Graphing the reciprocal trigonometric functions
Learn to graph the reciprocal trigonometric functions — secant, cosecant and cotangent — for Year 11 Specialist Mathematics in Queensland (QCAA). Each graph is built straight from its parent sine, cosine or tangent curve.
You will learn how every zero of the parent becomes a vertical asymptote and every peak becomes a turning point, then read off the period, range and asymptotes of sec, cosec and cot — and see how a coefficient stretches the range or shortens the period.
Theory
The graphs of the reciprocal trigonometric functions — secant, cosecant and cotangent — are sketched in Year 11 Specialist Mathematics (QCAA, Queensland) straight from their parent sine, cosine and tangent curves. This page shows how each zero of the parent becomes a vertical asymptote and each peak becomes a turning point, and reads off period, range and asymptotes.
The three reciprocal functions are \(\sec x=\dfrac{1}{\cos x}\), \(\operatorname{cosec} x=\dfrac{1}{\sin x}\) and \(\cot x=\dfrac{\cos x}{\sin x}=\dfrac{1}{\tan x}\). (In Australian schools cosecant is written cosec.) Their graphs are built directly from the parent \(\cos x\), \(\sin x\) and \(\tan x\) curves.
A vertical asymptote appears wherever the parent is zero, because you cannot divide by zero. So \(\sec x\) has asymptotes where \(\cos x=0\) (at \(x=\dfrac{\pi}{2}+k\pi\)), while \(\operatorname{cosec} x\) and \(\cot x\) have asymptotes where \(\sin x=0\) (at \(x=k\pi\)). As \(x\) nears an asymptote the graph shoots off to \(+\infty\) or \(-\infty\).
Where the parent reaches a peak of \(\pm 1\), the reciprocal has a turning point at \(\pm 1\). Because \(|\cos x|\le 1\) and \(|\sin x|\le 1\), both \(\sec x\) and \(\operatorname{cosec} x\) stay outside \((-1,1)\): their range is \((-\infty,-1]\cup[1,\infty)\) and they have no \(x\)-intercepts.
The period of \(\sec x\) and \(\operatorname{cosec} x\) is \(2\pi\), matching cosine and sine. The cotangent is different: it has period \(\pi\), its range is all real numbers, it does cross the axis (at \(x=\dfrac{\pi}{2}+k\pi\), where \(\cos x=0\)), and it decreases on every branch. A coefficient such as \(a\) in \(a\sec x\) scales the range, and a coefficient of \(x\) such as in \(\sec 2x\) divides the period.
The three functions and where each is undefined (its asymptotes):
Asymptotes of \(\sec x\) (where \(\cos x=0\)):
Asymptotes of \(\operatorname{cosec} x\) and \(\cot x\) (where \(\sin x=0\)):
Periods and ranges:
How to sketch a reciprocal trig graph
- Sketch the parent lightly: draw \(\cos x\) for \(\sec x\), or \(\sin x\) for \(\operatorname{cosec} x\) (for \(\cot x\), use \(\sin x\) to find the asymptotes and \(\cos x\) for the intercepts).
- Mark the asymptotes: draw a vertical dashed line at every zero of the parent — the reciprocal is undefined there.
- Mark the turning points: where the parent peaks at \(+1\) put a minimum at \(+1\); where it dips to \(-1\) put a maximum at \(-1\).
- Draw each branch curving from a turning point up (or down) towards the neighbouring asymptotes, then read off the period and range. Apply any coefficients last: a factor \(a\) scales the range, a factor of \(x\) divides the period.
Asymptotes occur where the denominator \(\sin x\) is zero:
| \(\operatorname{cosec} x\) | \(=\) | \(\dfrac{1}{\sin x}\) |
| \(\sin x\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0,\ \pi,\ 2\pi\) |
The period matches sine, and the reciprocal never enters \((-1,1)\):
| \(\text{period}\) | \(=\) | \(2\pi\) |
| \(|\operatorname{cosec} x|\) | \(\ge\) | \(1\) |
Period \(2\pi\); asymptotes \(x=0,\ \pi,\ 2\pi\); range \((-\infty,-1]\cup[1,\infty)\).
On this branch \(\cos x\) is negative; its greatest value is \(-1\), at \(x=\pi\):
| \(\sec x\) | \(=\) | \(\dfrac{1}{\cos x}\) |
| \(\cos x\) | \(=\) | \(-1\) |
| \(x\) | \(=\) | \(\pi\) |
Evaluate \(\sec x\) there to get the \(y\)-value:
| \(\sec\pi\) | \(=\) | \(\dfrac{1}{-1}\) |
| \(=\) | \(-1\) |
The local maximum is \((\pi,-1)\).
The coefficient \(2\) halves the base period; asymptotes need \(\sin 2x=0\):
| \(\cot 2x\) | \(=\) | \(\dfrac{\cos 2x}{\sin 2x}\) |
| \(\sin 2x\) | \(=\) | \(0\) |
| \(2x\) | \(=\) | \(0,\ \pi,\ 2\pi,\ 3\pi,\ 4\pi\) |
| \(x\) | \(=\) | \(0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2},\ 2\pi\) |
Count the asymptotes, and halve the cotangent period \(\pi\):
| \(\text{asymptotes}\) | \(=\) | \(5\) |
| \(\text{period}\) | \(=\) | \(\dfrac{\pi}{2}\) |
There are \(5\) vertical asymptotes and the period is \(\dfrac{\pi}{2}\).
Multiplying by \(3\) triples every output, so the range bound triples:
| \(3\operatorname{cosec} x\) | \(=\) | \(\dfrac{3}{\sin x}\) |
| \(|\operatorname{cosec} x|\) | \(\ge\) | \(1\) |
| \(|3\operatorname{cosec} x|\) | \(\ge\) | \(3\) |
On \((0,\pi)\), \(\sin x\) is greatest at \(1\), giving the least value:
| \(\text{on }(0,\pi):\ \sin x\) | \(\le\) | \(1\) |
| \(\dfrac{3}{1}\) | \(=\) | \(3\) |
Range \((-\infty,-3]\cup[3,\infty)\); least value on \((0,\pi)\) is \(3\).
Common pitfalls
Frequently asked questions
How do you sketch the graph of \(y=\sec x\)?
Sketch \(\cos x\) lightly, draw a vertical asymptote at each of its zeros (\(x=\dfrac{\pi}{2}+k\pi\)), put a minimum at \(+1\) where cosine peaks at \(+1\) and a maximum at \(-1\) where it dips to \(-1\), then curve each branch up or down towards the asymptotes.
Where are the asymptotes of \(y=\operatorname{cosec} x\)?
Wherever \(\sin x=0\), that is at \(x=0,\pi,2\pi,\dots\) (in general \(x=k\pi\)), because \(\operatorname{cosec} x=\dfrac{1}{\sin x}\) is undefined there.
What is the period of \(y=\cot x\)?
The period of \(\cot x\) is \(\pi\), the same as \(\tan x\). Do not use \(2\pi\); a coefficient of \(x\) divides it further, so \(\cot 2x\) has period \(\dfrac{\pi}{2}\).
Why do \(\sec x\) and \(\operatorname{cosec} x\) have no \(x\)-intercepts?
Because \(|\cos x|\le 1\) and \(|\sin x|\le 1\), their reciprocals satisfy \(|\sec x|\ge 1\) and \(|\operatorname{cosec} x|\ge 1\). The graph never enters \((-1,1)\), so it never touches the \(x\)-axis.
What is the range of \(y=\sec x\)?
The range is \((-\infty,-1]\cup[1,\infty)\); the same is true for \(\operatorname{cosec} x\). For \(a\sec x\) the bound scales to \(a\), giving \((-\infty,-a]\cup[a,\infty)\).
How does a number in front, like \(2\operatorname{cosec} x\), change the graph?
A factor \(a\) stretches the graph vertically, so the turning points move from \(\pm1\) to \(\pm a\) and the range becomes \((-\infty,-a]\cup[a,\infty)\). The asymptotes and period do not change.