The last two graphs, and the two that behave least like the rest. Both have period pi rather than two pi, both have range all real numbers, and each branch climbs or falls monotonically between consecutive asymptotes instead of turning around. Covers the proof that the period really is pi using the tangent sum identity, the geometric reason behind it, the different fundamental cycles and quarter marks the two functions use, and graphing transformed versions.
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. 804-809
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. 804-809 — the pages these objectives are drawn from
Warm-up
You know how the tangent behaves near an angle where the cosine vanishes. Plotting that behaviour is most of this lesson.
Discussion prompt
As an angle rises towards pi over two, what happens to its sine, to its cosine, and therefore to its tangent? Now answer the same three questions for an angle falling towards pi over two from above.
Hint: The sine is easy. The cosine is the interesting one, because its sign changes.
Answer:
Approaching from below: the sine rises to 1 and the cosine falls to zero through positive values, so the tangent runs to positive infinity.
Approaching from above: the sine is still near 1, but the cosine is now negative and heading to zero, so the tangent runs to negative infinity.
\[ \text{as } x \to \tfrac{\pi}{2}^-, \; \tan x \to +\infty; \qquad \text{as } x \to \tfrac{\pi}{2}^+, \; \tan x \to -\infty \]
So there is a vertical asymptote at pi over two, and the curve jumps from the top of the picture to the bottom. That single behaviour, repeated, is the whole tangent graph.
Concept
Secant and cosecant were reciprocals of bounded functions, which is why their ranges were bounded away from zero. Tangent and cotangent are quotients of two bounded functions, and a quotient of two small numbers can be anything at all — so their range is every real number.
\[ \tan(x) = \frac{\sin(x)}{\cos(x)}, \qquad \cot(x) = \frac{\cos(x)}{\sin(x)} \]
That difference in structure produces every difference in behaviour: an unbounded range, branches that climb rather than turn, and — for a reason the identity chapter supplies — a period of pi rather than two pi.
Figure (svg): The tangent curve over two full periods, rising from negative infinity to positive infinity between each pair of consecutive vertical asymptotes at the odd multiples of pi over two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 804-804
Section
Section 1
Concept
Between consecutive asymptotes the tangent rises steadily from negative infinity to positive infinity, crossing zero exactly once. It never turns around, so it has no maximum, no minimum and no amplitude.
That monotone behaviour is what makes the inverse tangent so well behaved in the next lesson: one branch already passes the horizontal line test, so no artificial restriction is needed beyond choosing which branch.
Figure (svg): The tangent curve over two full periods, rising from negative infinity to positive infinity between each pair of consecutive vertical asymptotes at the odd multiples of pi over two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 804-805
Picture it
Everything you learned about the first four graphs needs adjusting for these two.
Figure (svg): Two columns contrasting the sinusoids and their reciprocals with the tangent and cotangent
The right-hand column is the honest summary. These are not sinusoids with holes in them; they are a genuinely different kind of curve that happens to share the same building blocks.
Worked example
Divide the sine table by the cosine table and watch where it fails.
\[ \text{Tabulate } y = \tan(x) \text{ at the eighth marks on } [0, 2\pi]. \]
Divide sine by cosine at each mark
Why: At zero the sine is 0, so the tangent is 0.
\[ 0\text{ at } x = 0 \]
Continue to pi over four
Why: Both values are root two over two, so their quotient is 1.
\[ 1\text{ at } \frac{\pi}{4} \]
Reach pi over two
Why: The cosine is zero, so the quotient is undefined.
\[ \text{asymptote at } \frac{\pi}{2} \]
Continue past it
Why: At three pi over four the sine is positive and the cosine negative, so the tangent is negative.
\[ -1\text{ at } 3 \pi / 4 \]
Figure (svg): The solution to Worked example build the tangent table shown as a ladder of expressions, one row per legal move
\[ \tan: \; 0, \; 1, \; \text{asym}, \; -1, \; 0, \; 1, \; \text{asym}, \; -1, \; 0 \]
Verify: look for the repeat
Why: The values from zero to pi are 0, 1, asymptote, negative 1, 0 — and the values from pi to two pi are exactly the same list again. The pattern has repeated after only pi, not two pi, which is the first evidence that the period is halved.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 804-804
Sorting
Zeros come from the numerator, asymptotes from the denominator.
Sort into buckets
Sort each angle by what the tangent does there.
Worked example
A single branch is worth describing carefully, because every other branch is a copy of it.
\[ \text{Describe the branch of } y = \tan(x) \text{ on } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). \]
State the left-hand behaviour
Why: Just above negative pi over two the cosine is small and positive while the sine is near negative one.
State the crossing
Why: At zero the sine is zero so the tangent is zero.
\[ \text{crosses zero at } x = 0 \]
State the right-hand behaviour
Why: Just below pi over two the cosine is small and positive while the sine is near one.
State the monotonicity
Why: It increases throughout, never turning around.
Figure (svg): The solution to Worked example describe one branch precisely shown as a ladder of expressions, one row per legal move
\[ \text{on } \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right): \; \tan \text{ increases from } -\infty \text{ to } +\infty, \text{ crossing } 0 \text{ at } x = 0 \]
Verify: check the range claim
Why: A continuous function increasing from negative infinity to positive infinity takes every real value exactly once on the way. So this one branch already accounts for the entire range, and the other branches merely repeat it.
Trap
A student sketches the tangent as a wave that rises, peaks near the asymptote, and falls back — reasoning that every trigonometric graph so far has oscillated.
But a peak would mean a maximum, and the tangent has no maximum. It runs to infinity, which is not a value it reaches and turns back from.
Each branch is strictly increasing with no turning point anywhere. It approaches the asymptote without bound and simply stops existing there; the next branch starts again from negative infinity.
\[ \text{range}(\tan) = (-\infty, \infty) \]
The tell that catches this: a function with an unbounded range cannot have a maximum, and the range here is genuinely everything. The oscillating shape belongs to the four functions built from a bounded parent, and these two are not built that way.
Prediction
The tangent of 1.5 radians is about 14.1.
Predict first
Roughly what is the tangent of 1.6 radians?
Correct: About -34.
Why: Pi over two is about 1.571, so 1.5 is just below the asymptote and 1.6 is just above it. Crossing an asymptote flips the tangent from a large positive value to a large negative one, and the actual value at 1.6 is about negative 34.2. This is the single most surprising feature of the tangent graph and it is worth internalising: a tiny change in x near an asymptote produces an enormous change, including a sign change.
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 the tangent graph is wrong.
Survives elimination: B
Why: Statement B is false. A branch has no largest value: for any value it takes, it takes a larger one closer to the asymptote. Running to infinity is not the same as reaching a maximum, and confusing the two is the commonest misconception about this graph. The same argument rules out a minimum on each branch.
Socratic
The tangent and the sine share their zeros exactly.
Discussion prompt
Explain why, and say what the corresponding statement is for the tangent and the cosine.
Hint: A fraction is zero when what happens?
Answer:
A fraction is zero exactly when its numerator is zero and its denominator is not. The numerator of the tangent is the sine, so the tangent vanishes exactly where the sine does — provided the cosine is nonzero there, which it is, since cosine and sine never vanish together.
The corresponding statement for the cosine is about asymptotes rather than zeros: the tangent is undefined exactly where the cosine is zero. So the tangent inherits the sine's zeros as zeros and the cosine's zeros as asymptotes. That is the general rule for any quotient, and it also explains the cotangent, which swaps the two roles and therefore swaps its zeros and asymptotes.
Section
Section 2
Concept
The graph suggests a period of pi, but suggestion is not proof. The tangent sum identity settles it in one line, and a short argument shows that pi is not merely a period but the smallest one.
\[ \tan(x + \pi) = \frac{\tan x + \tan \pi}{1 - \tan x \tan \pi} = \frac{\tan x + 0}{1 - 0} = \tan x \]
Two steps are needed because a period is defined as the smallest positive shift that works. Showing something is a period is easy; showing nothing smaller works requires the second argument.
Figure (svg): The proof that the tangent has period pi, using the tangent sum identity with beta equal to pi so that the tangent of pi is zero and the expression collapses
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-805
Picture it
Behind the algebra there is a one-line picture, and it explains why the tangent is the odd one out.
Figure (svg): Two points diametrically opposite on the unit circle, showing that both coordinates change sign so their quotient is unchanged, which is why the tangent repeats after half a revolution
Half a revolution sends every point to the diametrically opposite one, negating both coordinates. Cosine and sine each notice; their quotient cannot, because both signs cancel.
Worked example
One application of the tangent sum identity.
\[ \text{Show that } \tan(x + \pi) = \tan(x). \]
Apply the tangent sum identity
Why: With beta equal to pi.
\[ \frac{\tan x + \tan \pi}{1 - \tan x \tan \pi} \]
Evaluate the tangent of pi
Why: The sine of pi is zero and the cosine is negative one, so the tangent is zero.
\[ \tan \pi = 0 \]
Substitute
Why: The numerator loses its second term and the denominator becomes 1.
\[ = \tan x / 1 \]
Conclude
Why: The value is unchanged, so pi is a period.
\[ = \tan x \]
Figure (svg): The solution to Worked example prove pi is a period shown as a ladder of expressions, one row per legal move
\[ \tan(x + \pi) = \tan(x) \text{ for all } x \text{ where both are defined} \]
Verify: test at pi over four
Why: The tangent of pi over four is 1, and the tangent of pi over four plus pi, which is five pi over four, is also 1. The values agree, as the identity says they must.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-805
Fill the middle
The geometric argument for the shorter period.
Fill in the blanks
\text-y \theta: \; \tan = \frac-x___ \qquad \text___ \theta + \pi: \; \tan = \frac___}___} = \frac______
Why: Adding pi sends the point to the diametrically opposite one, which negates both coordinates. The quotient of two negated quantities is unchanged, since the two minus signs cancel. Cosine and sine individually do change sign, which is why their periods are two pi while the tangent's is only pi.
Worked example
The second half of the proof, and the half that is usually skipped.
\[ \text{Show the period of } \tan \text{ is exactly } \pi, \text{ not smaller.} \]
Suppose p is a positive period
Why: That means the tangent of x plus p equals the tangent of x for every x.
\[ \text{assume } \tan(x + p) = \tan x \]
Test at a convenient value
Why: Putting x equal to zero specialises the assumption.
\[ \tan(p) = \tan(0) = 0 \]
Solve that equation
Why: The tangent vanishes exactly at the multiples of pi.
Take the smallest positive one
Why: That is pi itself.
\[ p = \pi \]
Figure (svg): The solution to Worked example show nothing smaller works shown as a ladder of expressions, one row per legal move
\[ \text{any period } p \text{ satisfies } \tan(p) = 0 \;\Longrightarrow\; p = \pi k \;\Longrightarrow\; p_{\min} = \pi \]
Verify: check that pi over two fails
Why: If pi over two were a period, the tangent of pi over four plus pi over two, which is three pi over four, would equal the tangent of pi over four. But those are negative one and one, which differ. So pi over two is genuinely not a period.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-805
Error analysis
A student reasons about the period of the tangent.
Annotate
On: \( \cos \text{ has period } 2\pi, \; \sin \text{ has period } 2\pi \;\Longrightarrow\; \tan \text{ has period } 2\pi \)
Inheriting a property from a parent function is safe for domains and for parity, but not for periods. A combination of two functions can repeat sooner than either of them does, and the tangent is the standard example.
Prediction
The tangent of some angle is 0.7.
Predict first
What is the tangent of that angle plus pi?
Correct: 0.7.
Why: Adding pi is adding exactly one period, so the tangent is unchanged. This is worth contrasting with the sine and cosine, both of which would change sign under the same shift. The distinctive behaviour of the tangent under a half-turn is the single fact that separates it from the other functions, and it is what makes tangent equations have one family of solutions where sine and cosine equations have two.
Discrimination
Only two of the six repeat after half a revolution.
Sort into buckets
Sort each function by its period.
Explain it to yourself
Proving the period required two separate arguments.
Discussion prompt
Explain why one argument was not enough, and what each of the two actually established.
Hint: What exactly does the word period mean?
Answer:
A period is defined as the smallest positive shift that leaves the function unchanged. The identity computation establishes only that pi is such a shift, which bounds the period above by pi — it does not rule out something smaller working too.
The second argument bounds it below. Assuming any period p and testing at zero forces p to be a multiple of pi, and the smallest positive multiple is pi. Together the two arguments pin the period to exactly pi. Establishing a minimum always needs an argument that no smaller value works, and that argument is a different kind of reasoning from a direct computation.
Section
Section 3
Concept
The cotangent is the cosine over the sine, so it takes its zeros from the cosine and its asymptotes from the sine — exactly the opposite of the tangent. And every branch falls rather than rises.
So the tangent's zeros are the cotangent's asymptotes and vice versa, which is the general relationship between any function and its reciprocal, seen here in its cleanest form.
Figure (svg): The cotangent curve over two full periods, falling from positive infinity to negative infinity between each pair of consecutive vertical asymptotes at the multiples of pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-806
Picture it
Compare with the tangent figure. The asymptotes have moved and every branch has been turned upside down.
Figure (svg): The cotangent curve over two full periods, falling from positive infinity to negative infinity between each pair of consecutive vertical asymptotes at the multiples of pi
Notice that the cotangent is not the tangent shifted — it is the tangent reflected as well. That is because reciprocating a rising function that passes through zero gives a falling one.
Worked example
Go back to the definition and read off both lists.
\[ \text{Find the zeros and asymptotes of } y = \cot(x). \]
Write it as a quotient
Why: Cotangent is the cosine over the sine.
\[ \cot = \cos / \sin \]
Zeros come from the numerator
Why: The cosine vanishes at the odd multiples of pi over two.
\[ \text{zeros at } \frac{\pi}{2} + \pi k \]
Asymptotes come from the denominator
Why: The sine vanishes at the multiples of pi.
Compare with the tangent
Why: The two lists are exactly swapped.
Figure (svg): The solution to Worked example locate the cotangent's features shown as a ladder of expressions, one row per legal move
\[ \text{zeros: } x = \tfrac{\pi}{2} + \pi k; \qquad \text{asymptotes: } x = \pi k \]
Verify: check at pi over two
Why: The cotangent there is the cosine of pi over two over the sine of pi over two, which is zero over one, namely zero. And the tangent at pi over two is undefined. The two functions genuinely swap behaviour at every one of these points.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-805
Comparison
Fill the blanks from memory. They agree on more than they differ on, but the differences are the memorable part.
Comparison matrix
| Property | tangent | cotangent |
|---|---|---|
| Zeros | multiples of pi | odd multiples of pi over 2 |
| Asymptotes | odd multiples of pi over 2 | multiples of pi |
| Direction of each branch | increasing | decreasing |
| Range | all real numbers | all real numbers |
| Period | pi | pi |
| Parity | odd | odd |
Three rows differ and three agree. Both differing rows about zeros and asymptotes are the same fact, since reciprocals always exchange those two.
Worked example
The direction follows from the reciprocal relationship rather than needing separate investigation.
\[ \text{Explain why each branch of } y = \cot(x) \text{ decreases.} \]
Take a branch of the tangent
Why: On the interval from zero to pi over two the tangent rises from 0 to infinity.
\[ \tan\text{ rises } 0\text{ to } \infty \]
Reciprocate
Why: The cotangent is one over the tangent wherever both exist.
\[ \cot = 1 / \tan \]
Track the reciprocal
Why: As the tangent rises from near zero to infinity, its reciprocal falls from infinity to near zero.
\[ \cot\text{ falls } \infty\text{ to } 0 \]
Extend to the whole branch
Why: The same reasoning on the other half gives a fall from zero to negative infinity.
Figure (svg): The solution to Worked example why the branches fall shown as a ladder of expressions, one row per legal move
\[ \tan \text{ increasing} \;\Longrightarrow\; \cot = \tfrac{1}{\tan} \text{ decreasing on each branch} \]
Verify: check two values
Why: The cotangent of pi over six is root three, about 1.73, and the cotangent of pi over three is root three over three, about 0.58. The angle increased and the value fell, as claimed.
Trap
\[ y = \cot(x) = \tan\left(x + \tfrac{\pi}{2}\right) \]
Assume a quarter-period shift relates them, as it did for secant and cosecant
Why: That pattern held for the previous pair, so it looks general.
Testing at x equal to pi over four: the cotangent is 1, and the tangent of three pi over four is negative 1. The two do not agree.
\[ \cot(x) = -\tan\left(x + \tfrac{\pi}{2}\right) = \tan\left(\tfrac{\pi}{2} - x\right) \]
A reflection is needed as well as a shift
Why: The cotangent falls where the tangent rises, and no pure translation can reverse a direction.
The second form is the cofunction identity, and it is the more illuminating one: the cotangent at an angle is the tangent at its complement, and complementing reverses direction because as the angle grows the complement shrinks. That is exactly why one climbs and the other falls.
Matching
Use the zeros and asymptotes of each function.
Match the pairs
Why: The tangent is zero where its numerator, the sine, vanishes and undefined where its denominator, the cosine, vanishes. The cotangent swaps both. Notice the pairing across the two functions: wherever one is zero the other is undefined, which is the general behaviour of a reciprocal pair and the reason their graphs interleave so neatly.
Prediction
The cotangent of some angle is 4.
Predict first
What is the tangent of that angle?
Correct: 1/4.
Why: The two functions are reciprocals wherever both are defined, so a cotangent of 4 means a tangent of one quarter. A large cotangent corresponds to a small tangent, which is exactly why one function's asymptotes are the other's zeros: an infinite value corresponds to a zero one. No quadrant information is needed, since a reciprocal preserves sign.
Counterexample
A student proposes: since the cotangent is the reciprocal of the tangent, their graphs are reflections of each other in the x-axis.
Discussion prompt
Give a specific counterexample and describe what the correct relationship is.
Hint: Reflection in the x-axis means negating. Is a reciprocal the same as a negative?
Answer:
\[ \tan\left(\tfrac{\pi}{3}\right) = \sqrt{3} \approx 1.73, \qquad \cot\left(\tfrac{\pi}{3}\right) = \tfrac{\sqrt{3}}{3} \approx 0.58 \]
Reflection in the x-axis would make the cotangent negative root three, about negative 1.73. It is not — it is positive 0.58, the reciprocal rather than the negative. The two operations agree only at 1 and negative 1.
The correct relationship is the cofunction identity: the cotangent of x is the tangent of pi over two minus x, which is a reflection in a vertical line combined with a shift, not a reflection in the x-axis. That composition is what reverses the direction of the branches while keeping the sign of the values.
Section
Section 4
Concept
The quarter-mark method still works, but the five marks are different from the ones used for a sinusoid, and different again between the two functions — because the natural cycle is bounded by asymptotes rather than by peaks.
The five marks still describe one complete cycle; they just describe a cycle with a different shape. Using the sinusoid marks here produces a graph covering two periods, which is a common and confusing error.
Figure (svg): The different quarter marks used for tangent and cotangent, tangent running from negative pi over two to pi over two and cotangent from zero to pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 805-807
Picture it
Both cycles have length pi, but they sit in different places on the axis.
Figure (svg): The different quarter marks used for tangent and cotangent, tangent running from negative pi over two to pi over two and cotangent from zero to pi
The shape of each cycle is the same: asymptote, negative value, zero, positive value, asymptote for the tangent, and the same reversed for the cotangent.
Worked example
Example 10.5.5, part 1. Same method, different quarter marks.
\[ \text{Graph one cycle of } f(x) = 1 - \tan\left(\frac{x}{2}\right) \text{ and state the period.} \]
Set the argument to the tangent quarter marks
Why: Those are negative pi over two through pi over two.
\[ \frac{x}{2} = -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2} \]
Solve each for x
Why: Multiply through by 2.
\[ x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \]
Substitute back into f
Why: The outer ones are undefined; the inner three give values.
\[ \text{undef}, 2, 1, 0, \text{undef} \]
Read the period
Why: The cycle runs from negative pi to pi.
\[ \text{period } = 2 \pi \]
Figure (svg): One cycle of the function one minus the tangent of x over two, running between asymptotes at negative pi and pi with a period of two pi
\[ \text{period } 2\pi; \quad \text{asymptotes } x = \pm\pi; \quad \text{points } \left(-\tfrac{\pi}{2}, 2\right), (0, 1), \left(\tfrac{\pi}{2}, 0\right) \]
Verify: check the direction of the branch
Why: The values go 2, then 1, then 0 as x increases, so the branch is falling. That is right: the coefficient of the tangent is negative 1, which reverses the usual rising behaviour.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 807-807
Sorting
The marks must span exactly one period.
Sort into buckets
Sort each set of quarter marks by which function it belongs to.
Worked example
Example 10.5.5, part 2. The cotangent marks start at zero rather than at negative pi over two.
\[ \text{Graph one cycle of } g(x) = 2\cot\left(\frac{\pi}{2}x + \pi\right) + 1 \text{ and state the period.} \]
Set the argument to the cotangent quarter marks
Why: Those are zero through pi.
\[ a r g = 0, \frac{\pi}{4}, \frac{\pi}{2}, 3 \pi / 4, \pi \]
Solve each for x
Why: Subtract pi and multiply by two over pi.
\[ x = -2, -\frac{3}{2}, -1, -\frac{1}{2}, 0 \]
Substitute back into g
Why: The outer ones are undefined; the inner three give values.
\[ \text{undef}, 3, 1, -1, \text{undef} \]
Read the period
Why: The cycle runs from negative two to zero.
\[ \text{period } = 2 \]
Figure (svg): The solution to Worked example graph a transformed cotangent shown as a ladder of expressions, one row per legal move
\[ \text{period } 2; \quad \text{asymptotes } x = -2, 0; \quad \text{points } \left(-\tfrac{3}{2}, 3\right), (-1, 1), \left(-\tfrac{1}{2}, -1\right) \]
Verify: check the direction and the period formula
Why: The values fall from 3 to 1 to negative 1, which is right for a cotangent with a positive coefficient. And the period is pi over the coefficient of x, which is pi divided by pi over two, namely 2 — matching the interval length.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 807-808
Error analysis
A student graphs y equals tangent x by setting the argument to the usual five values.
Annotate
On: \( x = 0, \; \tfrac{\pi}{2}, \; \pi, \; \tfrac{3\pi}{2}, \; 2\pi \)
The quarter marks must span one period of the function being graphed, and the period here is half what it was for the sinusoids. Using the wrong marks does not give a wrong graph so much as a graph of two cycles labelled as one, which then makes the stated period wrong by a factor of two.
Fill the middle
Find the asymptotes of one cycle of y equals tangent of the quantity 2x minus pi.
Fill in the blanks
2x - \pi = -\fracpi/43 pi/4 \;\Longrightarrow\; x = ___ \qquad 2x - \pi = \frac______ \;\Longrightarrow\; x = ___
Why: The tangent's asymptotes are where its argument is an odd multiple of pi over two, so setting the argument to negative pi over two and to pi over two gives the two ends of one cycle. Solving each gives pi over four and three pi over four, an interval of length pi over two — which matches the period formula, pi over the coefficient 2.
Prediction
A tangent function has argument 3x.
Predict first
What is its period?
Correct: pi over 3.
Why: For tangent and cotangent the period is pi divided by the coefficient of x, not two pi divided by it. So the period here is pi over three. This is the single most-missed detail when moving from the sinusoids to these two functions, since the same formula with two pi in it gives the wrong answer by a factor of two. The check is to solve for two consecutive asymptotes and measure the gap.
Edge cases
The tangent's fundamental cycle is usually taken to be from negative pi over two to pi over two.
Discussion prompt
Are the endpoints included in that interval? What would go wrong if they were, and what does this tell you about how the graph should be drawn there?
Hint: What is the tangent of pi over two?
Answer:
They are excluded. The tangent is undefined at both endpoints, so the interval is open at both ends and written with round brackets.
Including them would claim the function has values there, which it does not. On the graph, this means the curve must be drawn approaching each dashed asymptote without ever touching it, and with no dot at the end of the branch — a filled endpoint would assert a value that does not exist.
Contrast this with the fundamental cycle of the cosine, which is a closed interval because the cosine is defined at both ends and takes the same value at each. The bracket type is doing real work in both cases, and it is worth writing correctly.
Section
Section 5
Concept
Period, phase shift and vertical shift all apply as before, with the period formula adjusted. There is no amplitude, but there is a new question a sinusoid never raised: which way does each branch run?
The book notes that it omits a formal theorem for these, on the grounds that classical applications of tangent and cotangent are far rarer than those of sinusoids — and invites the reader to formulate one. Everything needed is above.
Figure (svg): One cycle of the function one minus the tangent of x over two, running between asymptotes at negative pi and pi with a period of two pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 808-808
Picture it
A negative coefficient turns a rising tangent into a falling one without moving a single asymptote.
Figure (svg): One cycle of the function one minus the tangent of x over two, running between asymptotes at negative pi and pi with a period of two pi
The zeros do move, because a vertical shift moves them; the asymptotes never do, because they are decided by the argument alone.
Worked example
Give every parameter that applies and name the direction.
\[ \text{Describe } y = -3\cot(2x) + 1. \]
Compute the period
Why: Pi over the coefficient of x.
\[ \text{period } = \frac{\pi}{2} \]
Locate the asymptotes
Why: Where the argument is a multiple of pi.
\[ 2 x = \pi k,\text{ so } x = \pi k / 2 \]
State the vertical shift
Why: The constant added outside.
\[ \text{up } 1 \]
Determine the direction
Why: The cotangent normally falls, and the negative coefficient reverses that.
Figure (svg): The solution to Worked example describe a transformed cotangent shown as a ladder of expressions, one row per legal move
\[ \text{period } \tfrac{\pi}{2}, \quad \text{asymptotes } x = \tfrac{\pi k}{2}, \quad \text{up } 1, \quad \text{branches increasing} \]
Verify: test two points
Why: At pi over eight the cotangent of pi over four is 1, so the value is negative 3 plus 1, which is negative 2. At three pi over eight the cotangent of three pi over four is negative 1, so the value is 3 plus 1, which is 4. The value rose as x rose, confirming the reversed direction.
Sorting
The formula depends on the function's own period.
Sort into buckets
Sort each function by which period formula applies.
Worked example
A tangent graph has no midline in the sinusoid sense, but a shifted one still has a level it crosses once per branch.
\[ \text{Where does } y = \tan(x) + 4 \text{ take the value } 4? \]
Set up the equation
Why: The added constant is exactly what is being asked for.
\[ \tan x + 4 = 4 \]
Simplify
Why: Subtracting 4 from both sides.
\[ \tan x = 0 \]
Solve
Why: The tangent vanishes at the multiples of pi.
\[ x = \pi k \]
Interpret
Why: Those are the zeros of the parent, raised to height 4.
Figure (svg): The solution to Worked example find where the graph crosses its midline shown as a ladder of expressions, one row per legal move
\[ x = \pi k, \quad k \in \mathbb{Z} \]
Verify: check that it is once per branch
Why: The asymptotes are at odd multiples of pi over two and the crossings at multiples of pi, so exactly one crossing sits between each consecutive pair of asymptotes. That is what must happen for a strictly increasing branch running from negative infinity to positive infinity: it hits every level exactly once.
Trap
\[ y = \tan(4x): \quad \text{period } = \frac{2\pi}{4} = \frac{\pi}{2} \]
Divide two pi by the coefficient, as for a sinusoid
Why: The formula is so familiar from the previous lesson that it transfers automatically.
But the tangent's own period is pi, not two pi. Compressing by a factor of 4 gives pi over four, and the answer above is twice too large.
\[ \text{period } = \frac{\pi}{4} \]
Divide the function's own period by the coefficient
Why: For tangent and cotangent that period is pi.
The reliable route avoids the formula entirely: solve for two consecutive asymptotes and take the gap. Here 4x equals negative pi over two and pi over two give x equal to negative pi over eight and pi over eight, a gap of pi over four. That method cannot be misremembered because it uses no formula at all.
Faded example
Describe y equals tangent of the quantity x over 3, plus 2.
Fill in the blanks
\text3 pi = \frac+- 3 pi/2___ = ___, \quad \text___ \frac______ = \pm\frac______, \text___ x = ___
Why: The coefficient of x is one third, so the period is pi divided by one third, which is three pi — a horizontal stretch. The asymptotes come from setting the argument to plus and minus pi over two, giving x equal to plus and minus three pi over two, which are three pi apart as the period requires. The vertical shift of 2 raises the zeros to height 2 but leaves the asymptotes untouched.
Prediction
A tangent graph has a negative coefficient in front of it.
Predict first
What changes compared with the ordinary tangent?
Correct: The branches fall instead of rising.
Why: A negative coefficient multiplies every output by a negative number, which reflects the graph in the x-axis and therefore reverses the direction of every branch. It changes nothing horizontal: the asymptotes are decided by the argument, which the coefficient does not touch, and the period is unaffected for the same reason. The zeros also stay put, since negating zero gives zero.
Real world
A security camera on a wall pans at a constant angular rate, and its beam sweeps along a straight corridor wall 4 metres away. The distance from the camera's perpendicular foot to the point the beam illuminates is 4 times the tangent of the pan angle.
Discussion prompt
Describe how that illuminated point moves as the camera pans steadily, and explain what the asymptote means physically. Why is this a genuinely tangent-shaped problem rather than a sinusoidal one?
Hint: Think about how fast the spot moves when the camera points nearly straight at the wall, versus nearly along it.
Answer:
\[ d(\theta) = 4\tan(\theta), \qquad -\tfrac{\pi}{2} < \theta < \tfrac{\pi}{2} \]
The spot moves slowly near the centre and extremely fast near the ends. At an angle of zero the camera points straight at the wall and the spot is at the foot; as the angle grows the spot accelerates away, and the tangent's steepening slope is exactly that acceleration.
The asymptote means that as the camera turns parallel to the wall the spot runs off to infinity, and at exactly parallel it never meets the wall at all. That is a real physical statement, not a mathematical artefact.
It is tangent-shaped rather than sinusoidal because the quantity being measured is a ratio of two lengths in a right triangle, opposite over adjacent, not a coordinate on a circle. Anything that measures how far along a straight line a rotating ray strikes is a tangent problem, and the unbounded range is the honest statement that a straight line goes on forever.
Comparison
Fill the blanks from memory. This table summarises the whole of Section 10.5.
Comparison matrix
| Function | Period | Range | Asymptotes at |
|---|---|---|---|
| cosine | two pi | from -1 to 1 | none |
| sine | two pi | from -1 to 1 | none |
| secant | two pi | size at least 1 | odd multiples of pi over 2 |
| cosecant | two pi | size at least 1 | multiples of pi |
| tangent | pi | all real numbers | odd multiples of pi over 2 |
| cotangent | pi | all real numbers | multiples of pi |
Read the asymptote column: there are only two patterns, and each is shared by exactly one reciprocal function and one quotient function — because both patterns are just the zeros of cosine and the zeros of sine.
Pattern
Whether you are graphing or describing a tangent or cotangent, the same five moves cover it.
The single most common error is using two pi in the period formula. Solving for two consecutive asymptotes and measuring the gap avoids the formula entirely and cannot be misremembered.
OpenStax Algebra and Trigonometry 2e, §8.2 Graphs of the Other Trigonometric Functions §8.2
Check
The period. Remember which base period applies.
Check your understanding
What is the period of y equals cotangent of 3x?
Answer: A
Why: The cotangent has base period pi, so compressing the argument by a factor of 3 gives a period of pi over 3. Checking by asymptotes: the cotangent is undefined where its argument is a multiple of pi, so 3x equals pi k gives x equal to pi k over 3, and consecutive asymptotes are pi over 3 apart.
Check
Asymptotes. Which parent supplies them?
Check your understanding
Where are the vertical asymptotes of y equals tangent x?
Answer: B
Why: The tangent is the sine over the cosine, so it is undefined where the cosine vanishes — at the odd multiples of pi over two. Those are the same asymptotes the secant has, since secant also has cosine downstairs.
Check
Range. Compare with the reciprocal functions.
Check your understanding
What is the range of y equals tangent x?
Answer: C
Why: Each branch increases continuously from negative infinity to positive infinity, so by continuity it takes every real value on the way — and takes each exactly once. There is no gap and no bound in either direction.
Real world
A road climbs at a constant gradient. Builders quote gradient as a percentage — the rise divided by the run, times 100 — while surveyors quote the angle of inclination.
Discussion prompt
Express the gradient percentage as a function of the angle, describe the shape of that function's graph, and explain what its vertical asymptote says about how steep a road can be.
Hint: Rise over run is which function of the angle?
Answer:
\[ g(\theta) = 100\tan(\theta), \qquad 0 \le \theta < \tfrac{\pi}{2} \]
The graph is one branch of a tangent, stretched by 100. It is nearly straight for small angles — a 5 degree slope is about 8.7 percent, and doubling the angle roughly doubles the percentage — but it curves upward sharply as the angle grows.
At 45 degrees the gradient is exactly 100 percent, which is worth knowing because it surprises people: 100 percent does not mean vertical, it means rising as fast as it advances. The asymptote at 90 degrees says that a vertical wall has no finite gradient percentage at all, which is why the notation breaks down for cliffs and why climbers use angles rather than percentages.
The non-linearity matters practically: the difference between a 10 and a 12 percent gradient is about one degree, while the difference between 100 and 102 percent is about a hundredth of a degree. Percentage is a poor scale near the steep end, and the tangent's shape is exactly why.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
How many times does the graph of y equals tangent x cross the horizontal line y equals 1000?
Correct: Once per period, so infinitely often.
\[ \tan(x) = 1000 \;\Longrightarrow\; x \approx 1.5698 + \pi k \]
Why: The range of the tangent is all real numbers, so 1000 is certainly attained. Each branch increases from negative infinity to positive infinity and is strictly increasing, so it crosses any horizontal line exactly once — and there are infinitely many branches. That is the crucial difference from a sinusoid, where a horizontal line inside the range is crossed twice per period. It is also why tangent equations have one family of solutions where sine and cosine equations have two.
Explain it
They have graphed cosine, sine, secant and cosecant and are expecting the tangent to look like another wave with gaps in it.
Discussion prompt
In no more than five sentences, tell them how the tangent differs and why. Give them the one structural reason behind all the differences.
Hint: What is the tangent built out of, and how is that different from a secant?
Answer:
A usable answer: the secant was one over a bounded function, so it could never be small — its values got pushed outward, away from zero. The tangent is one bounded function divided by another, and a small number divided by a small number can be anything at all, so the tangent takes every real value.
That is why it climbs steadily instead of turning around, why it has no amplitude and no maximum, and why the graph runs the full height of the picture between each pair of asymptotes. It also repeats twice as often, every pi rather than every two pi, because a half-turn flips the sign of both the top and the bottom and a fraction cannot see two cancelling sign changes.
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: The period is fixed by solving for two consecutive asymptotes and measuring the gap, which uses no formula at all. The quarter marks are fixed by remembering that both cycles are bounded by asymptotes, so they start where the function is undefined. Direction is fixed by two facts: tangent rises, cotangent falls, and a negative coefficient reverses either. Asymptote positions are fixed by going back to the quotient and asking which parent is downstairs. 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 of the page draw two full periods of the tangent, marking the asymptotes with dashed lines and the zeros with dots, and label the period. Underneath, draw two full periods of the cotangent the same way, and label its period. Beside the two graphs write a short comparison: zeros, asymptotes, direction, range, period and parity for each. In the middle right, write the two-step proof that the period of the tangent is pi — the identity computation and the argument that nothing smaller works — and beside it draw the unit circle with two diametrically opposite points, writing why their coordinate quotients are equal. In the bottom left, list the correct quarter marks for tangent and for cotangent, and note that the sinusoid marks would span two periods. In the bottom right, graph one cycle of y equals negative two tangent of the quantity x plus pi over four, showing your quarter-mark working. Finally, circle the two functions in your comparison that have period pi and write the single sentence explaining why they are the only two.
The sentence should be that both are quotients of the two coordinates, and a half-turn negates both coordinates, so a quotient of them is unchanged while each individually flips sign.
Recap
Five things, and the last one completes the graphing section.
| If the question says | Your first move is |
|---|---|
| State the period | Divide pi, not two pi, by the coefficient of x |
| Find the asymptotes | Solve for where the denominator's parent is zero |
| Graph one cycle of a tangent | Use the marks from -pi/2 to pi/2 |
| Graph one cycle of a cotangent | Use the marks from 0 to pi |
| Which way does the branch go | Tangent rises, cotangent falls, a minus reverses it |
That completes Section 10.5 and every graph in the course. The next section asks the opposite question: given a value, which angle produced it? That requires inverting these functions, and the graphs just drawn are exactly what determine whether an inverse can exist at all.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 804-809 — everything on these slides traces back here
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