The first graphs in the course with vertical asymptotes, and every feature of them inherited from cosine and sine by reciprocation. Covers where the asymptotes come from and how to determine which way each branch runs, the two values reciprocation leaves fixed, the range being everything of absolute value at least one, the absence of any amplitude, and the practical method of drawing the parent sinusoid first and reciprocating it point by point.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.5 Graphs of the Trigonometric Functions, pp. 800-804
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-804 — the pages these objectives are drawn from
Warm-up
You can graph the cosine curve. The secant graph is that curve with every height replaced by its reciprocal, and nothing else.
Discussion prompt
The cosine takes the values 1, then root two over two, then 0, then negative root two over two, then negative 1 across the first half of its cycle. Take the reciprocal of each. What goes wrong, and where?
Hint: Work through them one at a time and see which one you cannot do.
Answer:
\[ 1 \to 1, \quad \tfrac{\sqrt{2}}{2} \to \sqrt{2}, \quad 0 \to \text{undefined}, \quad -\tfrac{\sqrt{2}}{2} \to -\sqrt{2}, \quad -1 \to -1 \]
Four of the five reciprocate fine. The middle one does not exist, and that single failure is the whole difference between this lesson and the last one.
Notice also that 1 and negative 1 came back unchanged while root two over two, about 0.707, grew to about 1.414. Reciprocation fixes the extremes and pushes everything else outward, which is exactly what the graph will show.
Concept
The secant is one over the cosine and the cosecant is one over the sine. Every feature of their graphs — asymptotes, range, period, parity — is a consequence of that single fact applied to a curve you already know how to draw.
\[ \sec(x) = \frac{1}{\cos(x)}, \qquad \csc(x) = \frac{1}{\sin(x)} \]
So the practical method is never to plot these from scratch. Draw the parent sinusoid, mark its zeros as asymptotes, mark its peaks and troughs as the turning points of the new curve, and fill in between.
Figure (svg): One cycle of the secant curve with vertical asymptotes at pi over two and three pi over two, drawn above the dotted cosine curve it is the reciprocal of
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-800
Section
Section 1
Concept
The secant is undefined wherever the cosine is zero, which is at every odd multiple of pi over two. Near such a point the cosine is small, so its reciprocal is large — and the graph runs off to infinity.
Which way a branch runs is decided by the sign the parent approaches zero from. As x rises to pi over two, the cosine falls to zero through positive values, so the secant rises to positive infinity; just past pi over two the cosine is negative, so the secant comes up from negative infinity.
Figure (svg): A close-up of the secant curve either side of the asymptote at pi over two, showing it rising to positive infinity on the left and coming up from negative infinity on the right
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-800
Picture it
An asymptote has two sides, and they need not behave the same way.
Figure (svg): A close-up of the secant curve either side of the asymptote at pi over two, showing it rising to positive infinity on the left and coming up from negative infinity on the right
This is the one place where the sign of the parent matters more than its size. Asking which side of zero the cosine is on answers the question completely.
Worked example
Find them by solving for the zeros of the parent, never by inspecting the reciprocal.
\[ \text{Find all vertical asymptotes of } y = \sec(x) \text{ and of } y = \csc(x). \]
Set the parent function to zero for secant
Why: Secant is undefined exactly where the cosine vanishes.
\[ \cos x = 0 \]
Solve
Why: The cosine is zero on the y-axis, at pi over two plus any multiple of pi.
\[ x = \frac{\pi}{2} + \pi k \]
Do the same for cosecant
Why: Cosecant is undefined where the sine vanishes.
\[ \sin x = 0 \]
Solve
Why: The sine is zero on the x-axis, at every multiple of pi.
\[ x = \pi k \]
Figure (svg): The solution to Worked example locate the asymptotes shown as a ladder of expressions, one row per legal move
\[ \sec: \; x = \tfrac{\pi}{2} + \pi k \qquad \csc: \; x = \pi k, \quad k \in \mathbb{Z} \]
Verify: check one of each
Why: At pi over two the cosine is zero so the secant is undefined, and at pi the sine is zero so the cosecant is undefined. Meanwhile at pi over two the sine is 1, so the cosecant is defined and equals 1 — confirming that the two functions have different asymptotes.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-801
Matching
Every asymptote is a zero of something.
Match the pairs
Why: Secant and cosecant have asymptotes exactly at the zeros of their parents, which are the odd multiples of pi over two for the cosine and the multiples of pi for the sine. Cosine and sine themselves are defined everywhere, since neither has a denominator, so neither has an asymptote anywhere. That contrast is the entire structural difference between this lesson and the previous one.
Worked example
Two questions per asymptote: what does the parent do on each side?
\[ \text{Describe the behaviour of } y = \csc(x) \text{ on either side of } x = \pi. \]
Approach pi from the left
Why: Just below pi the sine is small and positive, since the sine is positive throughout quadrant two.
\[ \sin \to 0\text{ from above} \]
Reciprocate
Why: One over a small positive number is a large positive number.
\[ \csc \to + \infty \]
Approach pi from the right
Why: Just above pi the sine is small and negative, since quadrant three has negative sine.
\[ \sin \to 0\text{ from below} \]
Reciprocate
Why: One over a small negative number is a large negative number.
\[ \csc \to - \infty \]
Figure (svg): The solution to Worked example determine the direction of each branch shown as a ladder of expressions, one row per legal move
\[ \text{as } x \to \pi^-, \; \csc(x) \to +\infty; \qquad \text{as } x \to \pi^+, \; \csc(x) \to -\infty \]
Verify: test two nearby values
Why: The sine of 3.1 is about 0.0416, so the cosecant is about 24 — large and positive. The sine of 3.2 is about negative 0.0584, so the cosecant is about negative 17 — large and negative. The two sides genuinely run in opposite directions.
Trap
A student draws the secant graph with both branches at pi over two rising to positive infinity, on the grounds that the function is blowing up there.
But blowing up says nothing about the sign. The cosine changes sign as it passes through zero, so its reciprocal changes sign too.
On the left of pi over two the cosine is positive, so the secant is positive and rises to plus infinity. On the right the cosine is negative, so the secant is negative and comes up from minus infinity.
\[ \text{as } x \to \tfrac{\pi}{2}^-, \; \sec(x) \to +\infty; \qquad \text{as } x \to \tfrac{\pi}{2}^+, \; \sec(x) \to -\infty \]
The reliable habit: look at the parent curve either side of its zero. If the parent crosses from positive to negative, the reciprocal jumps from plus infinity to minus infinity. Only if the parent touched zero without crossing would both branches go the same way, and cosine and sine never do that.
Sorting
Ask what the parent does, then reciprocate.
Sort into buckets
Sort each approach by whether the function runs to positive or negative infinity.
Prediction
A function is one over some parent function, and the parent has a zero at x equal to 3.
Predict first
What must the reciprocal function do at x equal to 3?
Correct: It has a vertical asymptote there.
Why: A zero in the denominator makes the function undefined, and near it the denominator is small so the quotient is large. That is precisely a vertical asymptote. Note that this is a general fact about reciprocals rather than anything specific to trigonometry: every zero of a function becomes an asymptote of its reciprocal, and every asymptote of a function becomes a zero of its reciprocal.
Socratic
The cosine crosses the x-axis twice per cycle. The secant never does.
Discussion prompt
Explain why, using the definition rather than the picture.
Hint: What would it mean for the secant to be zero?
Answer:
\[ \sec(x) = 0 \;\Longrightarrow\; \frac{1}{\cos(x)} = 0 \]
A fraction is zero only when its numerator is zero, and this numerator is the constant 1, which is never zero. So the equation has no solutions and the secant is never zero.
Geometrically, the places where the cosine crosses zero are exactly the places where the secant is undefined, so the secant is not merely nonzero there — it is absent. A reciprocal turns zeros into holes, which is why neither secant nor cosecant has an x-intercept anywhere on its graph.
Section
Section 2
Concept
The cosine takes every value from negative one to one. Reciprocating that interval turns it inside out: values near zero become huge, values of size one stay put, and no value strictly between negative one and one is ever produced.
Theorem 10.24 — Secant and cosecant each have range all values of absolute value at least one, are continuous and smooth on their domains, and have period two pi. Secant is even and cosecant is odd.
\[ \text{range}(\sec) = \text{range}(\csc) = (-\infty, -1] \cup [1, \infty) \]
Figure (svg): A table of the four rules governing how reciprocation transforms a value: one goes to one, negative one to negative one, a small value to a large one, and zero to an asymptote
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 802-802
Picture it
Knowing what happens to each kind of value is enough to sketch either curve from its parent.
Figure (svg): A table of the four rules governing how reciprocation transforms a value: one goes to one, negative one to negative one, a small value to a large one, and zero to an asymptote
The two fixed points are what anchor the drawing. Mark every peak and trough of the parent first, because those points belong to both curves.
Worked example
This is exactly how the book constructs the graph. No new values are computed.
\[ \text{Build a table for } y = \sec(x) \text{ on } [0, 2\pi] \text{ from the cosine table.} \]
Start from the cosine values at the eighth marks
Why: Nine values, from 1 down to negative 1 and back.
\[ 1, \sqrt{2} / 2, 0, -\sqrt{2} / 2, -1,... \]
Reciprocate the nonzero ones
Why: One over root two over two is root two.
Mark the two failures as asymptotes
Why: The cosine is zero at pi over two and three pi over two.
\[ \text{asymptotes at } \frac{\pi}{2}, 3 \pi / 2 \]
Plot and join within each branch
Why: Three branches on this interval, each smooth.
Figure (svg): The solution to Worked example build the secant table by reciprocating shown as a ladder of expressions, one row per legal move
\[ \sec: \; 1, \; \sqrt{2}, \; \text{asym}, \; -\sqrt{2}, \; -1, \; -\sqrt{2}, \; \text{asym}, \; \sqrt{2}, \; 1 \]
Verify: check the two touching points
Why: At x equal to zero both the cosine and the secant are 1, and at x equal to pi both are negative 1. Those are the only two places on this interval where the curves meet, exactly as the fixed-point rule predicts.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-800
Comparison
Fill the blanks from memory. Every entry on the right is a consequence of the entry on the left.
Comparison matrix
| Property | cosine | secant |
|---|---|---|
| Domain | all real numbers | all except odd multiples of pi over 2 |
| Range | from -1 to 1 | everything of size at least 1 |
| Period | two pi | two pi |
| Even or odd | even | even |
| Amplitude | 1 | none; the range is unbounded |
The domain shrinks and the range turns inside out, but the period and the parity pass through unchanged. That is the general pattern for any function and its reciprocal.
Worked example
The period does not have to be found; it follows from the parent's.
\[ \text{Show that } \sec(x) \text{ has period } 2\pi. \]
Use the parent's periodicity
Why: The cosine repeats every two pi.
\[ \cos(x + 2 \pi) = \cos x \]
Reciprocate both sides
Why: Valid wherever both are defined, which is the same set of x.
\[ \sec(x + 2 \pi) = \sec x \]
Argue no smaller period works
Why: Two secants are equal exactly when their cosines are, so a smaller period for secant would give one for cosine.
Conclude
Why: The period is exactly two pi, inherited.
\[ \text{period } 2 \pi \]
Figure (svg): The solution to Worked example why the period is inherited shown as a ladder of expressions, one row per legal move
\[ \sec(x + 2\pi) = \frac{1}{\cos(x + 2\pi)} = \frac{1}{\cos(x)} = \sec(x) \]
Verify: check against the graph
Why: The fundamental cycle drawn above runs from zero to two pi and contains two asymptotes and two touching points. Continuing past two pi repeats that pattern exactly, so a period of two pi is what the picture shows.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-800
Error analysis
A student describes the graph of three secant x.
Annotate
On: \( y = 3\sec(x): \quad \text{amplitude } 3 \)
The book says this explicitly: since the ranges of secant and cosecant are unbounded, there is no amplitude associated with these curves. The coefficient is still worth naming — it is a vertical stretch — but calling it an amplitude claims something false about the range.
Sorting
The range is everything of absolute value at least one.
Sort into buckets
Sort each value by whether some angle has that secant.
Prediction
The cosine of some angle is 0.1.
Predict first
What is the secant of that angle?
Correct: 10.
Why: The secant is the reciprocal, and one over 0.1 is 10. This illustrates the outward push directly: a cosine close to zero produces a very large secant, which is what generates the asymptotes. The distractor 0.9 comes from subtracting from 1 rather than reciprocating, and the negative option would require the cosine itself to be negative.
Edge cases
The range of secant excludes everything strictly between negative one and one.
Discussion prompt
The values 1 and negative 1 are included. What is special about them, and what does that mean for the graph?
Hint: What happens when you reciprocate 1?
Answer:
\[ \frac{1}{1} = 1, \qquad \frac{1}{-1} = -1 \]
They are the fixed points of reciprocation — the only two real numbers equal to their own reciprocals. So wherever the cosine reaches 1 or negative 1, the secant reaches the same value.
On the graph this means the two curves touch at every maximum and minimum of the cosine, and nowhere else. Those touching points are the turning points of each secant branch, and marking them first is the fastest way to sketch the curve. It is also why the range is closed at 1 and negative 1 rather than open: those values are genuinely attained.
Section
Section 3
Concept
The quarter-mark method still works, and the sensible addition is to sketch the associated sinusoid in dotted lines before drawing anything else. Then the asymptotes and turning points are read off it.
For the associated curve, replace the secant by a cosine or the cosecant by a sine and change nothing else. That curve is not part of the answer, but drawing it makes the answer almost automatic.
Figure (svg): One cycle of the function one minus two secant of two x, with its two vertical asymptotes and the dotted cosine curve one minus two cosine of two x it reciprocates
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 802-803
Picture it
The dotted curve is a construction line. It is worth drawing every time and worth labelling as not part of the graph.
Figure (svg): One cycle of the function one minus two secant of two x, with its two vertical asymptotes and the dotted cosine curve one minus two cosine of two x it reciprocates
Notice how every feature of the solid curve is read off the dotted one: two zeros become two asymptotes, and the maximum and minimum become the turning points at negative one and three.
Worked example
Example 10.5.4, part 1. Quarter marks, then reciprocate.
\[ \text{Graph one cycle of } f(x) = 1 - 2\sec(2x) \text{ and state the period.} \]
Set the argument to the quarter marks and solve
Why: The argument is 2x.
\[ x = 0, \frac{\pi}{4}, \frac{\pi}{2}, 3 \pi / 4, \pi \]
Substitute each back into f
Why: Two of the five give undefined values.
\[ -1, \text{undef}, 3, \text{undef}, -1 \]
Mark the two failures as asymptotes
Why: They sit at pi over four and three pi over four.
Read the period off the interval
Why: The cycle runs from zero to pi.
\[ \text{period } = \pi \]
Figure (svg): The solution to Worked example one cycle of a transformed secant shown as a ladder of expressions, one row per legal move
\[ \text{period } \pi; \quad \text{asymptotes } x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}; \quad \text{points } (0,-1), \left(\tfrac{\pi}{2}, 3\right), (\pi, -1) \]
Verify: check against the associated cosine
Why: The associated curve is one minus two cosine of two x, which has maximum 3 and minimum negative 1 and crosses its midline of 1 at pi over four and three pi over four. Those crossings are the cosine's own zeros only if the midline were zero — here the asymptotes come from where the SECANT's argument makes cos 2x zero, namely those same two points. The extremes match the turning points exactly.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 802-803
Fill the middle
Find the asymptotes of one cycle of y equals secant of three x.
Fill in the blanks
3x = \fracpi/6pi/2 \;\Longrightarrow\; x = ___ \qquad 3x = \frac______ \;\Longrightarrow\; x = ___
Why: Asymptotes occur where the argument is an odd multiple of pi over two, since that is where the cosine of the argument vanishes. Solving each gives pi over six and pi over two. The period is two pi over three, and the two asymptotes are one third and one half of the way through it — a third and two thirds of the interval from zero to two pi over three, as expected.
Worked example
Example 10.5.4, part 2. Fix the sign of the coefficient before starting.
\[ \text{Graph one cycle of } g(x) = \csc(\pi - \pi x) - \tfrac{5}{3} \text{ and state the period.} \]
Set the argument to the quarter marks
Why: The argument is pi minus pi x.
\[ \pi - \pi x = 0, \frac{\pi}{2}, \pi, 3 \pi / 2, 2 \pi \]
Solve each for x
Why: The negative coefficient makes the x-values decrease.
\[ x = 1, \frac{1}{2}, 0, -\frac{1}{2}, -1 \]
Substitute back
Why: The cosecant is undefined where its argument is a multiple of pi.
\[ \text{undef}, -\frac{2}{3}, \text{undef}, -\frac{8}{3}, \text{undef} \]
Read the period
Why: The cycle runs from negative one to one.
\[ \text{period } = 2 \]
Figure (svg): The solution to Worked example a cosecant with a negative coefficient shown as a ladder of expressions, one row per legal move
\[ \text{period } 2; \quad \text{asymptotes } x = -1, 0, 1; \quad \text{points } \left(\tfrac{1}{2}, -\tfrac{2}{3}\right), \left(-\tfrac{1}{2}, -\tfrac{8}{3}\right) \]
Verify: check with the period formula
Why: Writing the argument as negative pi times the quantity x minus one, the coefficient of x has absolute value pi, so the period is two pi over pi, which is 2. That matches the interval length from negative one to one.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 803-803
Trap
A student sketches y equals secant x by drawing the cosine and putting asymptotes where the cosine peaks.
But a peak is where the cosine equals 1, and one over 1 is 1 — a perfectly ordinary value. Nothing goes wrong there at all.
The asymptotes go where the cosine is zero, and the peaks and troughs are where the two curves touch.
\[ \cos(x) = 0 \;\Longrightarrow\; \text{asymptote}; \qquad \cos(x) = \pm 1 \;\Longrightarrow\; \sec(x) = \pm 1 \]
The confusion is understandable because both are special points of the parent curve. The distinguishing question is simple: a reciprocal blows up where its denominator is zero, and a peak of the cosine is as far from zero as that function ever gets.
Ranking
Graph one cycle of a transformed secant.
Put in order
Why: The first three steps are identical to graphing a sinusoid, except that substituting back can now fail — and each failure is an asymptote rather than an error. Sketching the parent curve is what makes the last step easy, since every branch runs from an asymptote through a turning point to the next asymptote, and the turning points are the parent's extremes.
Prediction
One cycle of a secant graph is drawn.
Predict first
How many vertical asymptotes and how many branches does it contain?
Correct: Two asymptotes, three branches.
Why: The parent cosine crosses zero twice per cycle, so there are two asymptotes. Those two vertical lines cut one cycle into three pieces: a partial branch, a complete U-shaped branch, and another partial branch. The fundamental cycle drawn in the book shows exactly this — which is why the picture looks less symmetric than one might expect for a periodic function.
Explain it to yourself
The associated sinusoid is drawn dotted and is not part of the answer.
Discussion prompt
Explain what it is for, and why drawing it is worth the extra time rather than plotting the reciprocal points directly.
Hint: What does it tell you that a list of five points does not?
Answer:
The five points and two asymptotes tell you where the branches are anchored but not what shape they have. The parent curve supplies the shape: each secant branch is a U opening away from the parent's extreme, and its steepness follows from how fast the parent approaches zero.
It also serves as a check. If your secant curve ever crosses the parent anywhere except at the parent's peaks and troughs, something is wrong, since the two are equal only where the parent is 1 or negative 1. Having a curve to check against catches errors that a list of points cannot.
Section
Section 4
Concept
Period, phase shift and vertical shift all mean exactly what they meant for a sinusoid, because they describe horizontal and vertical placement rather than height. Amplitude does not survive, and asymptote positions take its place as the thing to state.
The last point in that list is worth pausing on: a vertical shift moves the turning points but leaves the asymptotes exactly where they were, because an asymptote is decided by the argument alone.
Figure (svg): Two columns separating the parameters that still make sense for a secant or cosecant graph from the one that does not
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 804-804
Picture it
Everything horizontal transfers unchanged. Only the vertical description has to be rethought.
Figure (svg): Two columns separating the parameters that still make sense for a secant or cosecant graph from the one that does not
This is a general pattern rather than a quirk of trigonometry: reciprocating a function preserves its horizontal structure exactly and completely rebuilds its vertical structure.
Worked example
Give the parameters that mean something and say plainly which one does not.
\[ \text{Describe } y = 4\sec\left(3x - \frac{\pi}{2}\right) + 2. \]
Read omega and compute the period
Why: The coefficient of x is 3.
\[ \text{period } = 2 \pi / 3 \]
Compute the phase shift
Why: Negative phi over omega, with phi equal to negative pi over two.
\[ \text{shift } = \frac{\pi}{6}\text{ right} \]
State the vertical shift
Why: The constant added at the end.
\[ \text{up } 2 \]
Locate the asymptotes instead of an amplitude
Why: Where the argument is an odd multiple of pi over two.
\[ 3 x - \frac{\pi}{2} = \frac{\pi}{2} + \pi k \]
Figure (svg): The solution to Worked example state everything that applies shown as a ladder of expressions, one row per legal move
\[ \text{period } \tfrac{2\pi}{3}, \quad \text{shift } \tfrac{\pi}{6} \text{ right}, \quad \text{up } 2, \quad \text{no amplitude} \]
Verify: find the turning points
Why: The associated cosine has extremes at 4 plus 2, which is 6, and negative 4 plus 2, which is negative 2. So the secant branches turn at heights 6 and negative 2, and the graph never takes any value strictly between them — the gap in the range is 8 units wide rather than 2, because of the stretch.
Sorting
An asymptote is decided by the argument alone.
Sort into buckets
Sort each transformation of y equals secant x.
Worked example
The asymptotes are decided entirely by the argument, so a vertical shift cannot touch them.
\[ \text{Compare the asymptotes and turning points of } y = \sec(x) \text{ and } y = \sec(x) + 3. \]
Find the asymptotes of the first
Why: Where the cosine of x is zero.
\[ x = \frac{\pi}{2} + \pi k \]
Find the asymptotes of the second
Why: Adding 3 does not change where the function is undefined.
\[ x = \frac{\pi}{2} + \pi k,\text{ the same} \]
Find the turning points of the first
Why: At heights 1 and negative 1.
\[ y = 1\text{ and } y = -1 \]
Find the turning points of the second
Why: Each raised by 3.
\[ y = 4\text{ and } y = 2 \]
Figure (svg): The solution to Worked example what a vertical shift does and does not move shown as a ladder of expressions, one row per legal move
\[ \text{asymptotes unchanged}; \quad \text{turning points } 1, -1 \;\to\; 4, 2 \]
Verify: check the range
Why: The first has range everything of size at least 1, so it misses the open interval from negative one to one. The second misses the open interval from 2 to 4 — the same gap, shifted up by 3. Notice the gap has moved entirely above the x-axis, so the shifted graph never takes a negative value at all.
Error analysis
A student describes the graph of cosecant x plus 2.
Annotate
On: \( \text{asymptotes at } y = 2 + \pi k \)
An asymptote of these graphs is a vertical line, located by where the function is undefined — which depends only on the argument. Nothing added outside the function can move it. The general rule: vertical transformations move horizontal features and leave vertical ones alone.
Faded example
Describe y equals negative cosecant of the quantity x over 2, minus 1.
Fill in the blanks
\text4 pi = \frac2 pi k___ = ___, \quad \text___ \frac______ = \pi k, \text___ x = ___
Why: The coefficient of x inside is one half, so the period is two pi divided by one half, which is four pi — the graph is stretched horizontally. The asymptotes come from setting the argument to a multiple of pi, giving x equal to two pi k, so they are spaced two pi apart, which is half a period as always for a cosecant. The negation flips the branches but moves nothing horizontally, and the shift of negative one lowers the turning points to zero and negative two.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about secant graphs is wrong.
Survives elimination: B
Why: Statement B is false. The asymptotes are located by solving for where the argument makes the parent function zero, and that equation involves only what is inside the function. Anything added or multiplied outside acts after the fact and cannot make a defined value undefined. The vertical shift moves the turning points and the entire range, but every asymptote stays exactly where it was.
Counterexample
A student proposes: since secant has no amplitude, the coefficient in front of it has no effect on the graph.
Discussion prompt
Give a specific counterexample and describe precisely what the coefficient does.
Hint: Compare the graphs of secant x and five secant x at a convenient point.
Answer:
\[ \sec(0) = 1 \qquad 5\sec(0) = 5 \]
The two graphs differ at every single point where they are defined, so the coefficient certainly has an effect. What it does is a vertical stretch: every value is multiplied by 5, so the turning points move from 1 and negative 1 out to 5 and negative 5.
The range changes accordingly, from everything of size at least 1 to everything of size at least 5 — a wider gap. So the coefficient controls the width of the gap in the range, which is a perfectly real and describable effect. It simply is not an amplitude, because there is no maximum for it to be half the distance to.
Section
Section 5
Concept
Since sine is cosine shifted a quarter period, cosecant is secant shifted a quarter period. Everything proved about one transfers to the other with that translation applied.
\[ \csc(x) = \sec\left(x - \tfrac{\pi}{2}\right) \]
Parity is the one property that does not simply translate, and it is inherited directly from the parent rather than from the shift.
Figure (svg): One cycle of the cosecant curve with vertical asymptotes at zero, pi and two pi, drawn above the dotted sine curve it is the reciprocal of
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 801-802
Picture it
Compare this with the secant figure from earlier. The shape is identical; only the horizontal placement differs.
Figure (svg): One cycle of the cosecant curve with vertical asymptotes at zero, pi and two pi, drawn above the dotted sine curve it is the reciprocal of
Every branch runs from an asymptote up or down through a turning point at plus or minus one and back to the next asymptote. There is exactly one branch between each consecutive pair of asymptotes.
Worked example
The cofunction identity does the work.
\[ \text{Show that } \csc(x) = \sec\left(x - \frac{\pi}{2}\right). \]
Write the right side in terms of cosine
Why: Secant is one over cosine.
\[ 1 / \cos(x - \frac{\pi}{2}) \]
Expand the cosine of a difference
Why: Matched terms, sign flips to a plus.
\[ \cos x \cos(\frac{\pi}{2}) + \sin x \sin(\frac{\pi}{2}) \]
Substitute the two quadrantal values
Why: The cosine of pi over two is zero and the sine is one.
\[ = \sin x \]
Conclude
Why: One over the sine is the cosecant.
\[ = \csc x \]
Figure (svg): The solution to Worked example verify the shift relationship shown as a ladder of expressions, one row per legal move
\[ \sec\left(x - \tfrac{\pi}{2}\right) = \frac{1}{\cos\left(x - \tfrac{\pi}{2}\right)} = \frac{1}{\sin(x)} = \csc(x) \]
Verify: check the asymptotes agree
Why: Secant has asymptotes at odd multiples of pi over two, and shifting right by pi over two moves them to multiples of pi — exactly where cosecant's asymptotes are. The shift relationship is consistent with both graphs.
Comparison
Fill the blanks from memory. They agree on more than they differ on.
Comparison matrix
| Property | secant | cosecant |
|---|---|---|
| Parent | cosine | sine |
| Asymptotes | odd multiples of pi over 2 | multiples of pi |
| Range | size at least 1 | size at least 1 |
| Period | two pi | two pi |
| Parity | even | odd |
| Defined at x = 0 | yes, equals 1 | no, asymptote there |
Three properties agree and three differ, and all three differences trace back to the quarter-period offset between cosine and sine.
Worked example
This is the one property that does not transfer through the shift.
\[ \text{Show that } \sec \text{ is even and } \csc \text{ is odd.} \]
Start with secant at a negated input
Why: Write it in terms of the cosine.
\[ \sec(-x) = 1 / \cos(-x) \]
Apply the even identity for cosine
Why: Cosine is even, so the negation disappears.
\[ = 1 / \cos(x) = \sec(x) \]
Do the same for cosecant
Why: Sine is odd, so a minus sign survives.
\[ \csc(-x) = \frac{1}{-\sin x} \]
Bring the minus out
Why: One over a negative is the negative of the reciprocal.
\[ = -\csc(x) \]
Figure (svg): The solution to Worked example parity of each shown as a ladder of expressions, one row per legal move
\[ \sec(-x) = \sec(x), \qquad \csc(-x) = -\csc(x) \]
Verify: check on the graphs
Why: The secant graph is symmetric about the y-axis, which is what even means. The cosecant graph has half-turn symmetry about the origin, which is what odd means — and the asymptote at zero is consistent with that, since an odd function that is defined at zero must vanish there, and this one is simply not defined there.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 802-802
Trap
\[ y = \csc(x): \quad \text{asymptotes at } x = \tfrac{\pi}{2} + \pi k \]
Reuse the asymptotes just learned for secant
Why: The two graphs look identical, so their descriptions get merged.
But the cosecant is defined at pi over two, where the sine is 1 and the cosecant is also 1. It is undefined at zero and pi instead.
\[ y = \csc(x): \quad \text{asymptotes at } x = \pi k \]
Go back to the parent every time
Why: Cosecant is one over sine, so its asymptotes are the zeros of the sine.
The two graphs really are the same shape, but they sit in different horizontal positions, and the asymptotes are the clearest marker of which is which. The quick test: the secant is defined at zero and the cosecant is not, since the cosine of zero is 1 and the sine of zero is 0.
Discrimination
One quick test distinguishes them at a glance.
Sort into buckets
Sort each described feature by which function it belongs to.
Prediction
You know the graph of the secant. You want the graph of the cosecant.
Predict first
What single transformation gets you there?
Correct: Shift right by pi over 2.
Why: The cosecant equals the secant of x minus pi over two, which is the secant graph translated right by a quarter period. This is the reciprocal of the relationship between the parents, and it must be, because reciprocation acts point by point and therefore commutes with any horizontal shift. Reflecting in either axis would change the parity, which is not what is wanted since the two functions have different parities for a different reason.
Explain it
A classmate is memorising the secant graph and the cosecant graph as two separate pictures.
Discussion prompt
Tell them in four sentences or fewer how to get both from one, and give them the one test for telling the two apart.
Hint: How many distinct shapes are actually involved?
Answer:
A usable answer: there is only one shape. Draw the cosine, put asymptotes at its zeros and touch points at its peaks and troughs, and you have the secant. For the cosecant do the same thing starting from the sine, which is the same curve moved a quarter period — so the cosecant graph is the secant graph moved a quarter period too.
The test for telling them apart: look at what happens at x equals zero. The secant has a point there, at height 1, because the cosine of zero is 1. The cosecant has an asymptote there, because the sine of zero is 0. One glance at the origin settles it.
Comparison
Fill the blanks from memory. Reciprocation preserves everything horizontal and rebuilds everything vertical.
Comparison matrix
| Feature | cosine and sine | secant and cosecant |
|---|---|---|
| Domain | everything | everything except the parent's zeros |
| Range | from -1 to 1 | everything of size at least 1 |
| Amplitude | half the vertical spread | none; the range is unbounded |
| Period | two pi over omega | two pi over omega, the same |
| Phase shift | negative phi over omega | negative phi over omega, the same |
| Asymptotes | none | at every zero of the parent |
Only two rows differ in substance, and both are about the vertical. The period and phase shift come through untouched because reciprocation acts on the output only.
Pattern
Whether you are graphing, describing, or locating asymptotes, the same five moves cover it.
When asked to describe rather than draw, the asymptote positions take the place amplitude would have occupied. Saying there is no amplitude is part of a complete answer rather than an omission from one.
OpenStax Algebra and Trigonometry 2e, §8.2 Graphs of the Other Trigonometric Functions §8.2
Check
Asymptotes. Go back to the parent.
Check your understanding
Where are the vertical asymptotes of y equals cosecant of 2x?
Answer: A
Why: The cosecant is undefined where its parent, the sine, is zero. Setting the argument 2x equal to a multiple of pi gives 2x equal to pi k, so x equals pi k over 2. The asymptotes are spaced pi over two apart, which is half the period of pi.
Check
Range. Reciprocation turns the interval inside out.
Check your understanding
What is the range of y equals 3 secant x?
Answer: C
Why: The secant takes every value of absolute value at least 1, and multiplying by 3 scales those to every value of absolute value at least 3. So the graph misses the open interval from negative 3 to 3, and the turning points sit at 3 and negative 3.
Check
Transformations. Which features does a vertical shift move?
Check your understanding
The graph of y equals secant x is shifted up by 4. What happens to its asymptotes and its turning points?
Answer: B
Why: The asymptotes are vertical lines located by where the cosine vanishes, which depends only on the argument, so adding 4 outside the function cannot move them. The turning points were at heights 1 and negative 1 and each rises by 4, to 5 and 3. Note that both are now positive, so the shifted graph never takes a negative value.
Real world
A lighthouse 500 metres offshore sweeps its beam at a constant rate. The distance from the lighthouse to the point where the beam strikes the straight shoreline is 500 times the secant of the angle the beam makes with the perpendicular to the shore.
Discussion prompt
Sketch how that distance behaves as the beam sweeps from pointing straight at the shore round towards parallel with it, and explain what the vertical asymptote means physically.
Hint: Consider what happens to the beam's intersection with the shore as the angle approaches a right angle.
Answer:
\[ d(\theta) = 500\sec(\theta), \qquad -\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2} \]
At an angle of zero the beam points straight at the nearest shore point and the distance is its minimum, 500 metres — the turning point of the secant branch. As the angle grows either way the distance increases, slowly at first and then very fast.
The asymptote at pi over two means that as the beam turns parallel to the shoreline, the point where it strikes runs away to infinity — and at exactly pi over two the beam never strikes the shore at all, which is why the function is undefined there rather than merely large. That is a genuine physical statement, not an artefact: a beam parallel to a straight shore has no intersection with it.
It is also why only one branch of the secant is used here. The domain is restricted to a single branch by the geometry, and the other branches would describe the beam pointing inland.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
The graph of y equals secant x is drawn. At how many points does it cross the x-axis?
Correct: Never.
\[ \sec(x) = 0 \;\Longrightarrow\; \frac{1}{\cos(x)} = 0 \;\Longrightarrow\; \text{no solutions} \]
Why: The secant is one over the cosine, and a fraction is zero only when its numerator is zero. The numerator here is the constant 1, so the secant is never zero and the graph never meets the x-axis. This is a general property of reciprocals: they turn the zeros of the parent into asymptotes and have no zeros of their own. The last option confuses the two — at an asymptote the function is undefined rather than zero, so the graph is not there at all.
Explain it
They have just seen the secant graph for the first time and find its shape completely unlike anything they have graphed before.
Discussion prompt
In no more than five sentences, tell them how to draw it starting from a curve they already know, and what the asymptotes are actually saying.
Hint: How many new things are they really being asked to learn?
Answer:
A usable answer: draw the cosine curve first, lightly. Wherever it touches 1 or negative 1, put a dot — the secant passes through those same points. Wherever the cosine crosses zero, draw a vertical dashed line, because you are about to divide by zero there.
Then, between each pair of dashed lines, draw a U-shaped curve through the dot, opening away from the x-axis and running up the dashed lines. The asymptotes are just the places where the cosine was zero, and dividing by something tiny is what makes the curve shoot off.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Asymptotes are fixed by always solving for where the parent function is zero, never by inspecting the reciprocal. Branch direction is fixed by asking which side of zero the parent is on, since the sign decides everything. Vertical shifts are fixed by remembering that an asymptote is a vertical line, so only horizontal changes can move it. Telling the two apart is fixed by one glance at x equal to zero, where the secant has a point and the cosecant has an asymptote. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Across the top half of the page draw one full cycle of the cosine curve lightly, then build the secant curve on top of it: mark the two zeros with dashed vertical lines, mark the peak and trough with dots, and draw the three branches. Label the asymptotes with their equations. Underneath, do exactly the same starting from the sine curve to build the cosecant, and label its asymptotes. To the right, write a short table comparing the two: parent, asymptote positions, range, period and parity. Below that table, write the four rules of reciprocation — what happens to 1, to negative 1, to a small value and to zero. In the bottom left, graph one cycle of y equals two secant of the quantity x minus pi, completely, showing the quarter-mark working. In the bottom right, write one sentence saying why these graphs have no amplitude and what you should state instead. Finally, circle the two points on your first graph where the cosine and the secant touch, and write why those are the only two.
The circled points are where the cosine equals 1 and negative 1, and they are the only ones because 1 and negative 1 are the only real numbers equal to their own reciprocals.
Recap
Five things, and the first one is the method that makes the other four easy.
| If the question says | Your first move is |
|---|---|
| Find the asymptotes | Set the argument to make the parent zero, and solve |
| Which way does the branch go | Check the sign of the parent on that side |
| State the amplitude | Say there is none, and give the asymptotes instead |
| Graph one cycle | Quarter marks, then sketch the parent dotted |
| Is this secant or cosecant | Look at what happens at x equal to zero |
Two functions remain. Tangent and cotangent also have asymptotes, but they are not reciprocals of a bounded function — they are quotients of two bounded ones, which gives them an unbounded range and a period of pi rather than two pi. That is the next lesson.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 800-804 — everything on these slides traces back here
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