the −1 shifts the graph down but does NOT move the asymptotes.(A)
x = π/2 + πk
That's for tan(x) with b=1. Here b=1/3, so period=3π ✗
(B)
x = π/6 + (π/3)k
Wrong period: that's period=π/3 ✗
✓ (C)
x = 3π/2 + 3πk
x/3=π/2+kπ → x=3π/2+3πk ✓. Period=3π ✓
(D)
x = π + 2πk
Wrong period ✗
📌 Example 4 — Graph of k (period = 2π, vertical shift down 2). Which expression fits?
1
Period: 2π = π/|b| → b = 1/2.2
Vertical shift: d = −2 (graph shifted down 2).3
b = 1/2 means tan(x/2) in the argument. So k(x) = tan(x/2) − 2.✓ (A)
tan(x/2) − 2
b=1/2→period=2π ✓, d=−2 ✓
(B)
tan(x/2) + 2
d=+2 shifts UP — graph shifts DOWN ✗
(C)
tan(2x) − 2
b=2→period=π/2, not 2π ✗
(D)
tan(2x) + 2
Both b and d are wrong ✗
Example 5 — Evaluating tan from the Unit Circle
tan θ = sin θ / cos θ = y/x at every angle
Method: always divide sin by cos
Look up (or recall) the unit circle point (cos θ, sin θ). Then divide: tan θ = sin θ ÷ cos θ. If cos θ = 0, the result is undefined. If sin θ = 0, tan θ = 0. Use ASTC signs to check.
tan(π/6)
(1/2) ÷ (√3/2)
1/√3 = √3/3
tan(π/4)
(√2/2) ÷ (√2/2)
1
tan(π/3)
(√3/2) ÷ (1/2)
√3
tan(5π/6)
(1/2) ÷ (−√3/2)
−1/√3 = −√3/3
tan(4π/3)
(−√3/2) ÷ (−1/2)
√3
tan(7π/4)
(−√2/2) ÷ (√2/2)
−1
tan(3π/2)
cos = 0
undefined
tan(π)
(0) ÷ (−1)
0
tan(π/2)
cos = 0
undefined
Q1
sin+, cos+
tan > 0 ✓
Q2
sin+, cos−
tan < 0 ✗
Q3
sin−, cos−
tan > 0 ✓
Q4
sin−, cos+
tan < 0 ✗
tan is positive in Q1 and Q3 · negative in Q2 and Q4 — "All Students Take Calculus" → Tan is the T in "Take"
Quick Reference
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🔑 Key Formulas
tan θ = sin θ / cos θ = y/x = slope
Period = π / |b| (NOT 2π / |b|!)
Asymptotes: bx = π/2 + kπ → solve x
tan(π/6)=1/√3 · tan(π/4)=1 · tan(π/3)=√3
tan(π/2)=undef · tan(π)=0 · tan(3π/2)=undef
⚠️ Common Mistakes
❌ Period of tan = 2π/|b|
tan's base period is π, not 2π. Period = π/|b|. E.g. tan(2x): period=π/2, not π.
❌ tan has amplitude = 1
tan has NO amplitude — it has no max or min. Range is all real numbers (−∞,+∞).
❌ tan is undefined at kπ
tan is undefined at π/2 + kπ (where cos=0). At kπ, tan=0 (sin=0 there).
❌ Vertical shift changes asymptotes
d only shifts the graph up/down. Asymptotes are vertical lines — they stay at the same x-values.