For $y = \tan x$ on $0 \le x \le 2\pi$, state the period and the equations of the asymptotes.

Period $\pi$; asymptotes $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$

Sketch $y = \sec x$ on $0 \le x \le 2\pi$: give its asymptotes, its range and the vertex of each U.

Asymptotes $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$; vertices $(0, 1)$, $(\pi, -1)$, $(2\pi, 1)$; range $y \le -1$ or $y \ge 1$

For $y = \csc x$ on $0 < x < 2\pi$, find the asymptotes, the local maximum and minimum, and the exact value of $\csc \frac{\pi}{6}$.

Asymptotes $x = 0, \pi, 2\pi$; local minimum $\left(\frac{\pi}{2}, 1\right)$, local maximum $\left(\frac{3\pi}{2}, -1\right)$; $\csc \frac{\pi}{6} = 2$

Find the period, the zeros and the asymptotes of $y = \tan 2x$ on $0 \le x \le \pi$.

Period $\frac{\pi}{2}$; zeros $x = 0, \frac{\pi}{2}, \pi$; asymptotes $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$

For $y = 3 \csc 2x - 1$, find the period, the asymptotes on $0 \le x \le \pi$, the range, and the vertex of each U in that interval.

Period $\pi$; asymptotes $x = 0, \frac{\pi}{2}, \pi$; range $y \le -4$ or $y \ge 2$; vertices $\left(\frac{\pi}{4}, 2\right)$ and $\left(\frac{3\pi}{4}, -4\right)$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Graphs of Tan, Sec, Csc and Cot

The other four trig functions are built from sine and cosine by dividing, and a division blows up wherever the bottom is $0$ . So each of these graphs is a set of vertical asymptotes with a curve between them, and the asymptotes are the first thing to find.

How It Works

Tangent. $\tan x = \frac{\sin x}{\cos x}$ , so the asymptotes are where $\cos x = 0$ : $x = \frac{\pi}{2} + n\pi$ . The zeros are the zeros of sine, $x = n\pi$ . Between two asymptotes the curve rises from $-\infty$ to $+\infty$ through its zero, and the pattern repeats every $\pi$ , not $2\pi$ . The range is all real numbers.

Cotangent is $\frac{\cos x}{\sin x}$ , so everything swaps: asymptotes at $x = n\pi$ , zeros at $x = \frac{\pi}{2} + n\pi$ , period $\pi$ , and each branch falls from $+\infty$ to $-\infty$ .

Secant and cosecant. $\sec x = \frac{1}{\cos x}$ , so draw $y = \cos x$ lightly first. Every zero of cosine becomes an asymptote, and every maximum or minimum of cosine becomes the vertex of a U shape: where $\cos x = 1$ , $\sec x = 1$ and the U opens upward; where $\cos x = -1$ , $\sec x = -1$ and it opens downward. The period stays $2\pi$ , and the range is $y \le -1$ or $y \ge 1$ , with nothing in between. Cosecant is the same construction from sine.

Transformations work the way they do for sine and cosine, with the asymptotes moving too. For $y = a \sec(k(x - d)) + c$ the period is $\frac{2\pi}{k}$ , and the asymptotes are the shifted, squeezed zeros of $\cos(k(x - d))$ . For $\tan(kx)$ the period is $\frac{\pi}{k}$ . The vertical stretch and shift change the range: the vertices sit at $c \pm |a|$ , so the gap in the range is between those two values.

The usual test question gives one of these in transformed form and asks for the period, the asymptotes in an interval, and the range. Find the asymptotes by solving for the angle where the parent function's denominator is $0$ , then solving that for $x$ .

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
For $y = \tan x$ on $0 \le x \le 2\pi$ , state the period and the equations of the asymptotes.
Show solution
1. The period of tangent is $\pi$ .
2. Asymptotes where $\cos x = 0$ : $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ in this interval.
Answer: Period $\pi$ ; asymptotes $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$
Easy
Sketch $y = \sec x$ on $0 \le x \le 2\pi$ : give its asymptotes, its range and the vertex of each U.
Show solution
1. $\sec x = \frac{1}{\cos x}$ , so the asymptotes are where $\cos x = 0$ : $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ .
2. Where $\cos x = 1$ , at $x = 0$ and $x = 2\pi$ , $\sec x = 1$ : upward U shapes with vertices $(0, 1)$ and $(2\pi, 1)$ . Where $\cos x = -1$ , at $x = \pi$ , $\sec x = -1$ : a downward U with vertex $(\pi, -1)$ .
3. Since $|\cos x| \le 1$ , $|\sec x| \ge 1$ : the range is $y \le -1$ or $y \ge 1$ .
Answer: Asymptotes $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ ; vertices $(0, 1)$ , $(\pi, -1)$ , $(2\pi, 1)$ ; range $y \le -1$ or $y \ge 1$
Medium
For $y = \csc x$ on $0 < x < 2\pi$ , find the asymptotes, the local maximum and minimum, and the exact value of $\csc \frac{\pi}{6}$ .
Show solution
1. $\csc x = \frac{1}{\sin x}$ . Inside the interval $\sin x = 0$ only at $x = \pi$ , and the interval's endpoints $0$ and $2\pi$ are asymptotes as well.
2. $\sin x$ peaks at $\left(\frac{\pi}{2}, 1\right)$ , so $\csc x$ has a local minimum there, $\left(\frac{\pi}{2}, 1\right)$ ; $\sin x$ bottoms out at $\left(\frac{3\pi}{2}, -1\right)$ , so $\csc x$ has a local maximum $\left(\frac{3\pi}{2}, -1\right)$ .
3. $\csc \frac{\pi}{6} = \frac{1}{\sin \frac{\pi}{6}} = \frac{1}{1/2} = 2$ .
Answer: Asymptotes $x = 0, \pi, 2\pi$ ; local minimum $\left(\frac{\pi}{2}, 1\right)$ , local maximum $\left(\frac{3\pi}{2}, -1\right)$ ; $\csc \frac{\pi}{6} = 2$
Medium
Find the period, the zeros and the asymptotes of $y = \tan 2x$ on $0 \le x \le \pi$ .
Show solution
1. The period of $\tan(kx)$ is $\frac{\pi}{k} = \frac{\pi}{2}$ .
2. Asymptotes where $\cos 2x = 0$ : $2x = \frac{\pi}{2} + n\pi$ , so $x = \frac{\pi}{4} + \frac{n\pi}{2}$ . In the interval: $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$ .
3. Zeros where $\sin 2x = 0$ : $2x = n\pi$ , so $x = 0, \frac{\pi}{2}, \pi$ .
Answer: Period $\frac{\pi}{2}$ ; zeros $x = 0, \frac{\pi}{2}, \pi$ ; asymptotes $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$
Test level
For $y = 3 \csc 2x - 1$ , find the period, the asymptotes on $0 \le x \le \pi$ , the range, and the vertex of each U in that interval.
Show solution
1. The period of $\csc(kx)$ is $\frac{2\pi}{k} = \pi$ , the same as $\sin 2x$ .
2. Asymptotes where $\sin 2x = 0$ : $2x = 0, \pi, 2\pi$ , so $x = 0, \frac{\pi}{2}, \pi$ .
3. $\csc 2x$ has range $y \le -1$ or $y \ge 1$ . Multiplying by $3$ makes it $y \le -3$ or $y \ge 3$ ; subtracting $1$ gives $y \le -4$ or $y \ge 2$ .
4. Vertices sit where $\sin 2x = \pm 1$ . At $x = \frac{\pi}{4}$ , $\sin 2x = 1$ and $y = 3 - 1 = 2$ . At $x = \frac{3\pi}{4}$ , $\sin 2x = -1$ and $y = -3 - 1 = -4$ .
Answer: Period $\pi$ ; asymptotes $x = 0, \frac{\pi}{2}, \pi$ ; range $y \le -4$ or $y \ge 2$ ; vertices $\left(\frac{\pi}{4}, 2\right)$ and $\left(\frac{3\pi}{4}, -4\right)$

Learn It on Martian Lab

Martian Lab teaches this in the Graphing Trigonometric Functions lesson of its MHF4U Trigonometric Functions unit. Watch it, then take the Trigonometric Functions practice tests to see if it stuck.

Watch the Graphing Trigonometric Functions lesson free

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Free Trigonometric Functions practice test, with solutions

More Trigonometric Functions topics: Solving Trig Equations · Graphing Sine and Cosine with Transformations · Sinusoidal Word Problems

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Proving Trig Identities · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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