B. Trigonometric Functions · Radian Measure and Trigonometric Functions

The Tangent Function

B2.2About 70 minutes

Learning goals

I can explain how the tangent ratio becomes the function f(x) = tan x, using the unit circle and a table of values.
I can explain why y = tan x has vertical asymptotes where cos x = 0.
I can sketch y = tan x and state its domain, range, period, zeros and asymptotes.
I can explain why the period of y = tan x is π, not 2π.

Words to know

tangent function
The function f(x) = tan x = sin x / cos x, with x in radians.

vertical asymptote
A vertical line x = a that the graph approaches but never touches; the y-values grow without bound near it.

undefined
Having no value. tan x is undefined where cos x = 0.

period of tan x
π. The graph repeats every π units.

odd function
A function with f(−x) = −f(x); its graph has rotational symmetry about the origin. sin x and tan x are odd.

Why a new kind of graph?

Sine and cosine are coordinates of a point on the unit circle, so they are trapped between −1 and 1. Tangent is different. It is a quotient:

tan x = sin x ÷ cos x = y ÷ x

When the point is near the top or bottom of the circle, x = cos x is tiny, so tan x is huge. When cos x = 0 exactly, tan x does not exist. That makes the graph of y = tan x look nothing like a wave.

Tangent as a slope

The terminal arm of θ passes through (0, 0) and (cos θ, sin θ), so its slope is sin θ / cos θ = tan θ. If you extend the terminal arm until it hits the vertical line x = 1, it meets that line at the height tan θ. That vertical segment is actually tangent to the circle, which is where the name comes from.

Extend the terminal arm to the line x = 1. By similar triangles, it meets that line at height tan θ. As θ approaches π/2, the arm becomes nearly vertical and the height grows without limit.

A table of values, using technology

Use a calculator in radian mode (or a graphing tool) to see what happens near π/2 ≈ 1.5708:

x	0	π/6 ≈ 0.52	π/4 ≈ 0.79	π/3 ≈ 1.05	1.5	1.55	1.57	1.58
tan x	0	0.58	1	1.73	14.10	48.08	1255.77	−108.65

As x approaches π/2 from the left, tan x shoots up toward +∞. Just past π/2 it is a huge negative number. The graph jumps across the vertical asymptote x = π/2.

Properties of y = tan x

Vertical asymptotes where cos x = 0: x = π/2 + nπ, n ∈ ℤ.
Domain: {x ∈ ℝ | x ≠ π/2 + nπ, n ∈ ℤ}.
Range: {y ∈ ℝ}. Every real number is a tangent value.
Zeros where sin x = 0: x = nπ, n ∈ ℤ.
Period: π. The point for x + π is directly opposite the point for x on the unit circle, (−cos x, −sin x), and the quotient of two negatives is the same as before. So tan(x + π) = tan x.
No maximum, no minimum and no amplitude.
Increasing on every interval between consecutive asymptotes.
Passes through (π/4, 1) and (−π/4, −1), halfway between a zero and an asymptote.
Odd: tan(−x) = −tan x, so the graph has rotational symmetry about the origin.

y = tan x for −π ≤ x ≤ 2π. Each branch rises from −∞ to +∞ between dashed asymptotes that are π apart, crossing the x-axis halfway between them.

Sketching tan x

Draw the asymptotes first (dashed). Then plot three points on each branch: the zero in the middle, and the points where y = −1 and y = 1, a quarter of the period (π/4) to each side. Make each branch bend toward its asymptotes without touching them.

Worked example

Example 1: Why the asymptotes are where they are

Explain, using sin x and cos x, why y = tan x has a vertical asymptote at x = 3π/2, and describe the graph on each side.

cos(3π/2) = 0 and sin(3π/2) = −1, so tan(3π/2) would be −1/0: undefined.
Just to the left (say x = 4.7), x is in quadrant III: sin x ≈ −1 and cos x is a tiny negative number, so tan x is a huge positive number.
Just to the right (x = 4.72, quadrant IV): sin x ≈ −1 and cos x is a tiny positive number, so tan x is a huge negative number.

The graph rises toward +∞ on the left of x = 3π/2 and comes up from −∞ on the right, the same pattern as at π/2.

Worked example

Example 2: State the properties on an interval

For 0 ≤ x ≤ 2π, state the zeros, the equations of the asymptotes and the x-values where tan x = 1.

Zeros where sin x = 0: x = 0, π, 2π.
Asymptotes where cos x = 0: x = π/2 and x = 3π/2.
tan x = 1 where sin x = cos x: x = π/4 and x = π/4 + π = 5π/4.

Zeros 0, π, 2π; asymptotes x = π/2, x = 3π/2; tan x = 1 at x = π/4, 5π/4.

Worked example

Example 3: Using the period

Given that tan 0.4 ≈ 0.4228, find tan(0.4 + π), tan(0.4 + 3π) and tan(−0.4) without a calculator.

The period is π, so adding any multiple of π does not change tangent: tan(0.4 + π) = tan(0.4 + 3π) ≈ 0.4228.
Tangent is odd: tan(−0.4) = −tan 0.4 ≈ −0.4228.

0.4228, 0.4228, −0.4228

Just for fun

Q: Why did y = tan x never make it to x = π/2? A: It kept getting closer and closer, but its plans were always asymptotic.

Hands-on activity

Measure the tangent with a ruler

You need: a sheet of paper, a ruler, a pencil, a protractor and a calculator.

Near the left edge of the paper, mark the origin O. Draw the positive x-axis to the right. Draw a vertical line 10 cm to the right of O (this is the line "x = 1" if 10 cm counts as 1 unit).
Using the protractor, draw terminal arms at 0.2, 0.4, 0.6, 0.8, 1.0 and 1.2 rad. (Convert first: 0.2 rad ≈ 11.5°, 0.4 ≈ 22.9°, 0.6 ≈ 34.4°, 0.8 ≈ 45.8°, 1.0 ≈ 57.3°, 1.2 ≈ 68.8°.) Extend each arm until it meets the vertical line.
Measure the height (in cm) where each arm meets the vertical line. Divide by 10 to get your estimate of tan θ.
Compare with the calculator in radian mode.
Plot your (θ, tan θ) points on a graph and sketch the curve. Try 1.4 rad (about 80.2°): what happens to the height?

What to look for: heights of about 2.0, 4.2, 6.8, 10.3, 15.6 and 25.7 cm (tan values 0.20, 0.42, 0.68, 1.03, 1.56, 2.57). At 1.4 rad the arm meets the line about 58 cm up, far off the page: the graph is heading for the asymptote at π/2 ≈ 1.57.

Your turn

Give the coordinates of the point on y = tan x at x = π/4.

Write the equations of the asymptotes of y = tan x for 0 ≤ x ≤ 2π.

List the zeros of y = tan x for 0 ≤ x ≤ 2π.

What is the period of y = tan x?

State the domain of y = tan x in set notation.

Find tan 1.56 with a calculator. Why is it so large?

Between two consecutive asymptotes, is y = tan x increasing or decreasing?

More practice: warm-up

tan 0 = ?

State the range of y = tan x.

tan(−π/4) = ?

Does y = tan x have an amplitude?

tan(π/2) = ?

More practice: on your own

Find all x in [0, 2π] with tan x = −1.

Explain why y = tan x has a vertical asymptote wherever cos x = 0, but y = sin x does not.

Using the graph, how many solutions does tan x = 5 have for 0 ≤ x ≤ 2π? Explain.

Given tan 1.2 ≈ 2.572, find tan(1.2 − π) and tan(−1.2).

A line through the origin makes an angle of π/3 with the positive x-axis. Find its slope and equation.

Challenge

Use the unit circle to show that tan(−x) = −tan x. What symmetry does this give the graph?

In the diagram of the terminal arm meeting x = 1 at (1, tan θ), find the distance from the origin to that point. Evaluate it for θ = π/3.

A rotating light is 50 m from a long straight wall. When the beam makes angle θ with the shortest line to the wall, the spot on the wall is d = 50 tan θ metres from the nearest point. Find d for θ = π/6, π/3 and 1.5. What happens as θ approaches π/2?

Check your understanding

Quiz

What is the period of y = tan x?

The vertical asymptotes of y = tan x are...

What is the range of y = tan x?

Which point lies on y = tan x?

y = tan x is undefined exactly where...

On −π/2 < x < π/2, the function y = tan x is...

For parents and teachers

Big idea (B2.2): tan x = sin x ÷ cos x, the slope of the terminal arm. It is undefined where cos x = 0, so its graph has vertical asymptotes at x = π/2 + nπ. It repeats every π, has no maximum or minimum, and its range is all real numbers.

How to help: Ask your teen to compute tan 1.5, tan 1.55 and tan 1.57 in radian mode and explain why the numbers explode. Ask "why is the period π and not 2π?" (opposite points on the unit circle give the same slope).

Common difficulties:

Drawing asymptotes at the zeros of sine (nπ) instead of the zeros of cosine.
Letting the curve touch or cross an asymptote.
Saying the period is 2π because sine and cosine have period 2π.
Giving tan x an amplitude.