Introduction
The tangent and reciprocal trigonometric functions are closely connected to sine and cosine, but their graphs have features that are easy to misread. Some have vertical asymptotes; some never cross the horizontal axis; and their periods are not all the same.
A reliable sketch begins with a few questions: Where is the function defined? What is its period? Where does it equal zero, if anywhere? Where does it have a local maximum or minimum? Answering these questions identifies the important points and gaps before you draw the curves between them.
Definitions and Graph Features
For angles \(x\) measured in radians, tangent is the quotient of sine by cosine. The reciprocal functions are defined by taking the reciprocal of sine or cosine, while cotangent is the reciprocal of tangent wherever that reciprocal is defined:
A function is undefined wherever its displayed denominator is zero. A vertical asymptote at \(x=a\) occurs when the function’s values grow without bound in magnitude as \(x\) approaches \(a\) from one or both sides. The asymptote is the vertical line \(x=a\), not a point on the graph.
A zero is an input where the function equals zero; on a graph, it is an \(x\)-intercept. A local maximum or local minimum is a function value that is respectively no smaller or no greater than nearby values. These local extrema are the turning points used to sketch secant and cosecant.
Periods, Zeros, Asymptotes, and Extrema
Sine and cosine repeat every \(2\pi\). Their zeros and the points where they equal \(1\) or \(-1\) therefore determine the key features of their quotients and reciprocals. The resulting facts are summarized below. In the table, \(k\) may be any integer.
| Function | Fundamental period | Zeros | Vertical asymptotes | Local extrema |
|---|---|---|---|---|
| \(\tan x\) | \(\pi\) | \(x=k\pi\) | \(x=\frac{\pi}{2}+k\pi\) | None |
| \(\cot x\) | \(\pi\) | \(x=\frac{\pi}{2}+k\pi\) | \(x=k\pi\) | None |
| \(\sec x\) | \(2\pi\) | None | \(x=\frac{\pi}{2}+k\pi\) | \(x=k\pi\), with value \((-1)^k\) |
| \(\csc x\) | \(2\pi\) | None | \(x=k\pi\) | \(x=\frac{\pi}{2}+k\pi\), with value \((-1)^k\) |
For tangent and cotangent, the period is \(\pi\), rather than \(2\pi\), because changing \(x\) by \(\pi\) changes the signs of both sine and cosine. Their quotient is unchanged. For secant and cosecant, the reciprocal also has period \(2\pi\), inherited from its denominator.
Why the Asymptotes Occur
The denominator explains the location of each vertical asymptote. For secant, \(\cos x=0\) at \(x=\frac{\pi}{2}+k\pi\). At each such angle, the numerator \(1\) stays nonzero while the denominator approaches zero, so the quotient grows without bound in magnitude. For cosecant, the same reasoning applies where \(\sin x=0\), namely at \(x=k\pi\).
For tangent, \(\cos x\) is the denominator. At \(x=\frac{\pi}{2}+k\pi\), the cosine is zero while the sine is \(1\) or \(-1\), so \(\tan x\) becomes unbounded in magnitude nearby. For cotangent, the denominator is \(\sin x\), which is zero at \(x=k\pi\); the numerator there is \(1\) or \(-1\). This is why cotangent has asymptotes at integer multiples of \(\pi\), not at the zeros of cosine.
The secant and cosecant extrema also follow directly from their definitions. At points where \(\cos x=1\) or \(-1\), secant equals \(1\) or \(-1\). Nearby, the cosine has the same sign but magnitude less than \(1\), so its reciprocal has magnitude greater than \(1\). Thus secant has a local minimum of \(1\) where \(\cos x=1\), and a local maximum of \(-1\) where \(\cos x=-1\). The same argument using sine gives the cosecant extrema: a local minimum of \(1\) where \(\sin x=1\), and a local maximum of \(-1\) where \(\sin x=-1\).
These observations also explain why secant and cosecant have no zeros: the reciprocal of a nonzero number cannot equal zero. Their ranges are \((-\infty,-1]\cup[1,\infty)\). Tangent and cotangent, by contrast, cross the horizontal axis at their listed zeros. On each interval between consecutive asymptotes, tangent increases from negative to positive values, while cotangent decreases from positive to negative values. Neither has a local maximum or minimum within such an interval.
A Reliable Sketching Process
Find the zeros of the function’s denominator and draw the corresponding vertical lines.
Use the table to place intercepts and, for secant or cosecant, the turning points.
Repeat the pattern at the correct interval, and place each branch on the side of the horizontal axis indicated by the function’s sign.
Do not connect across an asymptote. Make the curve approach the asymptote without touching it.
Worked Examples
Worked Example: Sketching Tangent on \([-\pi,\pi]\)
First, \(\tan x=\frac{\sin x}{\cos x}\), so it is undefined where \(\cos x=0\). In the interval \([-\pi,\pi]\), these inputs are \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). Draw vertical asymptotes at both values.
The zeros occur where \(\sin x=0\) and \(\cos x\ne0\). Thus the zeros on the interval are \(-\pi\), \(0\), and \(\pi\). The endpoints are included and are zeros because \(\tan(-\pi)=0\) and \(\tan(\pi)=0\).
Use exact intermediate points to see the scale and signs of the branches:
These values follow from the unit-circle sine and cosine values: for example, \(\tan(\frac{\pi}{4})=(\frac{\sqrt{2}}{2})/(\frac{\sqrt{2}}{2})=1\), and \(\tan(\frac{3\pi}{4})=(\frac{\sqrt{2}}{2})/(-\frac{\sqrt{2}}{2})=-1\). The other two follow in the same way from their sine and cosine signs.
Between \(-\pi\) and \(-\frac{\pi}{2}\), the branch rises from the zero at \(-\pi\) toward positive unbounded values. Between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), it rises from negative unbounded values, passes through \((0,0)\), and continues toward positive unbounded values. Between \(\frac{\pi}{2}\) and \(\pi\), it rises from negative unbounded values to the zero at \(\pi\). There are no local extrema.
Check: the asymptotes are precisely where the denominator \(\cos x\) is zero, and the listed zeros have \(\sin x=0\) with nonzero cosine. The points \(\pm\frac{\pi}{4}\) and \(\pm\frac{3\pi}{4}\) also give the stated values by direct substitution.
Worked Example: Sketching Secant on \([-\pi,\pi]\)
Since \(\sec x=\frac{1}{\cos x}\), it is undefined when \(\cos x=0\). On this interval the asymptotes are \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\). Secant has no zeros because its numerator is \(1\).
At the remaining key integer multiples of \(\pi\), compute the values:
The point \((0,1)\) is a local minimum. The two points \((-\pi,-1)\) and \((\pi,-1)\) are local maxima on the respective branches. On \((-\frac{\pi}{2},\frac{\pi}{2})\), cosine is positive, so secant is positive and forms a U-shaped branch with its lowest point at \((0,1)\). On the outer intervals, cosine is negative, so secant is negative; each branch lies at or below \(-1\) and falls without bound in magnitude near the asymptote.
Check: substituting the three marked inputs into \(1/\cos x\) gives \(-1,1,-1\), as shown. The graph has no \(x\)-intercepts, and its asymptotes agree with the zeros of the denominator.
Worked Example: Sketching Cosecant on \([0,2\pi]\)
Cosecant is \(\csc x=\frac{1}{\sin x}\), so it is undefined where \(\sin x=0\). On \([0,2\pi]\), those inputs are \(0\), \(\pi\), and \(2\pi\); draw asymptotes at all three. Cosecant has no zeros.
Its turning points occur where sine equals \(1\) or \(-1\):
On \((0,\pi)\), sine is positive, so the cosecant branch lies above the horizontal axis and has its local minimum at \((\frac{\pi}{2},1)\). It grows without bound in magnitude toward each endpoint asymptote. On \((\pi,2\pi)\), sine is negative, so the branch lies below the axis and has its local maximum at \((\frac{3\pi}{2},-1)\).
Check: the sine values at the two marked inputs are \(1\) and \(-1\), whose reciprocals are \(1\) and \(-1\). The asymptotes occur exactly at the zeros of the denominator, and neither branch crosses the axis.
Worked Example: Sketching Cotangent on \([0,2\pi]\)
Because \(\cot x=\frac{\cos x}{\sin x}\), its denominator vanishes at \(0\), \(\pi\), and \(2\pi\). These give the vertical asymptotes on the interval. The zeros occur when cosine is zero while sine is nonzero, so they are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).
Check the intercepts by substitution:
On each interval between asymptotes, cotangent decreases. It falls from positive unbounded values to negative unbounded values, crossing the axis at the midpoint: \(\frac{\pi}{2}\) on the first interval and \(\frac{3\pi}{2}\) on the second. It has no local extrema.
Check: both zeros have zero numerator and nonzero denominator. The asymptote inputs instead have zero denominator and nonzero numerator, so they are not points on the graph.
Common Mistake: Confusing the Denominators
A common error is to put the asymptotes of cotangent where \(\cos x=0\), because cosine appears in its definition. That uses the numerator rather than the denominator. For instance, at \(x=\frac{\pi}{2}\),
So \(x=\frac{\pi}{2}\) is an intercept, not an asymptote. The correction is to set the denominator \(\sin x\) equal to zero: \(\sin x=0\) at \(x=k\pi\), which gives cotangent’s vertical asymptotes. The same denominator check prevents the related error of placing cosecant asymptotes at zeros of cosine; cosecant’s denominator is sine.
Reading the Graphs as a Set
The four graphs are easier to remember when paired by their denominators. Tangent and secant both have cosine in the denominator, so both have asymptotes at \(x=\frac{\pi}{2}+k\pi\). Cotangent and cosecant both have sine in the denominator, so both have asymptotes at \(x=k\pi\). Within each pair, however, the zeros and extrema differ: tangent and cotangent can equal zero, while secant and cosecant cannot.
The most useful sketching habit is to mark the asymptotes before drawing any curve. Then place the zeros or extrema, check the sign on each interval, and use the period to repeat the pattern. This keeps separate branches from being accidentally joined and makes the graph’s structure visible.
Check Your Understanding
Answer these questions using the definitions and graph features above.
- What are the fundamental periods of tangent, cotangent, secant, and cosecant?
- Where are the vertical asymptotes of \(\sec x\), and which denominator determines them?
- Why does \(\csc x\) have no zeros?
- Find the zeros and vertical asymptotes of \(\cot x\) in \([0,2\pi]\).
- What are the local extrema of \(\sec x\) at \(x=0\) and \(x=\pi\), and what are their function values?
