Graphs the four derived trigonometric functions by reading consequences off the sine and cosine. Places the vertical asymptotes at the undefined points, explains the tangent's shorter period through its slope reading, sketches the secant and cosecant as reciprocals hugging the curves they invert, and applies the standard transformations to all four.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 6 — Periodic Functions
§6.2 Graphs of the Other Trigonometric Functions, pp. 739-763
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 739-763 — the pages these objectives are drawn from
Warm-up
Section 5.3 found where the tangent is undefined. That determines the graph before any point is plotted.
Discussion prompt
The tangent is the sine over the cosine. Where must its graph have vertical asymptotes?
Hint: Where does a fraction blow up?
Answer:
Wherever the denominator vanishes, which for the tangent is wherever the cosine is zero — at a quarter turn, three quarters, and every half turn from there.
Near those angles the numerator is close to plus or minus 1 while the denominator approaches zero, so the quotient grows without bound. That is a vertical asymptote.
So the graph's most striking feature is settled before anything is plotted. The asymptotes are §5.3's undefined points, and everything else about the shape fits between them.
Concept
The four derived functions are built from the sine and cosine by division, so their asymptotes sit at the underlying zeros and their sizes are determined by the underlying values.
\[ \tan=\frac{\sin}{\cos}, \; \sec=\frac{1}{\cos}, \; \csc=\frac{1}{\sin}, \; \cot=\frac{\cos}{\sin} \]
This makes the section a reading exercise rather than a memorisation one. Sketch the sine or cosine lightly, mark its zeros as asymptotes and its extremes as touch points, and the derived graph follows almost mechanically.
Figure (svg): A card summarising how a reciprocal graph relates to its original: touching at plus and minus one, an asymptote at each zero, and the sign preserved everywhere
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 739-745
Section
Section 1
Concept
The tangent has a vertical asymptote wherever the cosine vanishes, and between each pair of asymptotes it rises from far below to far above, crossing zero in the middle.
The absence of an amplitude is worth stating explicitly, because a coefficient in front of a tangent is still a vertical stretch — it just cannot be called an amplitude, since there is no maximum to measure. Questions asking for the amplitude of a tangent have no answer.
Figure (svg): The tangent graph over several periods, showing vertical asymptotes where the cosine vanishes and a complete branch between each pair
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 739-746
Picture it
Asymptotes in red, one complete branch between each pair.
Figure (svg): The tangent graph over several periods, showing vertical asymptotes where the cosine vanishes and a complete branch between each pair
Each branch is identical, which is what a period of pi means. The graph never turns and never repeats a value within a branch, which is why §6.3 can invert it on one branch.
Worked example
Set the denominator to zero.
\[ \text{Where are the asymptotes of } y=\tan x? \]
Identify the denominator
Why: The tangent is sine over cosine.
Find its zeros
Why: Where the point is on the vertical axis.
\[ \frac{\pi}{2}\text{ and } 3 \pi / 2 \]
Note the spacing
Why: They are pi apart.
Write the general form
Why: Starting from a quarter turn.
\[ \frac{\pi}{2} + \pi k \]
Figure (svg): The tangent graph over several periods, showing vertical asymptotes where the cosine vanishes and a complete branch between each pair
\[ x=\tfrac{\pi}{2}+\pi k \]
Verify: check the value just before an asymptote
Why: At 1.5 radians, just under pi over 2, the tangent is about 14; at 1.57 it is over 1200. The values grow without bound as the asymptote is approached, confirming it is genuinely an asymptote rather than a jump.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 740-742
Sorting
Only bounded ones do.
Sort into buckets
Sort each function.
Worked example
Three points and two asymptotes determine it.
\[ \text{Sketch } y=\tan x \text{ between } -\tfrac{\pi}{2} \text{ and } \tfrac{\pi}{2}. \]
Draw the two asymptotes
Why: At the ends of the interval.
\[ \text{at } -\frac{\pi}{2}\text{ and } \frac{\pi}{2} \]
Mark the zero
Why: Midway between them.
\[ \text{at } 0 \]
Mark the quarter points
Why: Where the tangent is plus and minus 1.
\[ \text{at } -\frac{\pi}{4}\text{ and } \frac{\pi}{4} \]
Join with a rising curve
Why: Steeply at both ends.
Figure (svg): The solution to Worked example sketch one branch shown as a ladder of expressions, one row per legal move
\[ \text{rises through } (-\tfrac{\pi}{4},-1), \; (0,0), \; (\tfrac{\pi}{4},1) \]
Verify: check the quarter points
Why: At pi over 4 the sine and cosine are equal, so their ratio is 1 — confirming that point. Those two points at plus and minus 1 are the easiest landmarks on a tangent branch, and they sit exactly halfway between the zero and each asymptote.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 742-746
Trap
\[ y=3\tan x \;\Longrightarrow\; \text{amplitude } 3 \]
Read the coefficient as an amplitude
Why: The number in front is taken to measure the vertical extent.
An amplitude of 3 is reported for the tangent.
The tangent has no amplitude, because it has no maximum or minimum — it is unbounded in both directions on every branch.
The 3 is a genuine vertical stretch, which makes the branches steeper, but there is no swing for it to be half of.
Amplitude applies only to bounded periodic functions. For the tangent, cotangent, secant and cosecant the coefficient is a stretch factor and nothing more.
Prediction
The tangent is the slope of the terminal side.
Predict first
What values does it take?
Correct: Every real number.
Why: A line through the origin can have any slope, and every slope is achieved by some terminal side. On each branch the tangent rises continuously from far below to far above, passing through every value exactly once — which is why §6.3 can invert it on a single branch.
Faded example
For the tangent, asymptotes occur where the cosine vanishes.
Fill in the blanks
\cos x = 0 \text2 x = \fracpi___}, \text___ ___ \text___
Why: The cosine first vanishes at a quarter turn, and it vanishes again every half turn — at three quarters, five quarters and so on. Those are the tangent's asymptotes, spaced pi apart, which is also its period.
Explain it to yourself
Each branch rises from far below to far above.
Discussion prompt
Explain why, using the sine and cosine values across one branch.
Hint: What do the numerator and denominator do across the interval?
Answer:
Across one branch the cosine goes from near zero, up to 1, and back to near zero, changing sign at each end. The sine stays close to plus or minus 1 near the ends and passes through zero in the middle.
So near the left end a number close to negative 1 is divided by a tiny positive number, giving a large negative result. In the middle the numerator is zero, giving zero. Near the right end a number close to 1 is divided by a tiny positive, giving a large positive result.
The branch therefore climbs continuously from far below to far above. Nothing about the shape has to be memorised — it follows from watching what the two underlying values do.
Section
Section 2
Concept
The tangent is the slope of the terminal side, and a half turn puts the terminal side on the same line pointing the other way — so the slope, and therefore the tangent, is unchanged.
The sine and cosine change under a half turn because both coordinates flip sign — the point moves to the opposite side of the origin. Their ratio is unaffected because both flips cancel, which is exactly why the tangent's period is half of theirs.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 746-750
Picture it
Four depend on the point and two on the line.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
The split is not arbitrary. Anything that survives negating both coordinates has the shorter period, and the tangent and cotangent are the two ratios where both signs cancel.
Worked example
Compare the values a half turn apart.
\[ \text{Show that } \tan(x+\pi)=\tan x. \]
Consider the effect of a half turn
Why: Both coordinates flip sign.
Write the tangent at the shifted angle
Why: Both parts negated.
\[ \frac{-\sin}{-\cos} \]
Cancel the signs
Why: Two minus signs cancel.
\[ \sin / \cos \]
Compare
Why: It is the original.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
\[ \tan(x+\pi)=\tan x \]
Verify: test numerically
Why: The tangent of pi over 4 is 1, and the tangent of five pi over 4 is also 1 — a half turn later and the same value. The sine and cosine at those two angles differ in sign, but their ratio does not, which is exactly the cancellation shown.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 747-748
Sorting
Two of the six differ from the rest.
Sort into buckets
Sort each function by its base period.
Worked example
The same formula as for the sine, with a different starting period.
\[ \text{Find the period of } y=\tan(3x). \]
Recall the tangent's base period
Why: Half a turn.
Apply the compression
Why: Divide by the coefficient.
\[ \frac{\pi}{3} \]
Interpret
Why: Three branches in the usual space.
Note the asymptote spacing
Why: It matches the period.
\[ \text{every } \frac{\pi}{3} \]
Figure (svg): The solution to Worked example find a transformed period shown as a ladder of expressions, one row per legal move
\[ \text{period}=\frac{\pi}{3} \]
Verify: check against the general rule
Why: The period of a transformed tangent is pi over B rather than 2 pi over B, because the tangent's base period is pi. Using the sine's formula here would give twice the right answer, so the base period has to be tracked separately for these two functions.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 748-750
Error analysis
A student computes a transformed tangent's period.
Annotate
On: \( y=\tan(2x) \;\Longrightarrow\; \text{period}=\frac{2\pi}{2}=\pi \)
Two of the six functions have a base period of pi and four have 2 pi. Using the wrong base doubles or halves every transformed period, so it is worth checking which family the function belongs to first.
Faded example
For the cotangent of 4x.
Fill in the blanks
\textpi = \frac4}___ = \frac______}
Why: The cotangent's base period is pi, so dividing by the coefficient 4 gives pi over 4. Using 2 pi as the base would double the answer, which is the standard error when transferring the sine's formula to these two functions.
Prediction
An angle is increased by pi.
Predict first
Which functions are unchanged?
Correct: The tangent and cotangent.
Why: A half turn negates both coordinates, so anything depending on a single coordinate flips sign while their ratio is unaffected. That is why the two ratios have the shorter period and the four functions built from single coordinates do not.
Explain it
Six functions and two different periods.
Discussion prompt
Explain to a classmate why the tangent repeats twice as often as the sine.
Hint: What does a half turn do to the terminal side and to its slope?
Answer:
A half turn puts the terminal side on the same line, pointing the opposite way. The point moves to the other side of the origin, so both its coordinates flip sign.
The sine is one coordinate, so it flips — it has not returned to its old value and needs another half turn. The tangent is a ratio of the two, and both flips cancel, so it is already back.
A good explanation adds the slope reading: the tangent is the line's slope, and a line has one slope regardless of direction. So the tangent depends on the line while the sine depends on the point, and there are twice as many points as lines.
Section
Section 3
Concept
The secant and cosecant are reciprocals of the cosine and sine, so their graphs are read off those curves: asymptotes at the zeros, touching at the extremes, and large where the original is small.
The last point is a striking consequence and is easy to overlook. Since the cosine is never larger than 1 in size, its reciprocal is never smaller than 1 — so the secant's range is everything outside the interval from negative 1 to 1, which is the complement of the cosine's range.
Figure (svg): The secant graph drawn over the cosine it inverts, showing the U-shaped branches touching the cosine at its peaks and troughs and rising to asymptotes at its zeros
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 750-757
Picture it
The dashed cosine determines everything about the solid secant.
Figure (svg): The secant graph drawn over the cosine it inverts, showing the U-shaped branches touching the cosine at its peaks and troughs and rising to asymptotes at its zeros
Each U sits inside a hump of the cosine, touching it at the peak and rising to the asymptotes at the zeros either side. Drawing the cosine first makes the secant a tracing exercise.
Worked example
Draw the sine first.
\[ \text{Sketch } y=\csc x \text{ over one period.} \]
Sketch the sine lightly
Why: As a guide.
Mark asymptotes at its zeros
Why: At 0, pi and 2 pi.
Mark the touch points
Why: Where the sine is 1 or negative 1.
\[ \text{at } \frac{\pi}{2}\text{ and } 3 \pi / 2 \]
Draw U-shaped branches
Why: Opening away from the axis.
Figure (svg): The secant graph drawn over the cosine it inverts, showing the U-shaped branches touching the cosine at its peaks and troughs and rising to asymptotes at its zeros
\[ \text{asymptotes at } 0,\pi,2\pi; \text{ touches at } (\tfrac{\pi}{2},1) \text{ and } (\tfrac{3\pi}{2},-1) \]
Verify: check a point between
Why: At pi over 6 the sine is one half, so the cosecant is 2 — larger than the touch value of 1, as it should be since the sine is smaller there. The reciprocal grows as the original shrinks, which is what pushes the branches up towards the asymptotes.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 751-754
Matching
Four rules, all about reciprocals.
Match the pairs
Why: None of these is specific to trigonometry — they are facts about reciprocals. That is why sketching the original first turns the derived graph into a mechanical tracing rather than a shape to recall.
Worked example
The reciprocal of a bounded quantity is unbounded away from zero.
\[ \text{Find the range of } y=\sec x. \]
Recall the cosine's range
Why: Between negative 1 and 1.
\[ | \cos | \le 1 \]
Take reciprocals
Why: A number at most 1 in size has a reciprocal at least 1.
\[ | \sec | \ge 1 \]
Note the excluded interval
Why: Nothing strictly between.
\[ \text{no values in } (-1, 1) \]
State the range
Why: Two unbounded pieces.
\[ (-\infty, -1) U [1, \infty] \]
Figure (svg): The solution to Worked example find the range shown as a ladder of expressions, one row per legal move
\[ (-\infty,-1]\cup[1,\infty) \]
Verify: check the endpoints are attained
Why: The secant equals 1 where the cosine equals 1, at angle zero, and negative 1 where the cosine does, at pi. So both endpoints are reached and the brackets are square. The excluded interval is exactly the cosine's range with its endpoints kept, which is the reciprocal relationship stated as sets.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 754-757
Trap
\[ \text{sketch the secant directly from memory of its shape} \]
Recall the secant's appearance and draw it
Why: The U-shaped branches are drawn from memory.
The asymptotes and touch points are placed by guesswork.
Draw the cosine first, lightly. Its zeros give the asymptotes and its extremes give the touch points, both exactly located.
Without the guide curve the branches end up in the wrong places, especially after a transformation has moved the underlying zeros.
The guide costs ten seconds and removes all the guesswork. It matters most for transformed functions, where memory of the standard shape is no help at all.
Prediction
The cosine never exceeds 1 in size.
Predict first
What values does the secant never take?
Correct: Anything strictly between -1 and 1.
Why: A reciprocal of something at most 1 in size is at least 1 in size, so the secant's values are always at or beyond 1. The excluded interval is the interior of the cosine's range, which is the reciprocal relationship expressed as a statement about ranges.
Faded example
The cosecant is the reciprocal of the sine.
Fill in the blanks
\sin x = 0 \textpi x = 0, asymptotes, 2\pi, \text______
Why: The sine vanishes at multiples of pi, and a reciprocal has an asymptote wherever its original vanishes. So the cosecant's asymptotes are at every multiple of pi, and its branches sit between them, one per hump of the sine.
Explain it to yourself
The reciprocal curve touches the original at certain places.
Discussion prompt
Explain where and why the two curves meet.
Hint: Which numbers are their own reciprocals?
Answer:
A number equals its own reciprocal exactly when it is 1 or negative 1, since one over 1 is 1 and one over negative 1 is negative 1.
So the secant equals the cosine precisely where the cosine takes one of those two values — at its peaks and troughs.
That is why the branches appear to sit inside the humps and touch at their tips. The touch points are the extremes of the original, which makes them the easiest landmarks to place when sketching, and they move with any transformation just as the original's extremes do.
Section
Section 4
Concept
Shifts, stretches and reflections work exactly as before. The only new care needed is using pi rather than 2 pi as the base period for the tangent and cotangent.
Moving the asymptotes with a shift is worth stating explicitly. A transformed tangent's asymptotes are found by setting its argument equal to a quarter turn rather than by taking the standard positions and hoping — the standard positions apply only to the untransformed function.
Figure (svg): The tangent graph over several periods, showing vertical asymptotes where the cosine vanishes and a complete branch between each pair
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 757-762
Picture it
The asymptote positions here are the ones a transformation moves.
Figure (svg): The tangent graph over several periods, showing vertical asymptotes where the cosine vanishes and a complete branch between each pair
After a shift or a compression these lines move, so they have to be recomputed from the argument rather than recalled from this picture.
Worked example
Period from the base pi, asymptotes from the argument.
\[ \text{Describe } y=2\tan(2x). \]
Read the outside coefficient
Why: A vertical stretch, not an amplitude.
\[ \text{stretch by } 2 \]
Compute the period
Why: Base pi over the coefficient.
\[ \frac{\pi}{2} \]
Find the asymptotes
Why: Set the argument to a quarter turn.
\[ 2 x = \frac{\pi}{2},\text{ so } x = \frac{\pi}{4} \]
Space them by the period
Why: Every pi over 2 from there.
\[ \frac{\pi}{4} + \pi k / 2 \]
Figure (svg): The solution to Worked example transform a tangent shown as a ladder of expressions, one row per legal move
\[ T=\tfrac{\pi}{2}, \; \text{asymptotes } x=\tfrac{\pi}{4}+\tfrac{\pi}{2}k \]
Verify: check the asymptote spacing equals the period
Why: The asymptotes are pi over 2 apart, which is the period — as it must be, since one branch lives between each consecutive pair. That equality is a reliable check that both were computed consistently.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 758-760
Faded example
For the tangent of x over 2.
Fill in the blanks
T = \frac2___ = ___\pi
Why: Dividing the base period pi by one half gives 2 pi, so the wave is stretched to twice its usual period. A coefficient below 1 stretches, exactly as for the sine, and the base period being pi is what distinguishes the tangent's calculation.
Worked example
Find the underlying cosine's zeros after the shift.
\[ \text{Find the asymptotes of } y=\sec\bigl(x-\tfrac{\pi}{4}\bigr). \]
Identify the underlying function
Why: The secant inverts the cosine.
Set the argument where the cosine vanishes
Why: At a quarter turn and its repeats.
\[ x - \frac{\pi}{4} = \frac{\pi}{2} + \pi k \]
Solve for x
Why: Add a quarter of pi.
\[ x = 3 \pi / 4 + \pi k \]
Check the shift moved them
Why: The standard positions were pi/2 + pi k.
\[ \text{moved right } \frac{\pi}{4} \]
Figure (svg): The secant graph drawn over the cosine it inverts, showing the U-shaped branches touching the cosine at its peaks and troughs and rising to asymptotes at its zeros
\[ x=\tfrac{3\pi}{4}+\pi k \]
Verify: confirm the shift
Why: The untransformed secant's asymptotes are at pi over 2 plus multiples of pi, and these are pi over 4 further right — matching the shift in the argument. Computing them from the argument rather than adjusting remembered positions is what makes this reliable.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 760-762
Error analysis
A student finds the asymptotes of a shifted tangent.
Annotate
On: \( y=\tan(x-\tfrac{\pi}{3}) \;\Longrightarrow\; \text{asymptotes at } \tfrac{\pi}{2}+\pi k \)
Asymptotes are part of the graph and move with every transformation. Computing them from the argument rather than recalling standard positions handles shifts and compressions in one step.
Sorting
Anything horizontal does.
Sort into buckets
Sort each transformation of a tangent.
Prediction
A transformed tangent has period pi over 4.
Predict first
How far apart are its asymptotes?
Correct: pi over 4, the same as the period.
Why: One complete branch of the tangent occupies exactly one period and sits between two consecutive asymptotes, so the spacing equals the period. Checking that they match is a quick verification that both were computed with the right base period.
Step zero
You are asked to graph a transformed tangent or secant.
Discussion prompt
What do you determine first, and what do you compute rather than recall?
Hint: Which feature dominates the picture?
Answer:
Find the asymptotes, because they frame the whole picture and everything else fits between them.
Compute them from the argument rather than recalling standard positions: set the argument equal to the value that makes the underlying function undefined, and solve.
That single method handles shifts, compressions and reflections at once, whereas adjusting remembered positions requires tracking each transformation separately. The argument carries all the horizontal information, which is why working from it is reliable.
Section
Section 5
Concept
The six functions fall into families: two bounded waves, two unbounded reciprocal curves, and two unbounded slope curves with the shorter period.
Organising them this way reduces six graphs to three shapes, each with a shifted partner. It also makes clear which properties transfer: anything true of the tangent has a cotangent analogue, and anything true of the secant has a cosecant one.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 745-763
Picture it
Four functions in one family and two in the other.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
The split is by whether the function depends on the point or on the line, which is the same distinction that gave two periods and is worth carrying rather than memorising the six values.
Worked example
Asymptotes, range and period narrow it down fast.
\[ \text{A graph has asymptotes every } \pi, \text{ range all reals, and rises on every branch. Which function?} \]
Use the asymptote spacing
Why: Every pi means period pi.
Use the range
Why: Both have all reals.
Use the direction
Why: The tangent rises; the cotangent falls.
Conclude
Why: It is the tangent.
Figure (svg): A comparison of the six functions' periods, showing that four have a period of a full turn and the tangent and cotangent have half of that
\[ y=\tan x \]
Verify: check the distinguishing feature
Why: The cotangent's branches fall from far above to far below, because its numerator and denominator are swapped relative to the tangent's. Direction is the only thing separating the two graphs, since they share their period, their range and their asymptote spacing.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 746-755
Sorting
Bounded waves, reciprocal curves, or slope curves.
Sort into buckets
Sort each function.
Worked example
Same construction, different underlying curve.
\[ \text{How do the secant and cosecant graphs differ?} \]
Identify the underlying functions
Why: Cosine and sine respectively.
Compare their zeros
Why: Different places.
Note the shift
Why: The sine is the cosine shifted.
Conclude
Why: Same shape, shifted.
Figure (svg): The solution to Worked example compare the two reciprocal graphs shown as a ladder of expressions, one row per legal move
\[ \csc x=\sec\bigl(x-\tfrac{\pi}{2}\bigr) \]
Verify: check the asymptotes
Why: The secant's are where the cosine vanishes, at a quarter turn and every half turn; the cosecant's are where the sine does, at zero and every half turn. Those two sets are a quarter turn apart, confirming the shift — which was inevitable since the sine and cosine are themselves shifted versions.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 755-763
Trap
\[ \text{six functions means six graphs to memorise} \]
Learn each curve independently
Why: Six shapes, six sets of asymptotes and six periods are committed to memory.
Six times as much is memorised as necessary.
There are three shapes, each with a shifted partner. The sine and cosine, the secant and cosecant, and the tangent and cotangent.
And all three shapes follow from the sine and cosine curves by arithmetic: reciprocals give one pair and the ratio gives the other.
Learn the sine curve and the reciprocal rules. Everything else is derived in seconds, and derived knowledge survives after memorised shapes have blurred together.
Matching
One feature separates each from its partner.
Match the pairs
Why: Within each pair the two differ only by which underlying function they use, and since the sine and cosine are shifted versions of each other, so are the two members of each pair. Three shapes and one shift covers all six.
Prediction
The tangent rises on every branch.
Predict first
What does the cotangent do?
Correct: Falls on every branch.
Why: The cotangent is the reciprocal of the tangent, and a reciprocal reverses whether a positive quantity is increasing. As the tangent climbs from small to large across a branch, its reciprocal falls from large to small — so the cotangent's branches descend.
Explain it
Six graphs is a lot to hold.
Discussion prompt
Explain to a classmate how to reduce them to something manageable.
Hint: How many genuinely different shapes are there?
Answer:
There are three shapes, each appearing twice as a shifted pair: the bounded wave, the reciprocal curve with U-shaped branches, and the rising or falling branches between asymptotes.
And all three come from the sine and cosine graphs. Take reciprocals for one pair and take the ratio for the other, using the rules about where a reciprocal has asymptotes and where it touches.
So what has to be held is one curve and a few reciprocal facts. A good explanation stresses that derived knowledge is more robust than memorised shapes, especially after a transformation has moved everything from where memory expects it.
Comparison
Fill the blanks from memory. Three shapes, two periods, one source.
Comparison matrix
| sine and cosine | secant and cosecant | tangent and cotangent | |
|---|---|---|---|
| bounded | yes, between -1 and 1 | no, but never within (-1, 1) | no, every real value |
| asymptotes | none | at the original's zeros | where the denominator vanishes |
| period | 2 pi | 2 pi | pi |
| has an amplitude | yes | no | no |
The third row is the one to watch when transforming, since using the wrong base period doubles or halves every answer that depends on it.
Pattern
Five steps, and the second is what makes the rest easy.
Step 2 is optional in principle and decisive in practice. The guide curve places every asymptote and touch point exactly, which is far more reliable than recalling a standard shape and adjusting it.
OpenStax Algebra and Trigonometry 2e, §8.2 Graphs of the Other Trigonometric Functions §8.2
Check
The base period differs.
Check your understanding
What is the period of the function tangent of 2x?
Answer: A
Why: The tangent's base period is pi, so dividing by the coefficient 2 gives pi over 2. Using 2 pi as the base would give pi, which is the standard error when transferring the sine's formula.
Check
Asymptotes sit at the underlying zeros.
Check your understanding
Where does the cosecant have vertical asymptotes?
Answer: A
Why: The cosecant is the reciprocal of the sine, so it is undefined and grows without bound wherever the sine vanishes — at every multiple of pi. Where the sine equals 1, the cosecant touches it rather than blowing up.
Check
Bounded or not.
Check your understanding
Which of these has an amplitude?
Answer: A
Why: Amplitude is half the distance between a maximum and a minimum, so it exists only for bounded functions. Only the sine and cosine among the six are bounded; the other four are unbounded and have no amplitude, though a coefficient still stretches them vertically.
Real world
The tangent's asymptotes have a physical meaning in any situation involving a line of sight.
Discussion prompt
A camera at ground level pans upward to track a rocket launching straight up. What happens to the tangent of the camera's angle?
Hint: What is the relationship between the angle and the rocket's height?
Answer:
The rocket's height is the horizontal distance times the tangent of the camera's elevation angle, by §5.4. So as the rocket climbs, the tangent grows.
As the angle approaches a quarter turn — the camera pointing straight up — the tangent grows without bound, which corresponds to the rocket being arbitrarily high. The asymptote is exactly that limit.
The physical reading is that a camera at a fixed distance can never quite point straight up while still tracking a finite height, and the rate at which it must pan accelerates as the rocket climbs. That accelerating pan is the asymptote made visible, and it is why tracking shots of launches get harder as they go.
Commit first
State your confidence along with your answer.
Predict first
Why do the tangent and cotangent have period pi rather than 2 pi?
Correct: Because they are slopes, and a line's slope is unchanged by a half turn.
Why: A half turn puts the terminal side on the same line pointing the other way, so both coordinates flip sign and their ratio is unchanged. The other four depend on a single coordinate and therefore flip, needing a full turn to return. Being unbounded and being undefined are consequences rather than causes.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate how to draw the secant graph without having memorised it.
Hint: What should be drawn first?
Answer:
Draw the cosine lightly first. Everything about the secant is read off it, because the secant is its reciprocal.
Then apply three reciprocal facts: an asymptote wherever the cosine crosses zero, a touch point wherever the cosine is 1 or negative 1, and large values wherever the cosine is small — with the sign preserved.
Joining those gives U-shaped branches sitting in the cosine's humps. None of this is trigonometry; it is what reciprocals do, which is why the method survives any transformation that moves the cosine's features around.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is where errors cluster, since standard asymptote positions get carried over after a transformation has moved them. The third is the technique that makes two of the four graphs require no memorisation at all.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Sketch the cosine lightly and draw the secant over it, marking the asymptotes at its zeros and the touch points at its extremes. Beside that, sketch two branches of the tangent with their asymptotes, and write why its period is pi. Underneath, list the six functions in three families with their periods and whether each has an amplitude.
If your secant was traced from the cosine rather than recalled, and your tangent's period is justified by the slope argument, both of the section's methods are on the page.
Recap
Five things, and none of them required memorising a new shape.
| if you remember one thing | it should be this |
|---|---|
| about asymptotes | they sit where the underlying denominator vanishes |
| about the period | pi for the two slopes, 2 pi for the other four |
| about reciprocals | sketch the original first; the rest is tracing |
| about transformations | recompute the asymptotes from the argument, never recall them |
Section 6.3 inverts these functions, which requires restricting each to one branch — and choosing that branch is what makes the inverse trigonometric functions return one answer where the equation has many.
OpenStax, Precalculus, §6.2 Graphs of the Other Trigonometric Functions §6.2, pp. 739-763 — everything on these slides traces back here
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