Unit 3 · Topic 3.8

3.8 The Tangent Function

The tangent function gives the slope of the terminal ray, so tan θ = sin θ / cos θ. Its graph repeats every π, has vertical asymptotes wherever cos θ = 0, and rises between each pair of asymptotes. You can transform it with the same a, b, c and d as sine and cosine.

Key terms

tangent
period of π
vertical asymptote
slope of the terminal ray

Tangent as a slope

For an angle θ in standard position, tan θ is the slope of the terminal ray. Slope is the change in y over the change in x between any two points on the ray, so using the origin and the unit-circle point (cos θ, sin θ) gives tan θ = sin θ / cos θ, wherever cos θ ≠ 0.

At θ = π/4 the ray rises at 45°, so the slope is 1: tan(π/4) = 1. At θ = 0 the ray is flat: tan 0 = 0. As θ approaches π/2, the ray gets steeper and steeper, and tan θ grows without bound.

Period and asymptotes

Turn a ray by half a revolution and you get a ray on the same line, pointing the opposite way. It has the same slope. So tan(θ + π) = tan θ, and the period of tangent is π, not 2π.

Where cos θ = 0, at θ = π/2 + kπ for any integer k, the ray is vertical and tan θ is undefined. The graph has vertical asymptotes there. Approaching π/2 from the left, tan θ → ∞; from the right, tan θ → −∞.

The range is all real numbers.

Shape between asymptotes

Between consecutive asymptotes, tangent is always increasing. On (−π/2, π/2) it rises from −∞ through 0 to ∞.

Its concavity changes in the middle: the graph is concave down on (−π/2, 0) and concave up on (0, π/2). So each branch has a point of inflection, at (kπ, 0) for the basic tangent.

On the unit circle, as θ increases from −π/2 to π/2, the terminal ray swings counterclockwise from pointing nearly straight down to pointing nearly straight up. Its slope keeps increasing, which is why tangent increases on each branch.

Tangent is an odd function: tan(−θ) = −tan θ, so its graph is symmetric about the origin.

Transformations

For f(θ) = a tan(b(θ + c)) + d:

Period: π/|b|.
Vertical dilation by |a|; if a < 0 the graph is reflected over the x-axis, so each branch decreases.
Phase shift: −c units.
Vertical shift: d units. The points of inflection, which sit on the line y = 0 for tan θ, move to the line y = d.
Asymptotes: where b(θ + c) = π/2 + kπ.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Exact tangent values
Find tan(2π/3), tan(5π/4) and tan(3π/2), if they exist.
Show the solutionHide the solution
1. Step 1: tan(2π/3): sin = √3/2 and cos = −1/2, so tan = (√3/2)/(−1/2) = −√3.
2. Step 2: tan(5π/4): sin = cos = −√2/2, so tan = 1. (A ray in quadrant III through (−1, −1) has slope 1.)
3. Step 3: tan(3π/2): cos(3π/2) = 0, so tan(3π/2) is undefined. The terminal ray is vertical.
Answer: tan(2π/3) = −√3, tan(5π/4) = 1, tan(3π/2) is undefined.
Example 2
Features of a transformed tangent
For g(θ) = 2 tan(θ/2) + 1, find the period, the asymptotes and the points of inflection.
Show the solutionHide the solution
1. Step 1: b = 1/2, so the period is π/(1/2) = 2π.
2. Step 2: Asymptotes where θ/2 = π/2 + kπ, so θ = π + 2πk: at θ = ±π, ±3π, ….
3. Step 3: Points of inflection where θ/2 = kπ, so θ = 2πk, at height g = 2 · 0 + 1 = 1. They lie on the line y = 1.
4. Step 4: a = 2 > 0, so each branch is increasing.
Answer: Period 2π; asymptotes θ = π + 2πk; points of inflection (2πk, 1).
Example 3
Trap: tangent's period formula
What is the period of h(θ) = tan(3θ)?
Show the solutionHide the solution
1. Step 1: The trap is to use the sine and cosine formula, 2π/|b|, and answer 2π/3.
2. Step 2: Tangent's basic period is π, so the period is π/|b| = π/3.
3. Step 3: Check: asymptotes are where 3θ = π/2 + kπ, so θ = π/6 + kπ/3. They are π/3 apart.
Answer: π/3.

Common mistakes

Using 2π/|b| for a tangent period. For tangent it's π/|b|.
Putting tangent's asymptotes at multiples of π. Those are its zeros; the asymptotes are where cos θ = 0.
Saying tangent is concave up everywhere it increases. Each branch switches from concave down to concave up at its point of inflection.

On the exam

Expect questions on tangent's period, asymptotes and increasing behavior, sometimes with a transformation, and exact values like tan(5π/6).
If asked why tangent has period π, say that rays half a turn apart lie on the same line, so they have the same slope.

Connected topics

Videos

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