None of the six circular functions is one-to-one, so none has an inverse until its domain is restricted. Covers why the restriction is necessary and how the intervals are chosen, the resulting properties of arccosine, arcsine, arctangent and arccotangent, the asymmetric cancellation rules that make arccos of cos x fail to be x outside a specific interval, and the substitution technique that turns any composition of a circular function with an inverse one into an ordinary triangle problem.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.6 The Inverse Trigonometric Functions, pp. 819-827
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. 819-827 — the pages these objectives are drawn from
Warm-up
In Lesson 10.2b you solved equations like cosine of theta equals one half and found infinitely many answers. This lesson asks a related question that has to have exactly one answer.
Discussion prompt
How many angles have a cosine of one half? Now: if you press the inverse cosine button on a calculator, how many numbers does it give you back? What has to happen to reconcile those two facts?
Hint: A function must give one output per input. What does that force?
Answer:
Infinitely many angles have a cosine of one half — two families of them. But a calculator returns exactly one, namely about 1.047, which is pi over three.
The reconciliation is that the calculator is not answering the question 'which angles have this cosine'. It is evaluating a different function, one that has been deliberately built to return a single specified answer out of the infinitely many candidates.
Which answer it returns is a convention, and this lesson is largely about that convention: what it is, why it was chosen, and what goes wrong when you forget it is there.
Concept
A function has an inverse only if it is one-to-one, and no periodic function is. So each circular function is cut down to an interval on which it is one-to-one, and it is that restricted function which gets inverted.
restricted domain — An interval chosen so that the function is one-to-one on it while still attaining its full range. The inverse is defined as the inverse of the restricted function, so the inverse's range is exactly that interval.
Three requirements decide the interval: the function must be one-to-one there, it must still reach every value in its range, and it should stay continuous and smooth. For cosine that gives zero to pi; for sine, negative pi over two to pi over two.
Figure (svg): The cosine curve over two periods with a horizontal line cutting it at many points, demonstrating that it fails the horizontal line test and therefore has no inverse
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 819-819
Section
Section 1
Concept
The horizontal line test fails spectacularly for every circular function: a horizontal line inside the range meets the graph infinitely often. So none of them has an inverse as it stands, and something has to be given up.
The same manoeuvre was used for the squaring function: it is not one-to-one on all real numbers, so the square root is defined as the inverse of the restriction to the non-negative numbers. Everything surprising about inverse trigonometric functions has an exact analogue there.
Figure (svg): The cosine curve over two periods with a horizontal line cutting it at many points, demonstrating that it fails the horizontal line test and therefore has no inverse
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 819-819
Picture it
Cosine and sine are restricted to different intervals, and that single fact accounts for their inverses having different ranges.
Figure (svg): The cosine curve with the portion from zero to pi highlighted, showing the restricted piece that is one-to-one, beside the sine curve with the portion from negative pi over two to pi over two highlighted
Cosine takes zero to pi because that is where it runs monotonically from 1 down to negative 1. Sine cannot use that interval, because sine is not one-to-one there — it rises and then falls.
Worked example
The intervals differ for a reason, and it is worth seeing the reason rather than memorising the difference.
\[ \text{Show that } \sin(x) \text{ is not one-to-one on } [0, \pi]. \]
Find two inputs in the interval with the same output
Why: The sine is symmetric about pi over two.
\[ \sin(\frac{\pi}{6})\text{ and } \sin(5 \pi / 6) \]
Evaluate both
Why: Both are one half, since the two angles have the same reference angle and both are in the upper half.
\[ \text{both equal } \frac{1}{2} \]
Conclude
Why: Two different inputs, one output, so the function is not one-to-one there.
Name the interval that works
Why: From negative pi over two to pi over two the sine rises steadily.
\[ [-\frac{\pi}{2}, \frac{\pi}{2}] \]
Figure (svg): The solution to Worked example why sine cannot use the same interval shown as a ladder of expressions, one row per legal move
\[ \sin\left(\tfrac{\pi}{6}\right) = \sin\left(\tfrac{5\pi}{6}\right) = \tfrac{1}{2} \]
Verify: check the alternative interval works
Why: On the interval from negative pi over two to pi over two the sine increases from negative 1 to 1 without ever repeating a value, and it attains every value in between. So it is one-to-one there and keeps the full range — both requirements met.
Sorting
The function must not repeat a value anywhere on the interval.
Sort into buckets
Sort each pairing by whether the function is one-to-one there.
Worked example
The same problem in a setting you have already met, which is worth recalling because the same surprises follow.
\[ \text{Why is } \sqrt{x^2} = |x| \text{ rather than } x\text{?} \]
Note that squaring is not one-to-one
Why: Both 2 and negative 2 square to 4.
Restrict to the non-negative numbers
Why: There squaring is one-to-one, and the square root inverts that restriction.
\[ \text{restrict to } x \ge 0 \]
Follow a negative input through
Why: Squaring negative 2 gives 4, and the square root of 4 is the positive root 2.
\[ \sqrt{(-2) ^{2}} = 2 \]
Name what happened
Why: The composition returned the representative from the restricted interval, not the original input.
Figure (svg): The solution to Worked example the analogous restriction for squaring shown as a ladder of expressions, one row per legal move
\[ \sqrt{(-2)^2} = \sqrt{4} = 2 \ne -2 \]
Verify: compare with the arccosine case
Why: Exactly the same thing will happen with arccos of cos of eleven pi over six: the answer will be pi over six, the representative from the interval zero to pi, rather than the original angle. The book makes this comparison explicitly, and it is the single most useful analogy in the lesson.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 823-823
Trap
\[ \arccos\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3} \;\text{ or }\; \tfrac{5\pi}{3} \;\text{ or }\; \tfrac{7\pi}{3} \;\text{ or } \ldots \]
List all the angles with the right cosine
Why: Every one of them does have cosine one half, so each looks like a legitimate answer.
But arccosine is a function, and a function returns exactly one output. Listing several is answering the equation-solving question from Lesson 10.2b, not evaluating this function.
\[ \arccos\left(\tfrac{1}{2}\right) = \tfrac{\pi}{3} \]
Give the unique angle in the range of arccosine
Why: That range is the interval from zero to pi, and pi over three is the only candidate inside it.
The distinction is worth stating precisely. Solving cosine of theta equals one half asks for every angle and has infinitely many answers. Evaluating arccosine of one half asks for one specific angle and has exactly one. They are different questions with different answers, and the restriction is what separates them.
Prediction
A different textbook restricts the cosine to the interval from pi to two pi instead.
Predict first
Would that produce a valid inverse function?
Correct: Yes, but its outputs would differ from the standard arccosine.
Why: On the interval from pi to two pi the cosine rises steadily from negative 1 back to 1, so it is one-to-one and attains the full range — both requirements are met. Inverting it would give a perfectly valid function, but it would return five pi over three where the standard arccosine returns pi over three. That is precisely why the choice has to be a stated convention: several intervals work, and everyone must agree on which one to use.
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 restricting a domain is wrong.
Survives elimination: B
Why: Statement B is false. The interval from pi to two pi works just as well as zero to pi: the cosine is one-to-one there and attains its full range, so inverting it gives a perfectly valid function. It simply gives different answers, which is why the standard interval has to be agreed on and stated rather than discovered. Calling the choice a convention is being precise, not vague.
Socratic
One of the three requirements on the restricted interval is that it still attains every value the original function took.
Discussion prompt
What would go wrong if the interval were chosen smaller, say cosine restricted to zero to pi over two?
Hint: What is the range of the cosine on that smaller interval, and what does that become for the inverse?
Answer:
On zero to pi over two the cosine is one-to-one, so an inverse would exist. But its range there is only from zero to one, not from negative one to one.
Reflecting swaps domain and range, so the resulting inverse would have domain only from zero to one. It could not answer the question 'which angle has cosine negative one half' at all, because negative one half would be outside its domain.
So the requirement is not aesthetic. A smaller interval gives an inverse that cannot accept some perfectly legitimate inputs, and the standard interval is the smallest one that avoids that.
Section
Section 2
Concept
The graph of an inverse is the graph of the restricted function reflected in the line y equals x. That reflection exchanges the coordinates, so the domain of one is the range of the other.
The notation used here is arccos and arcsin rather than cosine to the negative one. The book avoids the exponent notation deliberately, because with cosine squared meaning the square of the cosine, cosine to the negative one looks far too much like the secant.
| arccosine | arcsine | |
|---|---|---|
| Domain | from -1 to 1 | from -1 to 1 |
| Range | from 0 to pi | from -pi/2 to pi/2 |
| Endpoints | both included | both included |
| Parity | neither even nor odd | odd |
| Behaviour | decreasing | increasing |
Figure (svg): The restricted sine curve and the arcsine curve on the same axes, shown as reflections of each other in the line y equals x
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 819-820
Picture it
Every property of the inverse can be read off this picture rather than memorised.
Figure (svg): The restricted sine curve and the arcsine curve on the same axes, shown as reflections of each other in the line y equals x
The restricted sine runs from negative pi over two to pi over two horizontally and from negative one to one vertically. Its reflection runs from negative one to one horizontally and from negative pi over two to pi over two vertically — the two intervals have traded places.
Worked example
Example 10.6.1, parts a to d. Each answer must land in the right range.
\[ \text{Find } \arccos\left(\tfrac{1}{2}\right), \; \arcsin\left(\tfrac{\sqrt{2}}{2}\right), \; \arccos\left(-\tfrac{\sqrt{2}}{2}\right), \; \arcsin\left(-\tfrac{1}{2}\right). \]
Find the angle in [0, pi] with cosine one half
Why: Pi over three is in that interval and has the right cosine.
\[ \arccos(\frac{1}{2}) = \frac{\pi}{3} \]
Find the angle in [-pi/2, pi/2] with sine root two over two
Why: Pi over four qualifies.
\[ \arcsin(\sqrt{2} / 2) = \frac{\pi}{4} \]
Find the angle in [0, pi] with cosine negative root two over two
Why: Three pi over four is in quadrant two, where the cosine is negative.
\[ \arccos = 3 \pi / 4 \]
Find the angle in [-pi/2, pi/2] with sine negative one half
Why: Negative pi over six, since the range allows negative angles.
\[ \arcsin = -\frac{\pi}{6} \]
Figure (svg): The solution to Worked example four exact evaluations shown as a ladder of expressions, one row per legal move
\[ \tfrac{\pi}{3}, \quad \tfrac{\pi}{4}, \quad \tfrac{3\pi}{4}, \quad -\tfrac{\pi}{6} \]
Verify: check every answer is in range
Why: Pi over three and three pi over four are both between zero and pi, as arccosine requires. Pi over four and negative pi over six are both between negative pi over two and pi over two, as arcsine requires. An answer outside its function's range would be wrong no matter what its cosine or sine was.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 821-821
Sorting
Check each against the range of the function named.
Sort into buckets
Sort each claimed output by whether it could be correct.
Worked example
The two functions handle a negative input completely differently, and it is worth seeing side by side.
\[ \text{Compare } \arccos(-x) \text{ with } \arcsin(-x). \]
Note the range of arcsine is symmetric about zero
Why: From negative pi over two to pi over two.
Conclude arcsine is odd
Why: A negative input gives the negative of the answer.
\[ \arcsin(-x) = -\arcsin(x) \]
Note the range of arccosine is not symmetric about zero
Why: From zero to pi, entirely non-negative.
Find the correct relationship
Why: A negative input pushes the answer into the second half of the interval.
\[ \arccos(-x) = \pi - \arccos(x) \]
Figure (svg): The solution to Worked example a negative input to each shown as a ladder of expressions, one row per legal move
\[ \arcsin(-x) = -\arcsin(x), \qquad \arccos(-x) = \pi - \arccos(x) \]
Verify: test both at one half
Why: Arcsine of negative one half is negative pi over six, which is the negative of arcsine of one half — odd, as claimed. Arccosine of negative one half is two pi over three, and pi minus arccosine of one half is pi minus pi over three, which is two pi over three. Both relationships hold.
Error analysis
A student evaluates the arcsine of negative root three over two.
Annotate
On: \( \arcsin\left(-\frac{\sqrt{3}}{2}\right) = \frac{4\pi}{3} \)
Checking the answer against the range is a free test and it catches this class of error completely. If an arcsine output is outside negative pi over two to pi over two, or an arccosine output outside zero to pi, it is wrong regardless of what its sine or cosine happens to be.
Matching
Every answer must be between zero and pi.
Match the pairs
Why: Arccosine runs backwards through the interval: an input of 1 gives the left endpoint zero, an input of negative 1 gives the right endpoint pi, and an input of zero gives the midpoint pi over two. That reversal is because the restricted cosine is decreasing, so its inverse is decreasing too — a larger input gives a smaller output, which is worth noticing since arcsine behaves the opposite way.
Prediction
You evaluate the arccosine at a sequence of increasing inputs from negative one up to one.
Predict first
What happens to the outputs?
Correct: They decrease from pi to 0.
Why: The restricted cosine is decreasing on zero to pi, running from 1 down to negative 1, and reflecting a decreasing function in the diagonal gives another decreasing function. So arccosine of negative one is pi, arccosine of zero is pi over two, and arccosine of one is zero. Arcsine, by contrast, is increasing, because the restricted sine is. This difference in direction is worth remembering as a quick check on any answer.
Edge cases
The domain of both arccosine and arcsine is the closed interval from negative one to one.
Discussion prompt
Why is that the domain, and what happens if you try to evaluate the arcsine of 2?
Hint: The domain of an inverse is the range of the original.
Answer:
Reflecting in the diagonal makes the range of the restricted function become the domain of the inverse. The sine has range from negative one to one, so the arcsine has domain from negative one to one.
Evaluating arcsine of 2 would mean finding an angle whose sine is 2, and no such angle exists — the sine never exceeds one. So the expression is undefined, and a calculator returns an error rather than a number.
Contrast this with arctangent, whose domain is every real number, because the tangent's range is every real number. The domain of an inverse is always the range of the parent, which is why the four inverse functions have such different domains despite coming from similar-looking parents.
Section
Section 3
Concept
Composing a function with its inverse gives the identity, but only where the composition is defined and only in the direction the restriction permits. The two directions are genuinely not symmetric.
\[ \cos(\arccos(x)) = x, \quad -1 \le x \le 1; \qquad \arccos(\cos(x)) = x, \quad 0 \le x \le \pi \]
The safest habit is never to cancel by reflex. Check which direction you have and, for the inside-out direction, check the condition explicitly before applying it.
Figure (svg): A table of the two cancellation rules for each inverse pair, showing that one direction always works while the other requires the input to lie in the restricted interval
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 820-823
Picture it
Three always work and three are conditional, and the pattern is completely regular.
Figure (svg): A table of the two cancellation rules for each inverse pair, showing that one direction always works while the other requires the input to lie in the restricted interval
Every conditional rule has the same condition: the inner input must lie in the range of the outer inverse function. That is one sentence covering all three cases.
Worked example
Example 10.6.1, part e. The condition is satisfied, so the rule applies directly.
\[ \text{Find } \arccos\left(\cos\left(\frac{\pi}{6}\right)\right). \]
Identify the direction
Why: An inverse applied to the function, so the conditional rule.
Check the condition
Why: The inner input pi over six must lie between zero and pi.
\[ 0 \le \frac{\pi}{6} \le \pi,\text{ yes} \]
Apply the rule
Why: The condition holds, so the composition is the identity.
\[ = \frac{\pi}{6} \]
Note what was checked
Why: Skipping the check would have given the same answer here, but not always.
Figure (svg): The solution to Worked example when cancellation works shown as a ladder of expressions, one row per legal move
\[ \arccos\left(\cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6} \]
Verify: work it from the inside out
Why: The cosine of pi over six is root three over two, and the arccosine of root three over two is the angle in zero to pi with that cosine, namely pi over six. Both routes agree, which is what the condition guarantees.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 821-822
Discrimination
Check the direction, then check the condition.
Sort into buckets
Sort each composition by whether it simplifies to the inner input.
Worked example
Example 10.6.1, part f. This is the case worth remembering.
\[ \text{Find } \arccos\left(\cos\left(\frac{11\pi}{6}\right)\right). \]
Check the condition
Why: Eleven pi over six is about 5.76, which is well outside the interval from zero to pi.
Do not cancel
Why: The rule does not apply, so the answer is not eleven pi over six.
Evaluate the inner function
Why: Eleven pi over six is in quadrant four with reference angle pi over six.
\[ \cos = \sqrt{3} / 2 \]
Evaluate the outer function
Why: The angle in zero to pi with cosine root three over two.
\[ = \frac{\pi}{6} \]
Figure (svg): The solution to Worked example when cancellation fails shown as a ladder of expressions, one row per legal move
\[ \arccos\left(\cos\left(\frac{11\pi}{6}\right)\right) = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]
Verify: check the two angles share a cosine
Why: Eleven pi over six and pi over six both have cosine root three over two, since they are reflections of each other across the x-axis and cosine is even. The composition returned the one inside the restricted interval, exactly as the square root returns the non-negative root.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 822-822
Trap
\[ \arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6} \]
Cancel the inverse against the function
Why: The two operations look like they must undo each other, and in the other direction they always do.
But five pi over six is about 2.62, outside the range of arcsine, which stops at pi over two, about 1.57. So the claimed answer is not even a possible arcsine output.
\[ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \;\Longrightarrow\; \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \]
Check the condition first; when it fails, work strictly from the inside out
Why: Five pi over six is not in the restricted interval, so the rule does not apply.
The free check that catches every instance: is the claimed answer inside the range of the outer function? An arcsine answer above pi over two is impossible before you consider anything else, and the same test works for arccosine and arctangent.
Faded example
Evaluate the arcsine of the sine of five pi over four.
Fill in the blanks
\sin\left(\frac-sqrt2/2-pi/4\right) = ___ \;\Longrightarrow\; \arcsin(___) = ___
Why: Five pi over four is well outside the range of arcsine, so no cancellation is allowed and the composition must be evaluated inside out. The sine there is negative root two over two, and the angle in the range of arcsine with that sine is negative pi over four. Note that five pi over four and negative pi over four do share a sine, but only the latter is a legitimate arcsine output.
Prediction
You evaluate the arccosine of the cosine of x for some x between pi and two pi.
Predict first
What will the answer be?
Correct: two pi minus x.
Why: The condition fails, since x is beyond pi, so the composition returns the representative in zero to pi with the same cosine. That representative is two pi minus x, because x and two pi minus x are reflections across the x-axis and cosine is even. Checking with x equal to eleven pi over six: two pi minus that is pi over six, which is exactly the worked example's answer. This gives a formula covering the whole second half of the revolution at once.
Explain it to yourself
The book compares arccos of cos of eleven pi over six with the square root of negative two squared.
Discussion prompt
Explain the analogy carefully. What plays the role of the restricted interval in the square root case, and what plays the role of the representative?
Hint: What does the square root function refuse to return?
Answer:
Squaring is not one-to-one, so the square root is defined as the inverse of squaring restricted to the non-negative numbers. That restriction is the analogue of the interval zero to pi.
Feeding in negative two, squaring gives 4, and the square root returns 2 — the non-negative number with the same square. That is the representative from the restricted set, and it corresponds exactly to pi over six being the representative in zero to pi with the same cosine as eleven pi over six.
In both cases the composition is not the identity but a projection onto the restricted set. For the square root it is the absolute value; for arccosine it is a slightly more complicated folding, but the mechanism is identical.
Section
Section 4
Concept
The tangent has range all real numbers, so its inverse has domain all real numbers. And the tangent's vertical asymptotes reflect into horizontal asymptotes, which is what bounds the arctangent's range.
The endpoints are excluded here where they were included for arccosine and arcsine, and the reason is exact: the tangent never actually attains the values at plus or minus pi over two, because it is undefined there.
| arctangent | arccotangent | |
|---|---|---|
| Domain | all real numbers | all real numbers |
| Range | strictly between -pi/2 and pi/2 | strictly between 0 and pi |
| Endpoints | excluded | excluded |
| Asymptotes | horizontal at plus and minus pi/2 | horizontal at 0 and pi |
| Parity | odd | neither |
Figure (svg): The arctangent curve, increasing from a horizontal asymptote at negative pi over two to a horizontal asymptote at pi over two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 823-824
Picture it
Two closed intervals and two open ones, and the difference is not arbitrary.
Figure (svg): Two columns pairing each inverse function with its range, showing which are closed intervals and which are open
Arcsine and arctangent share the same interval of numbers but differ on the endpoints, because the sine attains its extremes and the tangent does not.
Worked example
Example 10.6.2, parts a to c. Range membership decides each answer.
\[ \text{Find } \arctan(\sqrt{3}), \; \operatorname{arccot}(-\sqrt{3}), \; \cot(\operatorname{arccot}(-5)). \]
Find the angle in the arctangent range with tangent root three
Why: Pi over three is between negative pi over two and pi over two.
\[ \arctan(\sqrt{3}) = \frac{\pi}{3} \]
Find the angle in the arccotangent range with cotangent negative root three
Why: The range is zero to pi, and five pi over six is in quadrant two where cotangent is negative.
\[ \operatorname{arccot} = 5 \pi / 6 \]
Recognise the third as an outside-in cancellation
Why: The cotangent applied to its own inverse always cancels.
\[ \cot(\operatorname{arccot}(x)) = x \]
Apply it
Why: The domain of arccotangent is every real number, so negative 5 is allowed.
\[ = -5 \]
Figure (svg): The solution to Worked example exact values for arctangent and arccotangent shown as a ladder of expressions, one row per legal move
\[ \arctan(\sqrt{3}) = \tfrac{\pi}{3}, \quad \operatorname{arccot}(-\sqrt{3}) = \tfrac{5\pi}{6}, \quad \cot(\operatorname{arccot}(-5)) = -5 \]
Verify: check each is in range
Why: Pi over three is strictly inside negative pi over two to pi over two; five pi over six is strictly inside zero to pi. And the third answer is not an angle at all but a value, which is right for an outside-in cancellation.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 825-825
Comparison
Fill the blanks from memory. Domain comes from the parent's range and vice versa.
Comparison matrix
| Function | Domain | Range |
|---|---|---|
| arccosine | -1 to 1, closed | 0 to pi, closed |
| arcsine | -1 to 1, closed | -pi/2 to pi/2, closed |
| arctangent | all real numbers | -pi/2 to pi/2, open |
| arccotangent | all real numbers | 0 to pi, open |
Notice the pattern: arccosine and arccotangent share the interval zero to pi, while arcsine and arctangent share negative pi over two to pi over two. The co-functions go together, which is not a coincidence.
Worked example
Example 10.6.2, part d. Name the inner expression and work with the angle.
\[ \text{Find } \sin\left(\arctan\left(-\frac{3}{4}\right)\right). \]
Name the inner expression
Why: Let t be the arctangent, so the tangent of t is negative three quarters.
\[ t = \arctan(-\frac{3}{4}) \]
Narrow down where t is
Why: The tangent is negative and t is in the arctangent range, so t is between negative pi over two and zero.
Find the cosecant from the cotangent
Why: The cotangent is negative four thirds, and one plus its square is the cosecant squared.
\[ \csc ^{2} t = \frac{25}{9} \]
Choose the sign and reciprocate
Why: Since t is negative, the sine is negative and so is the cosecant.
\[ \sin t = -\frac{3}{5} \]
Figure (svg): The solution to Worked example a composition needing the quadrant shown as a ladder of expressions, one row per legal move
\[ \sin\left(\arctan\left(-\frac{3}{4}\right)\right) = -\frac{3}{5} \]
Verify: check with a triangle
Why: A tangent of negative three quarters corresponds to the point four comma negative three, at radius five. The sine is the y-coordinate over the radius, which is negative three fifths — matching, and confirming that the angle is in quadrant four as the arctangent range requires.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 825-825
Error analysis
A student writes down the range of the arctangent.
Annotate
On: \( \text{range}(\arctan) = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
The bracket type is doing real work. Arcsine's endpoints are attained, at inputs of negative one and one; arctangent's are not, because the tangent has no largest value. Comparing the two is the fastest way to remember which is which.
Prediction
You evaluate the arctangent at larger and larger inputs: 10, then 100, then 1000.
Predict first
What happens to the outputs?
Correct: They approach pi over 2 without reaching it.
Why: The arctangent has a horizontal asymptote at pi over two, so its outputs creep towards about 1.5708 and never exceed it. The arctangent of 1000 is about 1.5698, already within a thousandth of the asymptote. This is the reflection of the tangent's vertical asymptote: as the tangent's input approaches pi over two its output runs to infinity, and reflecting swaps those roles exactly.
Fill the middle
Evaluate the arccotangent of negative one.
Fill in the blanks
\text3 pi/4 t \text___ (0, \pi) \text___ \cot t = -1 \;\Longrightarrow\; t = ___
Why: The cotangent is negative one where the coordinates are equal in size and opposite in sign, which happens in quadrants two and four. Only quadrant two lies inside the range of arccotangent, so the answer is three pi over four. Note that seven pi over four also has cotangent negative one but is outside the range, and negative pi over four is the answer arctangent of negative one would give — a different function with a different range.
Counterexample
A student proposes: since arctangent and arccotangent are inverses of reciprocal functions, arccotangent of x always equals arctangent of one over x.
Discussion prompt
Test the claim at a negative value and say what the correct statement is.
Hint: The book states this identity with a condition attached.
Answer:
\[ \operatorname{arccot}(-1) = \tfrac{3\pi}{4}, \qquad \arctan\left(\tfrac{1}{-1}\right) = \arctan(-1) = -\tfrac{\pi}{4} \]
Three pi over four is not negative pi over four, so the claim fails at negative inputs. The two answers differ by exactly pi.
The book states the identity as arccot of x equals arctan of one over x for x positive, and the condition is essential. The reason is that the two functions have different ranges: arccotangent's range is entirely positive while arctangent's is symmetric about zero, so for a negative input they cannot possibly agree. A restriction stated in a theorem is never decoration, and this is a clean example of what it is protecting against.
Section
Section 5
Concept
An expression such as the tangent of an arccosine looks intimidating and is not. Naming the inner expression t turns it into an ordinary question about a single angle, which a triangle or the Pythagorean identity answers.
That last step is the one most often skipped, and it matters: an expression like one minus two x squared is defined for every real number, yet the equivalence to cosine of twice an arcsine holds only from negative one to one.
Figure (svg): A right triangle labelled to evaluate the sine of the arccosine of x, with adjacent side x, hypotenuse one, and opposite side the square root of one minus x squared
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 822-823
Picture it
Every one of these compositions reduces to labelling a right triangle.
Figure (svg): A right triangle labelled to evaluate the sine of the arccosine of x, with adjacent side x, hypotenuse one, and opposite side the square root of one minus x squared
The hypotenuse is 1 whenever the inner function is a cosine or sine, because those are the values on the unit circle. For a tangent it is the two legs that are known and the hypotenuse must be computed.
Worked example
Example 10.6.1, part 2a. Name it, then use a triangle.
\[ \text{Write } \tan(\arccos(x)) \text{ as an algebraic expression and state where it is valid.} \]
Name the inner expression
Why: Let t be the arccosine of x, so the cosine of t is x and t is between zero and pi.
\[ \cos t = x, 0 \le t \le \pi \]
Find the sine from the Pythagorean identity
Why: Sine squared is one minus x squared; on zero to pi the sine is non-negative.
\[ \sin t = \sqrt{1 - x ^{2}} \]
Form the tangent
Why: Sine over cosine.
\[ \tan t = \sqrt{1 - x ^{2}} / x \]
State the validity
Why: The arccosine needs x between negative one and one, and the tangent needs x nonzero.
\[ x\text{ in } [-1, 0]\text{ or } (0, 1) \]
Figure (svg): The solution to Worked example the tangent of an arccosine shown as a ladder of expressions, one row per legal move
\[ \tan(\arccos(x)) = \frac{\sqrt{1 - x^2}}{x} \]
Verify: test at one half
Why: The arccosine of one half is pi over three, whose tangent is root three. The formula gives the root of three quarters over one half, which is root three over two times two, namely root three. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 822-823
Faded example
Write the sine of an arccosine as an algebraic expression.
Fill in the blanks
t = \arccos(x) \;\Longrightarrow\; \cos t = x, \; 0 \le t \le \pi \;\Longrightarrow\; \sin t = sqrt(1 - x^2), \textpositive ___
Why: The Pythagorean identity gives sine squared equal to one minus x squared, so the sine is plus or minus the square root. On the interval from zero to pi the sine is never negative, so the positive root is taken — and no case analysis is needed. This is one of the conveniences of the chosen restriction: the sine has a constant sign across the whole of arccosine's range.
Worked example
Example 10.6.1, part 2b. A double angle formula does the work.
\[ \text{Write } \cos(2\arcsin(x)) \text{ as an algebraic expression and state where it is valid.} \]
Name the inner expression
Why: Let t be the arcsine of x, so the sine of t is x.
\[ \sin t = x \]
Choose the double angle form
Why: The form containing only the sine avoids needing the cosine at all.
\[ \cos 2 t = 1 - 2 \sin ^{2} t \]
Substitute
Why: The sine of t is x.
\[ = 1 - 2 x ^{2} \]
State the validity
Why: The arcsine needs x between negative one and one, and nothing further was required.
\[ x\text{ in } [-1, 1] \]
Figure (svg): The solution to Worked example the cosine of twice an arcsine shown as a ladder of expressions, one row per legal move
\[ \cos(2\arcsin(x)) = 1 - 2x^2, \quad -1 \le x \le 1 \]
Verify: test at one half
Why: The arcsine of one half is pi over six, and twice that is pi over three, whose cosine is one half. The formula gives one minus two times one quarter, which is one half. They agree — and note the expression one minus two x squared is defined for every real number, yet the equivalence holds only on the stated interval.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 823-823
Trap
\[ \cos(2\arcsin(x)) = 1 - 2x^2 \quad \text{for all } x \]
Report the algebraic expression and stop
Why: The expression one minus two x squared is a perfectly ordinary polynomial defined everywhere, so no restriction seems needed.
But the left side is undefined for x equal to 2, since the arcsine of 2 does not exist, while the right side happily gives negative 7. The two sides are not the same function.
\[ \cos(2\arcsin(x)) = 1 - 2x^2, \quad -1 \le x \le 1 \]
Inherit the restriction from the domain of the inverse function
Why: The composition can only be evaluated where the inner inverse is defined.
The analogy the book draws is exact: the square root of x, squared, equals x — but only for x at least zero, even though x itself is defined everywhere. An algebraic simplification can widen the apparent domain, and stating the true one is part of the answer rather than an afterthought.
Ranking
Rewrite a composition as an algebraic expression.
Put in order
Why: Naming the inverse expression is what converts an unfamiliar composition into a familiar problem. Writing down both pieces of information — the value and the interval — matters because the interval is what resolves the sign later. And stating the validity last is part of the answer, not an optional extra, since the algebraic expression usually has a wider domain than the composition does.
Prediction
You rewrite the tangent of twice an arctangent as an algebraic expression.
Predict first
What restriction will appear beyond the domain of arctangent?
Correct: x cannot be plus or minus 1.
Why: The double angle formula for tangent has denominator one minus the tangent squared, which vanishes when the tangent of t is plus or minus one — that is, when t is plus or minus pi over four, which corresponds to x equal to plus or minus one. So the equivalence holds for every real x except those two values, and the resulting expression is two x over one minus x squared. This is exactly the extra exclusion the book highlights.
Edge cases
The expression for the tangent of an arccosine came out as the root of one minus x squared, over x.
Discussion prompt
What happens at x equal to zero, and what does that correspond to on the original composition?
Hint: Evaluate both sides at x equal to zero.
Answer:
The algebraic expression has a zero denominator at x equal to zero, so it is undefined there. Checking the original: the arccosine of zero is pi over two, and the tangent of pi over two is undefined as well.
\[ \tan(\arccos(0)) = \tan\left(\tfrac{\pi}{2}\right) \quad \text{undefined} \]
So the two sides fail together, which is exactly what should happen — the algebraic form has correctly inherited the exclusion rather than hiding it. When a simplification produces a new denominator, checking that its zeros correspond to genuine failures of the original is a good test that no step was invalid.
Comparison
Fill the blanks from memory. Lesson 10.2b did the left column; this lesson does the right.
Comparison matrix
| Solving cos t = c | Evaluating arccos(c) | |
|---|---|---|
| Question asked | which angles have this cosine | which single angle in [0, pi] has this cosine |
| Number of answers | infinitely many, in two families | exactly one |
| Answer form | families with an integer parameter | a single number |
| Role of the range | none; every solution counts | it selects the one answer |
Confusing these two is the deepest error available in this lesson. The equation has infinitely many solutions; the function has one value; and the restricted interval is exactly what converts one into the other.
Pattern
Whether you are evaluating, cancelling or rewriting, the same five moves cover it.
The single most useful habit is checking that the answer lies in the range. It costs nothing and it catches the entire family of errors that come from forgetting the restriction.
OpenStax Algebra and Trigonometry 2e, §8.3 Inverse Trigonometric Functions §8.3
Check
A plain evaluation. Check the range.
Check your understanding
What is the arcsine of negative root two over two?
Answer: B
Why: The arcsine returns the unique angle between negative pi over two and pi over two whose sine is the given value. Negative root two over two is negative, so the angle is negative, and negative pi over four has exactly that sine while lying inside the range.
Check
A composition. Check the condition before cancelling.
Check your understanding
What is the arccosine of the cosine of seven pi over four?
Answer: B
Why: Seven pi over four is outside the interval from zero to pi, so no cancellation is permitted. Working inside out, the cosine of seven pi over four is root two over two, and the angle in zero to pi with that cosine is pi over four.
Check
Ranges. The endpoints matter.
Check your understanding
Which of these is the range of the arctangent function?
Answer: B
Why: The tangent has vertical asymptotes at plus and minus pi over two, so it never attains a value there. Reflecting turns those into horizontal asymptotes for the arctangent, which therefore approaches plus and minus pi over two without ever reaching them. The interval is open at both ends.
Real world
A robot's position sensor reports x and y offsets from a reference point, and the control software must compute the heading angle. A programmer writes the heading as the arctangent of y over x.
Discussion prompt
Explain why that formula fails for points in quadrants two and three, and what the standard fix is. Relate the failure precisely to the range of the arctangent.
Hint: Compute the formula's answer for the point negative one comma negative one, and compare it with the true heading.
Answer:
\[ \text{point } (-1, -1): \quad \arctan\left(\frac{-1}{-1}\right) = \arctan(1) = \frac{\pi}{4} \]
The true heading is five pi over four, pointing down and to the left. The formula returned pi over four, pointing up and to the right — exactly the opposite direction.
The cause is the range. The arctangent only ever returns values strictly between negative pi over two and pi over two, which covers quadrants one and four only. It has no way to report a heading in quadrant two or three, because dividing y by x destroys the information about which of the two opposite directions the point lies in — both give the same quotient.
The standard fix is a two-argument function, written atan2 in almost every programming language, which takes y and x separately rather than as a quotient and so can see both signs. It returns a value over the full range from negative pi to pi. This is one of the few places where a mathematical convention causes a genuine and famous class of software bug, and knowing the range of the arctangent is what prevents it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is the arcsine of the sine of x always equal to x?
Correct: Only when x is between -pi/2 and pi/2.
\[ \arcsin\left(\sin\left(\tfrac{5\pi}{6}\right)\right) = \arcsin\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6} \ne \tfrac{5\pi}{6} \]
Why: The sine is inverted only on its restricted domain, which is that interval. For x inside it the cancellation is exact. For x outside it, the composition returns the representative of x in that interval instead: the arcsine of the sine of five pi over six is pi over six, not five pi over six. Note that the other direction, the sine of the arcsine of x, does always cancel — but only on the domain of the arcsine, from negative one to one. The two directions are genuinely not symmetric, and that asymmetry is the whole point of the restriction.
Explain it
They have just discovered that their calculator gives only one answer for an inverse sine and think it is broken.
Discussion prompt
In no more than five sentences, explain what the calculator is actually doing and why it is not a defect. Give them the analogy with square roots.
Hint: How many square roots does 9 have, and how many does the square root button return?
Answer:
A usable answer: the calculator is not broken, it is answering a narrower question than you asked. Infinitely many angles have a sine of one half, but a function has to give one answer, so mathematicians agreed in advance which one it would be — the one between negative ninety and ninety degrees.
Think about square roots. The number 9 has two square roots, 3 and negative 3, but the square root button returns only 3, because someone had to choose. Inverse sine is the same trick applied to angles. If you want all the angles, that is a different question and you get them by adding the other family and the multiples of a full turn yourself.
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 ranges are fixed by remembering that arccosine and arccotangent use zero to pi while arcsine and arctangent use negative pi over two to pi over two, with the tangent versions open at the ends. Cancellation is fixed by checking the direction first and then the condition. Algebraic rewrites are fixed by naming the inverse expression t and writing down both what you know about it and where it lives. Validity is fixed by inheriting the inverse's domain and then checking for any new denominator. 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 cosine curve with the portion from zero to pi highlighted, and beside it the sine curve with the portion from negative pi over two to pi over two highlighted, writing under each why that interval was chosen. Underneath, draw the restricted sine and the arcsine on one set of axes with the line y equals x dashed between them, showing the reflection. To the right, make a four-row table of the inverse functions with their domains and ranges, marking clearly which endpoints are included and which excluded. In the middle of the page, write the six cancellation rules in two columns, always and conditional, and box the condition each conditional one carries. In the bottom left, evaluate the arccosine of the cosine of five pi over three completely, showing why cancellation is not allowed. In the bottom right, rewrite the sine of an arctangent of x as an algebraic expression, drawing the triangle and stating the interval of validity. Finally, circle the one feature of the arctangent graph that the tangent's vertical asymptotes turned into, and name it.
The circled feature should be the horizontal asymptotes at plus and minus pi over two, and they are what makes the arctangent's range bounded despite the tangent's range being everything.
Recap
Five things, and the second one is the check that prevents most of the errors.
| If the question says | Your first move is |
|---|---|
| Evaluate arccos of a value | Find the angle in [0, pi] with that cosine |
| Evaluate arcsin of a value | Find the angle in [-pi/2, pi/2] with that sine |
| An inverse is applied to a function | Check whether the inner input is in the restricted interval |
| A function is applied to an inverse | It always cancels, on the inverse's domain |
| Rewrite as an algebraic expression | Name the inverse expression t and draw the triangle |
Four of the six are now inverted. The remaining two, secant and cosecant, are genuinely awkward: there is no agreement in the literature about which restriction to use, and the next lesson presents both of the competing conventions rather than pretending one is standard.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 819-827 — everything on these slides traces back here
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