The two circular functions whose inverses are genuinely not standardised. Covers why the secant is awkward to invert at all — its range has a gap, so the inverse's domain does too — and then presents both competing restrictions in full: the trigonometry-friendly one that keeps the arccosine's interval, and the calculus-friendly one chosen to make the tangent of an arcsecant a single formula rather than a piecewise one. The same worked examples are answered under each, so the differences are visible rather than described.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.6 The Inverse Trigonometric Functions, pp. 827-833
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 — §10.6 The Inverse Trigonometric Functions §10.6, pp. 827-833 — the pages these objectives are drawn from
Warm-up
Four of the six functions have been inverted. The last two are different, and it is worth seeing why before meeting the conventions.
Discussion prompt
The secant takes every value of absolute value at least one, and nothing in between. What does that force about the domain of any inverse of it? Is that a problem?
Hint: The domain of an inverse is the range of the original.
Answer:
\[ \text{range}(\sec) = (-\infty, -1] \cup [1, \infty) \;\Longrightarrow\; \text{domain}(\operatorname{arcsec}) = (-\infty, -1] \cup [1, \infty) \]
The domain has a hole in the middle. That by itself is not fatal — plenty of useful functions have disconnected domains — but it means the restricted piece of the secant has to be assembled from two separate intervals, one for each side of the gap.
And once you are choosing two intervals rather than one, there is more than one sensible way to do it. That is why this is the only place in the course where textbooks genuinely disagree, and why this lesson presents both answers rather than one.
Concept
Restricting the secant requires picking two intervals, and two different pairs are in common use. The trigonometry-friendly choice keeps the arccosine's interval; the calculus-friendly choice sacrifices that in order to make one particular formula come out without a case split.
Neither is more correct. What matters is knowing that both exist, being able to work in either, and stating which you are using — because for negative inputs they give different answers to the same question.
Figure (svg): The two competing restrictions of the secant side by side, the trigonometry-friendly one keeping the interval from zero to pi and the calculus-friendly one using zero to pi over two together with pi to three pi over two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 827-831
Section
Section 1
Concept
Reflecting in the diagonal turns the secant's range into the arcsecant's domain, and the secant's range is missing everything strictly between negative one and one. So the arcsecant simply cannot accept an input in that band.
\[ \text{domain}(\operatorname{arcsec}) = \{x : |x| \ge 1\} \]
Compare with the cosine, whose range is a single unbroken interval and which therefore has a single obvious restricted domain. The awkwardness here is inherited from the shape of the parent's range, not invented.
Figure (svg): The secant curve with the band between negative one and one shaded out, showing that no value in that band is ever attained and so no inverse can accept such an input
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 827-828
Picture it
The shaded band is what the secant never enters, and therefore what the arcsecant never accepts.
Figure (svg): The secant curve with the band between negative one and one shaded out, showing that no value in that band is ever attained and so no inverse can accept such an input
Notice the two branches on either side of the asymptote: one lies entirely above 1 and the other entirely below negative 1. Any restriction has to take one piece from each.
Worked example
Deciding whether an arcsecant expression is even defined comes first.
\[ \text{Which of } \operatorname{arcsec}(3), \; \operatorname{arcsec}(0.5), \; \operatorname{arcsec}(-1), \; \operatorname{arcsec}(0) \text{ are defined?} \]
Recall the domain condition
Why: The input must have absolute value at least 1.
\[ | x | \ge 1 \]
Test each input
Why: Three, one half, negative one, zero.
Handle the boundary case
Why: Negative one has absolute value exactly 1, so it is included.
\[ -1\text{ yes} \]
Handle zero
Why: Zero is inside the gap, and no secant is ever zero.
\[ 0\text{ no} \]
Figure (svg): The solution to Worked example what the domain excludes shown as a ladder of expressions, one row per legal move
\[ \operatorname{arcsec}(3) \; \checkmark, \quad \operatorname{arcsec}(-1) \; \checkmark, \quad \operatorname{arcsec}(0.5) \; \times, \quad \operatorname{arcsec}(0) \; \times \]
Verify: check the two failures against the parent
Why: A secant of one half would need a cosine of 2, which is impossible. A secant of zero would need one over a cosine to be zero, which cannot happen since the numerator is 1. Both failures are structural, not narrow misses.
Sorting
The condition is absolute value at least one.
Sort into buckets
Sort each input by whether arcsecant is defined there.
Worked example
The asymptote between them is not an inconvenience but an obstruction.
\[ \text{Why can the restricted secant not be a single interval containing } \tfrac{\pi}{2}\text{?} \]
Note the secant is undefined at pi over two
Why: The cosine vanishes there.
\[ \text{asymptote at } \frac{\pi}{2} \]
Consider an interval containing it
Why: Any such interval contains a point where the function does not exist.
Consider the values either side
Why: Just below pi over two the secant is huge and positive; just above, huge and negative.
Conclude
Why: Any restriction must take one piece from each branch, and the pieces are separated by the asymptote.
Figure (svg): The solution to Worked example why the two intervals cannot be joined shown as a ladder of expressions, one row per legal move
\[ \text{the branch below } \tfrac{\pi}{2} \text{ gives } \sec \ge 1; \text{ the branch above gives } \sec \le -1 \]
Verify: check both halves of the range are needed
Why: The domain of arcsecant includes both 3 and negative 3, so the restricted secant must attain both. One branch cannot do that, since each branch stays entirely on one side. Two intervals are genuinely forced.
Trap
\[ \operatorname{arcsec}\left(\tfrac{1}{2}\right) = \arccos(2) \]
Apply the reciprocal identity without checking the domain
Why: The identity arcsec of x equals arccos of one over x is genuinely useful and gets applied by reflex.
But both sides are undefined here. Arcsecant of one half fails because one half is inside the gap, and arccosine of 2 fails because 2 is outside the interval from negative one to one.
Check the domain first. An arcsecant input must have absolute value at least 1, and the identity is only worth applying once that is confirmed.
\[ \operatorname{arcsec}(2) = \arccos\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3} \quad \checkmark \]
Notice that the identity maps the domain correctly when it applies: an input of absolute value at least 1 has a reciprocal of absolute value at most 1, which is exactly the arccosine's domain. The two domains are reciprocals of each other, which is why the identity works at all — and why it fails on inputs that were never legitimate.
Prediction
The domain of arccosine is the interval from negative one to one, and the domain of arcsecant is everything of absolute value at least one.
Predict first
How are those two sets related?
Correct: They are complementary, apart from the two shared endpoints.
Why: Between them the two domains cover the whole real line, overlapping only at negative one and one. That is exactly what the reciprocal relationship predicts: taking reciprocals maps the interval from negative one to one onto everything of absolute value at least one, and fixes the two endpoints. It is also why the identity relating the two functions works, and why zero belongs to neither domain — zero has no reciprocal.
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 arcsecant's domain is wrong.
Survives elimination: B
Why: Statement B is false. The domain is the union of two separate rays, one running from negative infinity to negative one and one from one to infinity, with a gap between them. It is not an interval, and that disconnection is precisely what forces the restricted domain of the secant to be two intervals as well — which is where the two conventions come from.
Socratic
Restricting the cosine needed one interval. Restricting the secant needs two.
Discussion prompt
Explain the difference, tracing it back to a property of each function's range.
Hint: Is each range connected?
Answer:
The cosine's range is the single unbroken interval from negative one to one. To attain all of it, a restricted domain need only be one interval on which the cosine runs from one end of the range to the other — and zero to pi does exactly that.
The secant's range is two separate pieces, and the secant cannot pass from one to the other continuously, since doing so would require passing through the gap. So the restricted domain must contain one piece for each part of the range, and those two pieces are necessarily separated by an asymptote.
The general principle: the connectedness of the range dictates the connectedness of the restricted domain. Every awkwardness in this lesson follows from the secant's range having a hole in it.
Section
Section 2
Concept
The first convention restricts the secant to the same interval used for the cosine, zero to pi, with the point pi over two removed because the secant is undefined there. The cosecant is restricted analogously, to negative pi over two to pi over two with zero removed.
The great advantage is that the reciprocal identity holds without exception on the whole domain, so any arcsecant can be turned into an arccosine and evaluated. That is why this convention is called trigonometry friendly.
| arcsecant | arccosecant | |
|---|---|---|
| Domain | size at least 1 | size at least 1 |
| Range | 0 to pi, without pi/2 | -pi/2 to pi/2, without 0 |
| Reciprocal identity | arccos of 1 over x | arcsin of 1 over x |
| Valid for | every x in the domain | every x in the domain |
| Parity | neither | odd |
Figure (svg): The identity that converts an arcsecant into an arccosine of a reciprocal, with the condition attached under each convention
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 827-828
Picture it
This is the whole disagreement in one picture. Look at where the second highlighted piece sits in each.
Figure (svg): The two competing restrictions of the secant side by side, the trigonometry-friendly one keeping the interval from zero to pi and the calculus-friendly one using zero to pi over two together with pi to three pi over two
The trigonometry-friendly version keeps both pieces inside zero to pi. The calculus-friendly version moves the second piece out to beyond pi, which is what makes its range look strange and its formulas come out clean.
Worked example
Example 10.6.3, parts a and b. The reciprocal identity does all the work.
\[ \text{Find } \operatorname{arcsec}(2) \text{ and } \operatorname{arccsc}(-2), \text{ trigonometry friendly.} \]
Apply the reciprocal identity to the first
Why: Valid for every input of absolute value at least 1 under this convention.
\[ \operatorname{arcsec}(2) = \arccos(\frac{1}{2}) \]
Evaluate the arccosine
Why: The angle in zero to pi with cosine one half.
\[ = \frac{\pi}{3} \]
Apply the identity to the second
Why: Arccosecant becomes arcsine of the reciprocal.
\[ \operatorname{arccsc}(-2) = \arcsin(-\frac{1}{2}) \]
Evaluate the arcsine
Why: The angle in negative pi over two to pi over two with sine negative one half.
\[ = -\frac{\pi}{6} \]
Figure (svg): The solution to Worked example evaluate under this convention shown as a ladder of expressions, one row per legal move
\[ \operatorname{arcsec}(2) = \tfrac{\pi}{3}, \qquad \operatorname{arccsc}(-2) = -\tfrac{\pi}{6} \]
Verify: check each against the range
Why: Pi over three is in zero to pi and is not pi over two, so it is a legitimate arcsecant output. Negative pi over six is in negative pi over two to pi over two and is not zero, so it is a legitimate arccosecant output. And the secant of pi over three really is 2, since its cosine is one half.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 829-829
Matching
Use the reciprocal identity, then evaluate the arccosine or arcsine.
Match the pairs
Why: Each arcsecant becomes an arccosine of the reciprocal and each arccosecant an arcsine of the reciprocal. Arccosine of negative one half is two pi over three, which is in the arcsecant's range. Arcsine of negative one half is negative pi over six, which is in the arccosecant's range. Note the asymmetry in how the negatives behave: the arcsecant answer stays positive while the arccosecant answer goes negative, because their ranges sit differently relative to zero.
Worked example
Example 10.6.3, part d. The range supplies the sign.
\[ \text{Find } \cot(\operatorname{arccsc}(-3)), \text{ trigonometry friendly.} \]
Name the inner expression
Why: Let t be the arccosecant of negative three.
\[ \csc t = -3 \]
Locate t from the range and the sign
Why: The range is negative pi over two to pi over two without zero; a negative cosecant means t is negative.
\[ t\text{ in } [-\frac{\pi}{2}, 0] \]
Use the Pythagorean identity
Why: One plus cotangent squared is cosecant squared, which is 9.
\[ \cot ^{2} t = 8 \]
Choose the sign
Why: In that interval the cotangent is negative.
\[ \cot t = -2 \sqrt{2} \]
Figure (svg): The solution to Worked example a composition and a sign shown as a ladder of expressions, one row per legal move
\[ \cot(\operatorname{arccsc}(-3)) = -2\sqrt{2} \]
Verify: check with coordinates
Why: A cosecant of negative 3 means a sine of negative one third, and with t in quadrant four the cosine is positive, root eight over three. The cotangent is cosine over sine, which is root eight over three divided by negative one third, giving negative root eight, or negative two root two — matching.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 829-829
Error analysis
A student evaluates an arccosecant under the trigonometry-friendly convention.
Annotate
On: \( \operatorname{arccsc}(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3} \)
The pairing follows the parents: secant with cosine, cosecant with sine. This is the same crossed pairing from Lesson 10.3a, and it stays crossed all the way through to the inverse functions.
Faded example
Evaluate the arcsecant of negative root two, trigonometry friendly.
Fill in the blanks
\operatorname-sqrt2/2(-\sqrt3 pi/4) = \arccos\left(___\right) = ___
Why: The reciprocal of negative root two is negative one over root two, which rationalises to negative root two over two. The arccosine of that is the angle in zero to pi with that cosine, namely three pi over four. Checking: the cosine of three pi over four is negative root two over two, so its secant is negative root two.
Prediction
Under the trigonometry-friendly convention, arccosecant is odd.
Predict first
What does that tell you about the arccosecant of negative 5, given that the arccosecant of 5 is about 0.2014?
Correct: About -0.2014.
Why: An odd function negates its output when its input is negated, so the answer is the negative. This works because the trigonometry-friendly range, from negative pi over two to pi over two without zero, is symmetric about zero. Under the calculus-friendly convention the arccosecant is not odd at all, because its range is not symmetric — which is one of the concrete differences between the two.
Edge cases
The trigonometry-friendly range for arcsecant is zero to pi with pi over two removed.
Discussion prompt
Why is pi over two removed, and what does the arcsecant do as its input grows very large?
Hint: What is the secant of pi over two?
Answer:
Pi over two is removed because the secant is undefined there — the cosine vanishes, so no input to the arcsecant could ever produce it as an output.
\[ \text{as } x \to \infty, \; \operatorname{arcsec}(x) \to \tfrac{\pi}{2}^-; \qquad \text{as } x \to -\infty, \; \operatorname{arcsec}(x) \to \tfrac{\pi}{2}^+ \]
As the input grows in either direction the output approaches pi over two from one side or the other without ever reaching it. So pi over two is a horizontal asymptote approached from both sides, which is unusual and is a direct consequence of removing a single point from the middle of the range.
Section
Section 3
Concept
The second convention restricts the secant to zero to pi over two together with pi to three pi over two. The range looks stranger, but it buys one thing: the tangent of an arcsecant becomes a single formula rather than a piecewise one.
The cost is real: the reciprocal identity now holds only for positive inputs, so a negative input must be handled from the definition. That is the trade — a cleaner formula in calculus for a messier one in evaluation.
| arcsecant | arccosecant | |
|---|---|---|
| Range | 0 to pi/2, and pi to 3pi/2 | 0 to pi/2, and pi to 3pi/2 |
| Reciprocal identity | arccos of 1 over x | arcsin of 1 over x |
| Valid for | x at least 1 ONLY | x at least 1 ONLY |
| tan of arcsec x | root of x squared minus 1, always | |
| Parity | neither | neither |
Figure (svg): A comparison of the formula for the tangent of an arcsecant under each convention, showing that the trigonometry-friendly one needs a piecewise definition while the calculus-friendly one does not
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 830-831
Picture it
This is the entire motivation, and it is worth seeing before the convention seems arbitrary.
Figure (svg): A comparison of the formula for the tangent of an arcsecant under each convention, showing that the trigonometry-friendly one needs a piecewise definition while the calculus-friendly one does not
In calculus, a trigonometric substitution replaces the root of x squared minus one with a tangent. Under the first convention that substitution splits into two cases; under the second it does not. Integration tables are written for the second.
Worked example
Example 10.6.4, parts a and b. One is unchanged and one is not.
\[ \text{Find } \operatorname{arcsec}(2) \text{ and } \operatorname{arccsc}(-2), \text{ calculus friendly.} \]
Check the identity's condition for the first
Why: The input 2 is at least 1, so the identity applies.
\[ \operatorname{arcsec}(2) = \arccos(\frac{1}{2}) = \frac{\pi}{3} \]
Check the condition for the second
Why: Negative 2 is not at least 1, so the identity does NOT apply.
Go back to the definition
Why: Find t in the range with cosecant equal to negative 2.
Locate it
Why: A cosecant of negative 2 means a sine of negative one half, and the only such t in that range is seven pi over six.
\[ = 7 \pi / 6 \]
Figure (svg): The solution to Worked example the same two evaluations, calculus friendly shown as a ladder of expressions, one row per legal move
\[ \operatorname{arcsec}(2) = \tfrac{\pi}{3}, \qquad \operatorname{arccsc}(-2) = \tfrac{7\pi}{6} \]
Verify: compare with the other convention
Why: The first answer, pi over three, is identical under both conventions. The second differs: trigonometry friendly gave negative pi over six and calculus friendly gives seven pi over six. Both angles have cosecant negative 2, and they differ by four pi over three — each convention picks the representative inside its own range.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 832-832
Discrimination
Under the calculus-friendly convention it needs a positive input at least 1.
Sort into buckets
Sort each case by whether the identity may be used, calculus friendly.
Worked example
Example 10.6.4, part d. The same question, and the sign comes out the other way.
\[ \text{Find } \cot(\operatorname{arccsc}(-3)), \text{ calculus friendly.} \]
Name the inner expression and locate t
Why: A negative cosecant, and this convention puts such angles in the second piece.
\[ t\text{ in } (\pi, 3 \pi / 2) \]
Use the Pythagorean identity
Why: The same computation as before.
\[ \cot ^{2} t = 8 \]
Choose the sign from the interval
Why: Between pi and three pi over two, both coordinates are negative, so the cotangent is positive.
\[ \cot t > 0 \]
State the answer
Why: The positive root this time.
\[ \cot t = 2 \sqrt{2} \]
Figure (svg): The solution to Worked example the composition, calculus friendly shown as a ladder of expressions, one row per legal move
\[ \cot(\operatorname{arccsc}(-3)) = 2\sqrt{2} \]
Verify: compare with the trigonometry-friendly answer
Why: That one gave negative two root two. Both are correct answers to the question as each convention poses it, because the two conventions are asking about different angles — one in quadrant four and one in quadrant three. This is the clearest possible demonstration that the convention is not a formality.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 832-832
Trap
\[ \text{calculus friendly: } \operatorname{arcsec}(-2) = \arccos\left(-\tfrac{1}{2}\right) = \tfrac{2\pi}{3} \]
Apply the identity as it was used under the other convention
Why: The identity looks the same in both theorems, and the condition attached to it is easy to skim past.
But two pi over three is not in the calculus-friendly range, which is zero to pi over two together with pi to three pi over two. So the answer cannot be right under this convention.
Under the calculus-friendly convention the identity is stated for x at least 1 only. For a negative input, go back to the definition.
\[ \sec(t) = -2, \; t \in [\pi, \tfrac{3\pi}{2}) \;\Longrightarrow\; t = \tfrac{4\pi}{3} \]
The book flags this explicitly with a footnote inviting you to compare the two theorems on exactly this point. The identity looks identical in both and its condition does not, which makes it the single most dangerous line in the lesson.
Fill the middle
Evaluate the arcsecant of negative 2 under the calculus-friendly convention.
Fill in the blanks
\sec t = -2, \; t \in [\pi, \tfrac-1/24 pi/3) \;\Longrightarrow\; \cos t = ___ \;\Longrightarrow\; t = ___
Why: A secant of negative 2 means a cosine of negative one half. Between pi and three pi over two the angle with that cosine is four pi over three, since its reference angle is pi over three and it lies in quadrant three. Under the trigonometry-friendly convention the answer would have been two pi over three instead — the same cosine, a different representative.
Prediction
Under the calculus-friendly convention, tangent of arcsecant of x equals the root of x squared minus one, with no cases.
Predict first
Why does no case split arise?
Correct: Because the tangent is always positive on that range.
Why: The range consists of zero to pi over two and pi to three pi over two, and the tangent is non-negative on both of those — quadrant one and quadrant three are exactly the two where the tangent is positive. So taking the positive square root is always correct and no sign analysis is needed. Under the trigonometry-friendly convention the second piece lies in quadrant two, where the tangent is negative, which is what forces the case split there.
Explain it to yourself
The second convention exists to make one formula in calculus come out without cases.
Discussion prompt
Explain what a convention is buying and what it is spending, and say whether it is possible to have both advantages at once.
Hint: Look at where each convention's second interval sits and which quadrant that is.
Answer:
The calculus-friendly convention buys a tangent of constant sign across its whole range, by putting the second interval in quadrant three where the tangent is positive, matching quadrant one. It spends the reciprocal identity for negative inputs, because quadrant three does not contain the angles the arccosine would return.
The two cannot both be had. Matching arccosine forces the second interval into quadrant two, and quadrant two has a negative tangent. The tangent's sign and the arccosine's range genuinely conflict, which is exactly why two conventions exist rather than one: they are the two consistent ways to resolve a real tension, and each discipline picked the one that suited its formulas.
Section
Section 4
Concept
The two conventions agree completely on positive inputs and disagree on every negative one. Knowing that narrows the whole problem: if every input in a question is positive, the convention does not matter at all.
So the practical rule is short: work positive if you can, and state your convention when you cannot.
Figure (svg): Two columns separating the results that are the same under both conventions from those that differ
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 828-833
Picture it
Everything on the left is safe to use without stating a convention. Everything on the right is not.
Figure (svg): Two columns separating the results that are the same under both conventions from those that differ
Notice that the differing items are the ones most likely to appear in an exam question, which is precisely why the convention has to be stated.
Worked example
Positive input throughout, so no ambiguity arises.
\[ \text{Find } \tan(\operatorname{arcsec}(3)) \text{ under each convention.} \]
Locate t under the first convention
Why: A positive secant puts t in zero to pi over two.
\[ t\text{ in } [0, \frac{\pi}{2}] \]
Locate t under the second
Why: Also zero to pi over two, since both conventions share that piece.
Use the Pythagorean identity
Why: One plus tangent squared is secant squared, which is 9.
\[ \tan ^{2} t = 8 \]
Choose the sign
Why: The tangent is positive in quadrant one under either convention.
\[ \tan t = 2 \sqrt{2} \]
Figure (svg): The solution to Worked example a composition where the conventions agree shown as a ladder of expressions, one row per legal move
\[ \tan(\operatorname{arcsec}(3)) = 2\sqrt{2} \quad \text{either way} \]
Verify: confirm the agreement is structural
Why: Both conventions use the interval from zero to pi over two for inputs at least 1, so any question with only positive inputs cannot distinguish them. That is worth knowing, since it means most routine practice is convention-independent.
Sorting
They agree exactly when every input involved is positive.
Sort into buckets
Sort each evaluation by whether both conventions give the same answer.
Worked example
The same expression at a negative input, worked both ways.
\[ \text{Find } \tan(\operatorname{arcsec}(-3)) \text{ under each convention.} \]
Locate t, trigonometry friendly
Why: A negative secant puts t in pi over two to pi, which is quadrant two.
\[ t\text{ in } (\frac{\pi}{2}, \pi) \]
Find the tangent there
Why: The tangent is negative in quadrant two.
\[ \tan t = -2 \sqrt{2} \]
Locate t, calculus friendly
Why: A negative secant puts t in pi to three pi over two, which is quadrant three.
\[ t\text{ in } [\pi, 3 \pi / 2] \]
Find the tangent there
Why: The tangent is positive in quadrant three.
\[ \tan t = +2 \sqrt{2} \]
Figure (svg): The solution to Worked example a composition where they disagree shown as a ladder of expressions, one row per legal move
\[ \text{trig: } -2\sqrt{2}; \qquad \text{calculus: } +2\sqrt{2} \]
Verify: check both are genuinely correct
Why: Both angles have secant negative 3, so both are legitimate answers to their own convention's question. The magnitudes agree because the Pythagorean identity cannot see a sign; only the quadrant differs, and the quadrant is exactly what the convention chooses.
Error analysis
A student evaluates a composition, using the reciprocal identity and then a range.
Annotate
On: \( \operatorname{arcsec}(-4) = \arccos\left(-\tfrac{1}{4}\right) \approx 1.823, \quad \text{so } \tan \approx +\sqrt{15} \)
Pick a convention at the start of a problem and stay in it. Mixing them produces an answer that is wrong under both, which is worse than picking either one — and it is a genuinely easy mistake, since each individual step looks defensible.
Comparison
Fill the blanks from memory. Only the negative half is in dispute.
Comparison matrix
| Trigonometry friendly | Calculus friendly | |
|---|---|---|
| arcsec range | 0 to pi, without pi/2 | 0 to pi/2 and pi to 3pi/2 |
| arcsec(2) | pi over 3 | pi over 3 |
| arcsec(-2) | two pi over 3 | four pi over 3 |
| Reciprocal identity valid for | all x with |x| at least 1 | x at least 1 only |
| tan(arcsec x) | piecewise | root of x squared minus 1 |
The second row is the point: for positive inputs the two columns are identical, and every difference in the table involves a negative one.
Prediction
Two students evaluate the arcsecant of negative 4 under the two different conventions.
Predict first
How will their answers be related?
Correct: They will differ by pi.
Why: The trigonometry-friendly convention places the angle in quadrant two and the calculus-friendly one places it in quadrant three, and those two are exactly a half-turn apart for a given secant value. Concretely, arccosine of negative one quarter is about 1.823, and the calculus-friendly answer is about 4.965, which is 1.823 plus pi. The relationship is a useful practical shortcut for converting between the conventions.
Explain it
A classmate has found two different answers for arcsec of negative 2 in two different textbooks and thinks one of them must contain an error.
Discussion prompt
In four sentences or fewer, explain what is really going on and what they should do about it.
Hint: Is there a fact of the matter here?
Answer:
A usable answer: neither book is wrong. Infinitely many angles have a secant of negative 2, and someone has to choose which one the arcsecant function returns — but unlike arcsine and arccosine, mathematicians never agreed on a single choice for arcsecant. One book returns the angle in quadrant two and the other the one in quadrant three.
What to do: check which range the book states, use that one consistently throughout a problem, and if you are writing something for someone else, say which convention you are using. For positive inputs it makes no difference at all, so this only matters when a negative shows up.
Section
Section 5
Concept
Calculators provide arcsine, arccosine and arctangent, and nothing else. Getting an arcsecant, arccosecant or arccotangent requires converting to one of those three, and the conversion has conditions.
The negative arccotangent is worth practising because the failure is silent: a calculator will happily return the arctangent of the reciprocal and give an answer that is off by exactly pi.
Figure (svg): The reference-angle method for finding the arccotangent of a negative number on a calculator that only has arctangent
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 833-834
Picture it
This picture is the method for every negative input to an arccotangent.
Figure (svg): The reference-angle method for finding the arccotangent of a negative number on a calculator that only has arctangent
The acute angle alpha is found from the positive value, which the identity handles. Then the answer is pi minus alpha, because the arccotangent's range puts the angle in quadrant two.
Worked example
Example 10.6.5, parts a and b. Each converts to a button that exists.
\[ \text{Approximate } \operatorname{arccot}(2) \text{ and } \operatorname{arcsec}(5) \text{ to four decimal places.} \]
Convert the arccotangent
Why: The input is positive, so the identity applies.
\[ \operatorname{arccot}(2) = \arctan(\frac{1}{2}) \]
Evaluate in radian mode
Why: A calculator gives this directly.
\[ = 0.4636 \]
Convert the arcsecant
Why: The input is at least 1, so both conventions permit the identity.
\[ \operatorname{arcsec}(5) = \arccos(\frac{1}{5}) \]
Evaluate
Why: Again a direct calculator step.
\[ = 1.3694 \]
Figure (svg): The solution to Worked example three straightforward approximations shown as a ladder of expressions, one row per legal move
\[ \operatorname{arccot}(2) \approx 0.4636, \qquad \operatorname{arcsec}(5) \approx 1.3694 \]
Verify: sanity-check both
Why: An arccotangent of 2 should be a small positive angle, since a large cotangent means a shallow angle — and 0.4636 is about 26.6 degrees. An arcsecant of 5 means a cosine of 0.2, so an angle just past 78 degrees, and 1.3694 radians is 78.5 degrees. Both are plausible.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 833-833
Sorting
Each identity carries a condition on the sign of the input.
Sort into buckets
Sort each conversion by whether it can be applied directly.
Worked example
Example 10.6.5, part c. This is the one where the direct route fails.
\[ \text{Approximate } \operatorname{arccot}(-2) \text{ to four decimal places.} \]
Note the identity does not apply
Why: It is stated for positive inputs only, and this one is negative.
Locate the answer
Why: The arccotangent range is zero to pi, and a negative cotangent puts the angle in quadrant two.
\[ \frac{\pi}{2} < t < \pi \]
Find the reference angle
Why: By the Reference Angle Theorem the reference angle has cotangent 2, which is arccot(2).
\[ \alpha = \arctan(\frac{1}{2}) \]
Place it in quadrant two
Why: The angle is pi minus its reference angle.
\[ t = \pi - \arctan(\frac{1}{2}) \]
Figure (svg): The solution to Worked example the negative arccotangent shown as a ladder of expressions, one row per legal move
\[ \operatorname{arccot}(-2) = \pi - \arctan\left(\tfrac{1}{2}\right) \approx 2.6779 \]
Verify: check by the alternative route
Why: The book gives a second method: the arctangent of negative one half is about negative 0.4636, in quadrant four, and the quadrant two angle with the same tangent is exactly pi more, giving pi minus 0.4636, about 2.6779. Both routes agree, and both land in the arccotangent's range.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 834-834
Trap
\[ \operatorname{arccot}(-2) = \arctan\left(-\tfrac{1}{2}\right) \approx -0.4636 \]
Apply the identity regardless of sign
Why: It worked for positive inputs and nothing about the calculator complains.
But the arccotangent's range is zero to pi, entirely positive, so a negative answer cannot be right. The calculator has silently answered a different question.
\[ \operatorname{arccot}(-2) = \pi + \arctan\left(-\tfrac{1}{2}\right) \approx 2.6779 \]
Add pi to place the answer in the correct range
Why: The tangent and cotangent both have period pi, so adding pi finds the angle with the same value in the other half.
The check that catches this every time: is the answer inside the range of the function you asked for? An arccotangent answer must be between zero and pi. A negative result means the wrong branch was returned, and adding pi fixes it.
Faded example
Approximate the arccotangent of negative 3.
Fill in the blanks
\operatornamepi(-3) = 2.8199 - \arctan\left(\tfrac______\right) \approx ___ - 0.3217 \approx ___
Why: A negative cotangent puts the angle in quadrant two, within the arccotangent's range of zero to pi. The reference angle has cotangent 3, so it is the arctangent of one third, about 0.3217. Subtracting from pi gives about 2.8199, which is between pi over two and pi as required.
Estimation
You need the arcsecant of 1.02, a number just above 1.
Predict first
Roughly what should the answer be?
Correct: Close to 0.
Why: An arcsecant of exactly 1 is zero, since the secant of zero is 1. So an input just above 1 gives an angle just above zero — about 0.198 radians here. Estimating before computing catches a great many errors: a common wrong instinct is that a large-looking input should give a large angle, but for arcsecant the small inputs near 1 give small angles and it is the very large inputs that approach pi over two.
Real world
A spreadsheet has ATAN and ATAN2 but no ACOT. An engineer needs the arccotangent of a column of values, some positive and some negative, to convert slope ratios into angles.
Discussion prompt
Write a formula that works for the whole column, and explain what would go wrong with the naive version.
Hint: What does the two-argument arctangent do that the one-argument version cannot?
Answer:
The naive formula, arctangent of one over x, fails in two ways: it returns a negative angle for negative inputs, which is outside the arccotangent's range, and it divides by zero when x is zero, where the arccotangent is perfectly well defined and equals pi over two.
\[ \operatorname{arccot}(x) = \operatorname{atan2}(1, x) \]
The two-argument form takes the numerator and denominator separately, so it never divides and can see both signs. It returns a value in the right range for every input including zero. This is the same fix as the heading problem in the previous lesson, and it is the standard answer whenever a quotient inside an inverse trigonometric function destroys information the two components separately retain.
Comparison
Fill the blanks from memory. Only the last two have a convention problem.
Comparison matrix
| Function | Domain | Range, trigonometry friendly |
|---|---|---|
| arccosine | -1 to 1 | 0 to pi |
| arcsine | -1 to 1 | -pi/2 to pi/2 |
| arctangent | all reals | -pi/2 to pi/2, open |
| arccotangent | all reals | 0 to pi, open |
| arcsecant | size at least 1 | 0 to pi, without pi/2 |
| arccosecant | size at least 1 | -pi/2 to pi/2, without 0 |
The first four ranges are universally agreed. The last two are agreed only within a convention, and this table shows one of the two in use.
Pattern
Whether you are evaluating exactly or approximating, the same five moves cover it.
For a negative arccotangent on a calculator, the reliable route is always to find the reference angle from the positive value and then place it: pi minus that angle, since the range is zero to pi.
OpenStax Algebra and Trigonometry 2e, §8.3 Inverse Trigonometric Functions §8.3
Check
Domain first. Is it even defined?
Check your understanding
Which of these is undefined?
Answer: C
Why: The domain of arcsecant is everything of absolute value at least 1, and 0.5 lies strictly inside the excluded band. No angle has a secant of 0.5, since that would need a cosine of 2.
Check
A positive input, so the convention does not matter.
Check your understanding
What is the arcsecant of root two?
Answer: B
Why: The reciprocal identity gives arccosine of one over root two, which is arccosine of root two over two, and that is pi over four. Both conventions agree here because the input is positive. Checking: the cosine of pi over four is root two over two, so its secant is root two.
Check
A negative input, so the convention matters.
Check your understanding
Under the calculus-friendly convention, what is the arccosecant of negative 2?
Answer: B
Why: The calculus-friendly range for arccosecant is zero to pi over two together with pi to three pi over two. A cosecant of negative 2 means a sine of negative one half, and the angle in that range with that sine is seven pi over six, in quadrant three.
Real world
An integral table gives the antiderivative of one over x times the root of x squared minus one as arcsecant of the absolute value of x, plus a constant. A different table gives it as arcsecant of x with no absolute value.
Discussion prompt
Explain how both tables can be correct, and say what a user of either table needs to check before trusting it.
Hint: Which convention makes the derivative of arcsecant come out without an absolute value?
Answer:
Under the calculus-friendly convention the arcsecant is increasing on both pieces of its range and its derivative comes out as one over x times the root of x squared minus one, with no absolute value needed. That table is written for that convention.
Under the trigonometry-friendly convention the second piece lies in quadrant two, where the function is decreasing relative to the first piece, and the derivative picks up an absolute value to compensate. That table is written for this convention.
What a user must check is which range the table's arcsecant assumes, usually stated in the preface or an appendix and almost always skipped. Using a calculus-friendly antiderivative with a trigonometry-friendly arcsecant button gives an answer that is wrong by a sign on the negative branch — and since definite integrals over a negative interval are exactly where that branch is used, the error is not hypothetical.
This is the practical reason the book refuses to pick a convention and presents both. A convention that changes the sign of a real answer is not a formality, and a course that hid one of them would leave you unable to read half the literature.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two textbooks give different values for arcsec(-2). What follows?
Correct: They are using different conventions, and both may be correct.
\[ \sec\left(\tfrac{2\pi}{3}\right) = \sec\left(\tfrac{4\pi}{3}\right) = -2 \]
Both angles have secant negative 2. Each convention returns the one inside its own range.
Why: Infinitely many angles have a secant of negative 2, and the arcsecant returns one of them by convention. Unlike arcsine and arccosine, no single convention became universal for arcsecant, so two books can give two pi over three and four pi over three respectively and both be internally consistent. Note that the fourth option is genuinely wrong: the secant itself is defined identically everywhere, and only the choice of restricted domain differs. The disagreement is about which answer to return, not about what the function is.
Explain it
They have grasped arcsine and arccosine and are unsettled to find that arcsecant apparently has two different definitions.
Discussion prompt
In no more than five sentences, reassure them without pretending the problem away. Tell them what to do in practice.
Hint: Is this a defect in mathematics or in bookkeeping?
Answer:
A usable answer: it is a bookkeeping problem, not a mathematical one. For arcsine and arccosine everyone happened to agree on which answer to return; for arcsecant they did not, because the two natural choices each make one thing tidy and the other messy. One choice keeps the arccosine relationship simple, the other makes a formula in calculus come out without cases.
In practice: for positive inputs it makes no difference at all, so most of your work is unaffected. When a negative shows up, look at which range your book states, use it throughout, and if you are writing something down for anyone else, say which one you used.
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 domain is fixed by remembering that a secant is never small, so its inverse never accepts a small input. The trigonometry-friendly evaluations are fixed by the reciprocal identity, which works everywhere there. The calculus-friendly ones are fixed by checking the identity's condition and falling back on the definition for negatives. The calculator case is fixed by finding the reference angle from the positive value and placing it, then checking the answer is between zero and pi. 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
At the top of the page draw the secant curve over two periods and shade the horizontal band between negative one and one, writing beside it what that band means for the arcsecant's domain. Below, draw the secant curve twice more, side by side, and on the first highlight the trigonometry-friendly restriction while on the second highlight the calculus-friendly one, labelling the range under each. To the right, make a table with four rows — arcsec(2), arcsec(-2), the reciprocal identity's condition, and tan(arcsec x) — and fill in both columns, circling every entry where the two disagree. In the bottom left, evaluate cot(arccsc(-3)) completely under both conventions, showing why the sign differs. In the bottom right, approximate arccot(-4) on a calculator that has only arctangent, showing the reference-angle step. Finally, write one sentence stating the practical rule about when the convention matters.
The sentence should say that the two conventions agree for every positive input and disagree for every negative one, so the convention only needs stating when a negative input appears.
Recap
Five things, and the fourth is the one that keeps you out of trouble.
| If the question says | Your first move is |
|---|---|
| Evaluate arcsec of a positive number | Reciprocal identity; both conventions agree |
| Evaluate arcsec of a negative number | Check which convention is stated first |
| A calculator answer looks wrong | Check it lies in the function's range |
| Approximate arccot of a negative | Reference angle from the positive, then pi minus it |
| An input is between -1 and 1 | Arcsecant is undefined there; say so |
All six functions are now inverted. The next lesson puts the inverse functions to work: approximating with a calculator in earnest, and using them to solve equations whose answers are not special angles — which is the bridge into Section 10.7.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 827-833 — everything on these slides traces back here
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