Year 12 Maths - Methods (Unit 3 & Unit 4) Circular (trigonometric) functions

The tangent function

Q: What is the period of y = tan x?

The tangent function repeats every pi radians, so its period is pi (not 2 pi like sine and cosine). The graph is a series of identical branches spaced pi apart, one between each pair of neighbouring asymptotes.

Q: Where are the asymptotes of y = tan x?

Since tan x = sin x over cos x, the function is undefined wherever cos x = 0, that is at x = pi/2 + k pi for any integer k. The graph has a vertical asymptote at each, for example x = -pi/2, pi/2 and 3 pi/2.

Q: Where does y = tan x cross the x-axis?

The tangent curve is zero wherever sin x = 0, that is at the integer multiples of pi: x = 0, pi, -pi and so on. Each x-intercept sits exactly halfway between two neighbouring asymptotes.

Q: How do you find the period of y = a tan nx?

Divide the base period pi by n to get pi/n. For example y = tan 2x has period pi/2, so its branches and asymptotes are twice as close as those of y = tan x. The factor a does not change the period.

Q: Does the tangent function have an amplitude?

No. The range of the tangent function is all real numbers, so it has no maximum or minimum and no amplitude. In y = a tan nx the number a is a dilation factor from the x-axis that changes the steepness, not a height.

Q: How do you solve tan x = a?

Find one solution x = arctan a, then add whole multiples of the period: x = arctan a + k pi. Keep those in the required domain. For example tan x = 1 gives x = pi/4 + k pi, so on 0 to 2 pi the solutions are pi/4 and 5 pi/4.

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Theory

The tangent function \(y=\tan x\) has period \(\pi\), vertical asymptotes at \(x=\dfrac{\pi}{2}+k\pi\) and \(x\)-intercepts at the multiples of \(\pi\). Its range is all real numbers, so it has no amplitude. A transformed graph \(y=a\tan n(x-e)\) has period \(\dfrac{\pi}{n}\), a dilation factor \(a\) that changes the steepness, and asymptotes shifted by \(e\).

The tangent function is defined by \(\tan x=\dfrac{\sin x}{\cos x}\). It is undefined wherever \(\cos x=0\), which happens at \(x=\dfrac{\pi}{2}+k\pi\) for every integer \(k\); the graph has a vertical asymptote at each of these values. Between consecutive asymptotes the curve climbs from \(-\infty\) to \(+\infty\), so the range is all real numbers and there is no maximum, minimum or amplitude.

The function repeats every \(\pi\) radians, so its period is \(\pi\) — half that of \(\sin\) and \(\cos\). It is zero wherever \(\sin x=0\), giving \(x\)-intercepts at the multiples of \(\pi\) \((x=0,\pm\pi,\pm2\pi,\dots)\), each sitting exactly halfway between two asymptotes. A useful key point is \(\tan\dfrac{\pi}{4}=1\), and by symmetry \(\tan\left(-\dfrac{\pi}{4}\right)=-1\).

A general transformed tangent graph is \(y=a\tan n(x-e)\). The factor \(n\) compresses the graph horizontally so the period becomes \(\dfrac{\pi}{n}\) and the asymptotes crowd closer together; \(a\) is a dilation from the \(x\)-axis that steepens or flattens each branch (it is not an amplitude); and \(e\) is a horizontal translation that shifts every asymptote and intercept by \(e\).

Key idea. \(y=\tan x\) has period \(\pi\), asymptotes at \(x=\dfrac{\pi}{2}+k\pi\), and zeros at \(x=k\pi\). For \(y=a\tan n(x-e)\) the period shrinks to \(\dfrac{\pi}{n}\) and the asymptotes are the solutions of \(n(x-e)=\dfrac{\pi}{2}+k\pi\).

\(y=\tan x\): period \(\pi\), asymptotes at \(x=\pm\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\), zeros at \(x=0,\pi\)

\(y=\tan 2x\): the period is \(\dfrac{\pi}{2}\), so asymptotes at \(x=\dfrac{\pi}{4},\dfrac{3\pi}{4}\) sit twice as close

The definition and the two features that fix the shape of \(y=\tan x\):

\[\tan x=\frac{\sin x}{\cos x} \qquad \text{period}=\pi \qquad \text{asymptotes: } x=\frac{\pi}{2}+k\pi\]

\tan x = \frac{\sin x}{\cos x}

For the transformed graph \(y=a\tan n(x-e)\), the period and the asymptote/intercept locations:

\[\text{period}=\frac{\pi}{n} \qquad n(x-e)=\frac{\pi}{2}+k\pi \ \text{(asymptotes)} \qquad n(x-e)=k\pi \ \text{(zeros)}\]

period = \frac{π}{n}

To solve a tangent equation, use the period \(\pi\) to generate every solution:

\[\tan x=a \ \Longrightarrow\ x=\tan^{-1}a+k\pi, \quad k\in\mathbb{Z}\]

x = \tan^{- 1} a + k π

General solution. Because the period is \(\pi\), consecutive solutions of \(\tan x=a\) differ by \(\pi\) — add \(k\pi\) (not \(2k\pi\)) and keep those inside the required domain.

How to sketch \(y=a\tan n(x-e)\) or solve a tangent equation

Read off the transformations. Identify the dilation factor \(a\), the factor \(n\) (which gives period \(\dfrac{\pi}{n}\)), and the translation \(e\).
Locate the asymptotes. Solve \(n(x-e)=\dfrac{\pi}{2}+k\pi\) for the \(x\)-values in the domain; draw a dashed vertical line at each.
Mark the intercepts. Solve \(n(x-e)=k\pi\) — each zero sits midway between neighbouring asymptotes.
Plot a key point. A quarter-period from an intercept the height is \(\pm a\); this fixes the steepness of each increasing branch.
To solve \(\tan(\cdot)=a\), take \(\tan^{-1}a\), add \(k\pi\), then substitute back and keep the solutions inside the given interval.

Widen-the-window shortcut. When solving \(\tan nx=a\) on \(0\le x\le c\), first let \(u=nx\) and solve on \(0\le u\le nc\) — this catches every solution before you divide by \(n\).

Example 1 — features of \(y=\tan x\)

State the period of \(y=\tan x\), and the equations of its asymptotes for \(0\le x\le 2\pi\).

Solution

Period — the tangent function repeats every \(\pi\):

period

\(=\)

\(\pi\)

Asymptotes — where \(\cos x=0\), i.e. \(x=\dfrac{\pi}{2}+k\pi\) in \([0,2\pi]\):

\(x\)	\(=\)	\(\dfrac{\pi}{2}\)
\(x\)	\(=\)	\(\dfrac{3\pi}{2}\)

\(\therefore\) period \(=\pi\); asymptotes \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\)

x = \frac{π}{2}, \frac{3 π}{2}

Example 2 — solving \(\tan x=1\)

Solve \(\tan x=1\) for \(0\le x\le 2\pi\).

Solution

Base solution — \(\tan^{-1}1=\dfrac{\pi}{4}\), then add \(k\pi\):

\(x\)

\(=\)

\(\dfrac{\pi}{4}+k\pi\)

Keep the values in \([0,2\pi]\) — take \(k=0\) and \(k=1\):

\(x\)	\(=\)	\(\dfrac{\pi}{4}\)
\(x\)	\(=\)	\(\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}\)

\(\therefore\) \(x=\dfrac{\pi}{4}\) or \(x=\dfrac{5\pi}{4}\)

x = \frac{π}{4}, \frac{5 π}{4}

Example 3 — period of \(y=a\tan nx\)

For \(y=2\tan 3x\), state the period and the asymptotes nearest the origin.

Solution

Period — divide the base period \(\pi\) by \(n=3\):

period

\(=\)

\(\dfrac{\pi}{n}=\dfrac{\pi}{3}\)

Asymptotes — solve \(3x=\dfrac{\pi}{2}+k\pi\):

\(3x\)	\(=\)	\(\pm\dfrac{\pi}{2}\)
\(x\)	\(=\)	\(\pm\dfrac{\pi}{6}\)

Note the factor \(a=2\) steepens the branches but leaves the period unchanged.

\(\therefore\) period \(=\dfrac{\pi}{3}\); nearest asymptotes \(x=\pm\dfrac{\pi}{6}\)

period = \frac{π}{3}

Example 4 — solving \(\tan 2x=\sqrt{3}\)

Solve \(\tan 2x=\sqrt{3}\) for \(0\le x\le \pi\).

Solution

Substitute \(u=2x\) and widen the domain — if \(0\le x\le\pi\) then \(0\le u\le 2\pi\):

\(\tan u\)	\(=\)	\(\sqrt{3}\)
\(u\)	\(=\)	\(\dfrac{\pi}{3}+k\pi\)

Take \(u\) in \([0,2\pi]\) — \(k=0\) and \(k=1\):

\(u\)

\(=\)

\(\dfrac{\pi}{3},\ \dfrac{4\pi}{3}\)

Divide by \(2\) to return to \(x\):

\(x\)

\(=\)

\(\dfrac{\pi}{6},\ \dfrac{2\pi}{3}\)

\(\therefore\) \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{2\pi}{3}\)

x = \frac{π}{6}, \frac{2 π}{3}

Common pitfalls

The tangent function has no amplitude. Its range is all real numbers, so there is no maximum or minimum. In \(y=a\tan nx\) the number \(a\) is a dilation factor that changes the steepness of the branches, not a height.

The period is \(\pi\), not \(2\pi\). Consecutive solutions of \(\tan x=a\) differ by \(\pi\); adding \(2k\pi\) misses half of them. For \(y=a\tan nx\) the period is \(\dfrac{\pi}{n}\), not \(\dfrac{2\pi}{n}\).

Widen the interval before dividing. When solving \(\tan nx=a\), first find every value of \(nx\) across the enlarged interval \([0,nc]\); dividing too early loses solutions such as the second one in \(\tan 2x=\sqrt{3}\).

Frequently asked questions

What is the period of y = tan x?

The tangent function repeats every \(\pi\) radians, so its period is \(\pi\) — not \(2\pi\) like sine and cosine. The graph is a series of identical branches spaced \(\pi\) apart, one between each pair of asymptotes.

Where are the asymptotes of y = tan x?

Since \(\tan x=\dfrac{\sin x}{\cos x}\), the function is undefined where \(\cos x=0\), i.e. at \(x=\dfrac{\pi}{2}+k\pi\). There is a vertical asymptote at each, for example \(x=-\dfrac{\pi}{2},\dfrac{\pi}{2},\dfrac{3\pi}{2}\).

Where does y = tan x cross the x-axis?

The curve is zero where \(\sin x=0\), i.e. at the multiples of \(\pi\): \(x=0,\pm\pi,\pm2\pi,\dots\) Each intercept sits exactly halfway between two neighbouring asymptotes.

How do you find the period of y = a tan nx?

Divide the base period \(\pi\) by \(n\) to get \(\dfrac{\pi}{n}\). For example \(y=\tan 2x\) has period \(\dfrac{\pi}{2}\), so its branches are twice as close. The factor \(a\) does not change the period.

Does the tangent function have an amplitude?

No. The range is all real numbers, so there is no maximum, minimum or amplitude. In \(y=a\tan nx\) the number \(a\) is a dilation factor from the \(x\)-axis that changes the steepness.

How do you solve tan x = a?

Find one solution \(x=\tan^{-1}a\), then add \(k\pi\): \(x=\tan^{-1}a+k\pi\). Keep those in the domain. For example \(\tan x=1\) gives \(x=\dfrac{\pi}{4}+k\pi\), so on \([0,2\pi]\) the solutions are \(\dfrac{\pi}{4}\) and \(\dfrac{5\pi}{4}\).

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The tangent function

📖 Theory

How to sketch \(y=a\tan n(x-e)\) or solve a tangent equation

Common pitfalls

Frequently asked questions

Theory