Where are the asymptotes of y equals tan t?

At t equals pi over two plus multiples of pi, where cosine of t is zero and the function is undefined. They are drawn as vertical dashed lines.

Where does y equals tan t cross the t-axis?

At the multiples of pi, that is t equals zero, pi, two pi and so on, where sine of t is zero. These sit midway between the asymptotes.

Where are the asymptotes of y equals tan t?

At t equals pi over two plus multiples of pi, where cosine of t is zero and the function is undefined. They are drawn as vertical dashed lines.

Where does y equals tan t cross the t-axis?

At the multiples of pi, that is t equals zero, pi, two pi and so on, where sine of t is zero. These sit midway between the asymptotes.

Year 11 Methods (Unit 1 & 2) Trigonometric Functions

The Tangent Function

Q: What is the period of the tangent function?

The period of y equals tan t is pi. For y equals a tan of n t it is pi divided by n.

Q: Does the coefficient a change the asymptotes?

No. In y equals a tan of n t only n affects the period and asymptote positions. The a stretches the curve vertically.

20 practice questions 1 video lesson Theory + worked examples

Understand the tangent graph for Queensland Year 11 Mathematical Methods (QCAA). Unlike the smooth waves of sine and cosine, the tangent curve climbs steeply and breaks at regular gaps, repeating every pi radians.

You will learn to locate the intercepts where the curve crosses the axis, draw the vertical asymptotes where tangent is undefined, adjust the period when the angle is scaled, and solve tangent equations over a domain in the QCAA course.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the tangent function \(y=\tan t\) has period \(\pi\), crosses the \(t\)-axis at every multiple of \(\pi\), and has vertical asymptotes at \(t=\dfrac{\pi}{2}+k\pi\) where cosine is zero. For \(y=a\tan nt\) the period becomes \(\dfrac{\pi}{n}\). This page covers the shape, the asymptotes, and solving \(\tan t=k\).

The tangent function is \(\tan t=\dfrac{\sin t}{\cos t}\). Wherever \(\cos t=0\) it is undefined, producing a vertical asymptote. These occur at \(t=\dfrac{\pi}{2}+k\pi\) and are drawn as red dashed lines the curve approaches but never touches.

Between consecutive asymptotes the curve rises from very negative to very positive, so the range is all real numbers. It has period \(\pi\): the whole picture repeats every \(\pi\) units. The \(t\)-intercepts are at \(t=k\pi\), where \(\sin t=0\).

For \(y=a\tan nt\), the factor \(n\) changes the period to \(\dfrac{\pi}{n}\) and moves the asymptotes to \(t=\dfrac{\pi}{2n}+\dfrac{k\pi}{n}\). The \(a\) stretches the curve vertically but does not move the intercepts or asymptotes.

Asymptotes come from \(\cos=0\). For \(y=\tan t\) they sit at \(t=\dfrac{\pi}{2}+k\pi\); one period \(\pi\) later they repeat.

\(y=\tan t\): period \(\pi\), \(t\)-intercepts at \(k\pi\) (gold), red dashed asymptotes at \(\tfrac{\pi}{2}+k\pi\).

\(y=\tan 2t\): period \(\dfrac{\pi}{2}\), so the asymptotes are twice as close.

For \(y=a\tan nt\):

\[\text{period}=\dfrac{\pi}{n}\]

period = \frac{π}{n}

\[\text{asymptotes:}\ t=\dfrac{\pi}{2n}+\dfrac{k\pi}{n}\]

t = \frac{π}{2 n} + \frac{k π}{n}

\[t\text{-intercepts:}\ t=\dfrac{k\pi}{n}\]

t = \frac{k π}{n}

Solving \(\tan t=k\): tangent has period \(\pi\), so once you have one solution \(t_0\), all others are \(t_0+k\pi\).

How to work with \(y=a\tan nt\)

Period: compute \(\dfrac{\pi}{n}\); the graph repeats every \(\dfrac{\pi}{n}\).
Asymptotes: draw red dashed lines at \(t=\dfrac{\pi}{2n}+\dfrac{k\pi}{n}\); the curve never crosses them.
Intercepts: mark \(t\)-intercepts at \(t=\dfrac{k\pi}{n}\), midway between asymptotes.
Solve: for \(\tan t=k\), find one solution, then add multiples of the period \(\pi\) to reach every solution in the domain.

Example 1 — Features of the tangent curve

State the period, \(t\)-intercepts and asymptotes of \(y=\tan t\) on \([0,2\pi]\).

Solution

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

\(\text{period}\)

\(=\)

\(\pi\)

\(t\)-intercepts — where \(\sin t=0\):

\(t\)

\(=\)

\(0,\ \pi,\ 2\pi\)

Asymptotes — where \(\cos t=0\):

\(t\)

\(=\)

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

Period \(\pi\); intercepts \(0,\pi,2\pi\); asymptotes \(t=\dfrac{\pi}{2},\dfrac{3\pi}{2}\).

period = π

Example 2 — Period of a transformed tangent

Find the period and asymptotes of \(y=\tan 2t\) on \([0,\pi]\).

Solution

Period — \(n=2\):

\(\text{period}\)	\(=\)	\(\dfrac{\pi}{n}\)
	\(=\)	\(\dfrac{\pi}{2}\)

Asymptotes — \(t=\dfrac{\pi}{2n}+\dfrac{k\pi}{n}\):

\(t\)	\(=\)	\(\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
	\(=\)	\(\dfrac{\pi}{4},\ \dfrac{3\pi}{4}\)

Period \(\dfrac{\pi}{2}\); asymptotes \(t=\dfrac{\pi}{4},\dfrac{3\pi}{4}\).

period = \frac{π}{2}

Example 3 — Solving a tangent equation

Solve \(\tan t=1\) for \(t\in[0,2\pi]\).

Solution

Reference angle — the acute angle with tangent \(1\):

\(\theta\)

\(=\)

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

Add the period — tangent repeats every \(\pi\):

\(t\)	\(=\)	\(\dfrac{\pi}{4},\ \dfrac{\pi}{4}+\pi\)
	\(=\)	\(\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\)

\(t=\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\).

t = \frac{π}{4}, \frac{5 π}{4}

Example 4 — A vertical stretch and dilation

For \(y=2\tan 3t\), state the period and the first positive asymptote.

Solution

Period — \(n=3\) (the \(2\) does not affect it):

\(\text{period}\)

\(=\)

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

First positive asymptote — \(t=\dfrac{\pi}{2n}\):

\(t\)	\(=\)	\(\dfrac{\pi}{2\times 3}\)
	\(=\)	\(\dfrac{\pi}{6}\)

Period \(\dfrac{\pi}{3}\); first asymptote at \(t=\dfrac{\pi}{6}\).

period = \frac{π}{3}

Common pitfalls

Using \(2\pi\) for the period. Tangent has period \(\pi\), not \(2\pi\); sine and cosine have period \(2\pi\).

Drawing through an asymptote. The curve approaches the red dashed line but never crosses it; each branch is separate.

Thinking \(a\) moves the asymptotes. In \(y=a\tan nt\), only \(n\) changes the period and asymptote spacing; \(a\) just stretches vertically.

Missing the second solution. \(\tan t=k\) has two solutions in \([0,2\pi]\), one period \(\pi\) apart.

Frequently asked questions

What is the period of the tangent function?

The period of \(y=\tan t\) is \(\pi\). For \(y=a\tan nt\) it is \(\dfrac{\pi}{n}\).

Where are the asymptotes of y = tan t?

At \(t=\dfrac{\pi}{2}+k\pi\), where \(\cos t=0\) and the function is undefined. They are drawn as vertical dashed lines.

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

At the multiples of \(\pi\): \(t=0,\pi,2\pi,\dots\), where \(\sin t=0\). These sit midway between the asymptotes.

How do you solve tan t = k over a domain?

Find one solution from the reference angle, then add multiples of the period \(\pi\) until you have every solution in the domain.

Does the coefficient a change the asymptotes?

No. In \(y=a\tan nt\) only \(n\) affects the period and asymptote positions; \(a\) stretches the curve vertically.

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Further Symmetry Properties And The Pythagorean Identity

The Tangent Function

📖 Theory

How to work with \(y=a\tan nt\)

Common pitfalls

Frequently asked questions

Theory