Builds the inverse trigonometric functions by restricting each domain until the horizontal line test passes. Explains why each standard restriction was chosen, evaluates inverses exactly and with a calculator, composes them with the original functions, and shows how to recover the other angle a problem may want.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 6 — Periodic Functions
§6.3 Inverse Trigonometric Functions, pp. 764-780
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 764-780 — the pages these objectives are drawn from
Warm-up
This looks like a question with one answer, and that is exactly the difficulty.
Discussion prompt
Which angle has a sine of one half?
Hint: How many can you find?
Answer:
Thirty degrees is one. So is a hundred and fifty degrees, since the sine is positive in the second quadrant too.
And adding a full turn to either gives another, endlessly. So there are infinitely many angles with sine one half.
A function must return exactly one output, so no function can answer this question as asked. The inverse sine exists only after a decision about which of the infinitely many angles to return.
Concept
The trigonometric functions are periodic and therefore fail the horizontal line test badly. Each inverse is built by first cutting the domain down to an interval where the function is one-to-one.
restricted domain — a deliberately chosen piece of a function's domain on which it is one-to-one, so that an inverse can be defined
\[ \arcsin: [-1,1]\to\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], \quad \arccos: [-1,1]\to[0,\pi] \]
The restrictions are conventions, but universal ones — every calculator and textbook uses the same three. Knowing them is what lets you predict what a calculator will return and recognise when the answer you want is a different angle with the same value.
Figure (svg): A card giving the three standard domain restrictions and the resulting ranges of the inverse sine, inverse cosine and inverse tangent
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 764-770
Section
Section 1
Concept
A periodic function repeats every value infinitely often, so a horizontal line meets its graph infinitely many times and there is no way to undo it without first choosing a piece.
This is the same situation as the square root in §3.8, where the parabola was restricted to its right half. The difference is only that a periodic function fails the test infinitely often rather than twice, which makes the choice more visible but not different in kind.
Figure (svg): The sine graph with a horizontal line crossing it many times, showing that infinitely many angles share one sine value
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 764-769
Picture it
The dashed line at one half meets the sine at every red dot, and beyond both edges.
Figure (svg): The sine graph with a horizontal line crossing it many times, showing that infinitely many angles share one sine value
This is the whole problem in one picture. The inverse cannot exist until all but one of those crossings has been ruled out by a restriction.
Worked example
Before restricting, see how bad it is.
\[ \text{How many angles satisfy } \sin\theta=\tfrac{1}{2}? \]
Find one solution
Why: A special angle.
\[ \frac{\pi}{6} \]
Find the second in one turn
Why: The sine is positive in the second quadrant too.
\[ 5 \pi / 6 \]
Add full turns
Why: Each gives another solution.
\[ +2 \pi k \]
Count
Why: Two per turn, endlessly.
Figure (svg): The sine graph with a horizontal line crossing it many times, showing that infinitely many angles share one sine value
\[ \theta=\tfrac{\pi}{6}+2\pi k \;\text{ or }\; \tfrac{5\pi}{6}+2\pi k \]
Verify: check the second angle
Why: The sine of five pi over six is the sine of pi over six by the reference angle, and the second quadrant keeps the sine positive — so it is one half, confirming the second family. Two families per period is what makes a single-valued inverse impossible without a choice.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 765-767
Prediction
The cosine of an angle is one half.
Predict first
How many angles satisfy this?
Correct: Infinitely many.
Why: There are two in each full turn — one in the first quadrant and one in the fourth, where the cosine is also positive — and every full turn added to either gives another. Periodicity always makes the count infinite, which is why a restriction is needed before an inverse can exist.
Worked example
What makes an interval a legitimate choice?
\[ \text{Why is } \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \text{ chosen for the sine?} \]
Check one-to-oneness
Why: The sine rises steadily across it.
Check the range is complete
Why: It runs from -1 up to 1.
Check it is centred
Why: Symmetric about zero.
Conclude
Why: It satisfies all three.
Figure (svg): The solution to Worked example choose a restriction shown as a ladder of expressions, one row per legal move
\[ \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] \]
Verify: check that symmetry is preserved
Why: On this interval the sine of a negative angle is the negative of the sine, so the inverse inherits that odd symmetry — arcsin of a negative input is the negative of arcsin of the positive one. A restriction like zero to pi would have lost that, which is one reason it is not the choice made.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 767-770
Trap
\[ \sin\theta=\tfrac{1}{2} \;\Longrightarrow\; \theta=\arcsin\tfrac{1}{2}=\tfrac{\pi}{6} \]
Report pi over 6 as the solution
Why: The inverse is applied and its output is taken to be the complete answer.
The other solutions, including five pi over six, are lost.
The inverse returns one angle by design, the one inside its restricted range. That is what makes it a function.
Solving an equation usually wants all the angles, which means taking the inverse's answer and generating the rest from symmetry and periodicity.
Evaluating an inverse and solving an equation are different tasks. The inverse is a tool used in the first step of the second, not the whole of it.
Sorting
It must be one-to-one and reach every output.
Sort into buckets
Sort each interval.
Faded example
The inverse gives pi over 6; find the other one in the first turn.
Fill in the blanks
\theta = \pi - \frac65} = \frac___\pi}___
Why: Subtracting from pi reflects across the vertical axis into the second quadrant, where the sine is also positive and has the same size. That reflection is the standard way to recover the solution the inverse function does not return.
Explain it to yourself
The restriction is not an inconvenience but a requirement.
Discussion prompt
Explain why an inverse cannot exist without one.
Hint: What does a function have to do?
Answer:
A function assigns exactly one output to each input. That is the definition, and it is not negotiable.
But infinitely many angles share each sine value, so a rule that returned 'the angle with this sine' would have infinitely many outputs for a single input. That is not a function.
Restricting the domain removes all but one of the candidates, making the assignment single-valued. The restriction is what makes the inverse a function at all, so the arbitrariness of the choice is the price of having an inverse rather than a flaw in it.
Section
Section 2
Concept
The inverse sine returns angles in the right half of the circle, the inverse cosine in the top half, and the inverse tangent strictly between the two vertical asymptotes.
The arctangent's endpoints are excluded because the tangent has asymptotes there rather than values, so no input produces those outputs. Its range is open where the arcsine's is closed, and its domain is every real number rather than an interval.
| function | range | on the circle |
|---|---|---|
| arcsin | from -pi/2 to pi/2 | the right half |
| arccos | from 0 to pi | the top half |
| arctan | strictly between -pi/2 and pi/2 | the right half, open |
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 770-776
Picture it
Two half-circles, overlapping in the first quadrant.
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
For a positive input both land in the first quadrant and appear to behave identically. The difference only shows for a negative input, where the arcsine goes below the axis and the arccosine goes past a quarter turn.
Worked example
Find the angle in the top half with that cosine.
\[ \text{Evaluate } \arccos\left(-\tfrac{1}{2}\right). \]
Identify the reference angle
Why: The cosine of pi over 3 is one half.
\[ \frac{\pi}{3} \]
Note the sign
Why: Negative cosine means the left half.
Apply the range
Why: Only the second is in the top half.
Compute the angle
Why: Subtract the reference from pi.
\[ \pi - \frac{\pi}{3} \]
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
\[ \arccos\left(-\tfrac{1}{2}\right)=\tfrac{2\pi}{3} \]
Verify: substitute back
Why: The cosine of two pi over three is negative one half, since its reference angle is pi over three and the second quadrant makes the cosine negative. And two pi over three lies between zero and pi, so it is inside the required range.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 771-774
Matching
Three functions, three intervals.
Match the pairs
Why: The arctangent's endpoints are excluded because the tangent is undefined there, so no input produces them. The other two include their endpoints because the sine and cosine are perfectly well defined at the ends of their restricted domains.
Worked example
Here the two ranges visibly differ.
\[ \text{Evaluate } \arcsin\left(-\tfrac{\sqrt{2}}{2}\right). \]
Identify the reference angle
Why: The sine of pi over 4 is root two over two.
\[ \frac{\pi}{4} \]
Note the sign
Why: Negative sine means below the axis.
Apply the range
Why: Only the fourth is in the right half.
Write it as a negative angle
Why: Measured clockwise.
\[ -\frac{\pi}{4} \]
Figure (svg): The solution to Worked example evaluate an inverse sine with a negative input shown as a ladder of expressions, one row per legal move
\[ \arcsin\left(-\tfrac{\sqrt{2}}{2}\right)=-\tfrac{\pi}{4} \]
Verify: check the range
Why: Negative pi over four lies between negative pi over two and pi over two, so it is legitimate. Reporting seven pi over four instead — the same terminal side measured the other way — would be outside the range and therefore wrong, even though the sine is the same.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 774-776
Error analysis
A student evaluates an inverse cosine of a negative number.
Annotate
On: \( \arccos\left(-\tfrac{\sqrt{3}}{2}\right)=-\tfrac{\pi}{6} \)
The two ranges only differ for negative inputs, which is exactly when this error appears. Checking whether the answer lies inside the stated range catches it every time.
Prediction
The input to an arccosine is negative.
Predict first
What can be said about the output angle?
Correct: It is between pi/2 and pi.
Why: The arccosine's range is zero to pi, and within that the cosine is negative only past a quarter turn. So a negative input always produces a second-quadrant angle, and never a negative one — the arccosine has no negative outputs at all.
Faded example
The tangent of pi over 3 is root three.
Fill in the blanks
\arctan(-\sqrt3) = -\frac4___}, \text___ ___
Why: The arctangent's range is the open right half, so a negative input gives a negative angle in the fourth quadrant. Like the arcsine and unlike the arccosine, the arctangent is odd — negating the input negates the output.
Step zero
You are asked to evaluate an inverse trigonometric function of a negative number.
Discussion prompt
What do you determine first?
Hint: Two things need settling before any arithmetic.
Answer:
Two things at once: the reference angle from the size of the input, and the quadrant from the sign of the input together with the function's range.
The reference angle is a special-value recall. The quadrant is the part that depends on which inverse it is, since the three ranges cover different halves of the circle.
Then combine them: the reference angle gives the size and the quadrant gives how to place it. Doing the quadrant first prevents the commonest error, which is applying one function's sign convention to another.
Section
Section 3
Concept
Each inverse graph is the restricted original reflected across the line where output equals input, so their domains and ranges are exchanged.
The arctangent's horizontal asymptotes are the reflected images of the tangent's vertical ones. Reflection turns a vertical asymptote into a horizontal one, which is why the arctangent levels off at plus and minus a quarter turn rather than growing.
Figure (svg): The restricted sine and its inverse drawn together, mirror images across the diagonal line y equals x
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 776-781
Picture it
The teal curve reflected in the dashed diagonal gives the pink one.
Figure (svg): The restricted sine and its inverse drawn together, mirror images across the diagonal line y equals x
Reading the axes shows the exchange directly: the teal curve runs across from negative pi over two to pi over two and up from negative one to one, and the pink one does the opposite.
Worked example
Read them off the original's restriction.
\[ \text{State the domain and range of } \arccos. \]
Recall the restricted cosine
Why: Domain zero to pi.
\[ [0, \pi] \]
Recall its outputs
Why: From -1 to 1.
\[ [-1, 1] \]
Swap them
Why: Reflection exchanges the roles.
State the result
Why: Domain from the outputs, range from the domain.
Figure (svg): The restricted sine and its inverse drawn together, mirror images across the diagonal line y equals x
\[ \text{domain }[-1,1],\; \text{range }[0,\pi] \]
Verify: check an endpoint
Why: The arccosine of 1 is zero and the arccosine of negative 1 is pi, so both range endpoints are attained at the domain endpoints. That is the reflection working exactly: the original's endpoints become the inverse's, with the axes exchanged.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 777-779
Sorting
The domain is the original's range.
Sort into buckets
Sort each expression.
Worked example
Reflection changes vertical into horizontal.
\[ \text{Why does } \arctan \text{ level off at } \pm\tfrac{\pi}{2}? \]
Recall the tangent's asymptotes
Why: Vertical, at a quarter turn.
\[ x = \pm \pi / 2 \]
Reflect across the diagonal
Why: Vertical lines become horizontal.
\[ y = \pm \pi / 2 \]
Interpret
Why: The output approaches but never reaches them.
Note the domain
Why: Every real number is now allowed.
Figure (svg): The solution to Worked example explain the arctangent's asymptotes shown as a ladder of expressions, one row per legal move
\[ y=\pm\tfrac{\pi}{2} \text{ are horizontal asymptotes} \]
Verify: check a large input
Why: The arctangent of a thousand is about 1.5698, just under pi over two, which is about 1.5708. Larger inputs get closer without reaching it, confirming the asymptote. This is why the arctangent is used to compress an unbounded quantity into a bounded angle.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 779-781
Trap
\[ \arcsin \text{ has domain all real numbers, like } \sin \]
Carry the sine's domain over to the inverse
Why: Since the sine accepts any angle, the arcsine is assumed to accept any number.
An input of 2 is treated as legitimate.
Reflection swaps domain and range. The arcsine's domain is the sine's range, which is from negative 1 to 1 — nothing outside.
There is no angle whose sine is 2, so the arcsine of 2 is undefined. A calculator returns an error, not a number.
Every inverse's domain is the original's range. For the arctangent that gives all real numbers, since the tangent's range is everything — which is why only that one accepts any input.
Prediction
The input grows very large.
Predict first
What happens to the output?
Correct: It approaches pi/2 without reaching it.
Why: The tangent's vertical asymptote reflects into a horizontal one for the inverse, so the output levels off. Reaching pi over two exactly would require a terminal side pointing straight up, whose tangent is undefined rather than large — so the value is approached but never attained.
Faded example
Reflection swaps domain and range.
Fill in the blanks
\text-1\arcsin = \text1\sin = [___, ___]
Why: The sine's outputs run from negative one to one, so those are the only legal inputs to its inverse. This is why a calculator errors on arcsin of 2, and it is the same for the arccosine.
Explain it
Inverse graphs are reflections across the diagonal.
Discussion prompt
Explain to a classmate why that reflection swaps domain and range.
Hint: What does reflecting across that line do to a point?
Answer:
Reflecting across the line where output equals input sends a point to the point with its two coordinates exchanged. That is exactly what the reflection does, geometrically.
The inverse's job is to undo the original, so if the original sent an input to an output, the inverse sends that output back to the input. Each point's coordinates swap, which is the reflection.
So the set of first coordinates becomes the set of second coordinates and vice versa — the domain and range exchange. A good explanation notes that this makes every domain and range fact about an inverse a restatement of a fact about the original, with nothing new to memorise.
Section
Section 4
Concept
Applying a function and its inverse returns the input, but only when the input is in the right interval. Outside it, the composition returns the equivalent angle that is inside.
The asymmetry between the two directions is worth pausing on. Going out and back through the inverse always works because the inverse's outputs are all inside the restricted domain by construction; going the other way can start outside it, and then the round trip lands somewhere else.
Figure (svg): A card giving the three standard domain restrictions and the resulting ranges of the inverse sine, inverse cosine and inverse tangent
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 781-786
Picture it
These intervals decide whether a composition cancels.
Figure (svg): A card giving the three standard domain restrictions and the resulting ranges of the inverse sine, inverse cosine and inverse tangent
If the inner angle is inside the relevant interval the composition cancels; if not, the answer is the angle inside the interval with the same value.
Worked example
The inner angle is outside the range.
\[ \text{Evaluate } \arcsin\left(\sin\tfrac{5\pi}{6}\right). \]
Check the inner angle's range
Why: Five pi over six exceeds pi over two.
Evaluate the inner sine
Why: The reference angle is pi over 6.
\[ \frac{1}{2} \]
Apply the arcsine
Why: Find the angle in the right half.
\[ \frac{\pi}{6} \]
Compare
Why: Not the original angle.
Figure (svg): The sine graph with a horizontal line crossing it many times, showing that infinitely many angles share one sine value
\[ \arcsin\left(\sin\tfrac{5\pi}{6}\right)=\tfrac{\pi}{6} \]
Verify: check both angles have the same sine
Why: Both have sine one half, so the arcsine cannot distinguish them and must return the one in its range. The composition returns the equivalent angle inside the range rather than the angle you started with, which is the general behaviour whenever the input is outside.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 782-784
Sorting
Check whether the inner value is in range.
Sort into buckets
Sort each composition.
Worked example
Draw the triangle the inverse describes.
\[ \text{Simplify } \sin(\arccos x) \text{ for } 0\le x\le 1. \]
Name the inner angle
Why: Call it theta.
\[ \theta = \arccos x \]
Draw the triangle
Why: Adjacent x, hypotenuse 1.
\[ \cos \theta = x \]
Find the opposite side
Why: By the Pythagorean theorem.
\[ \sqrt{1 - x ^{2}} \]
Read the sine
Why: Opposite over hypotenuse.
\[ \sqrt{1 - x ^{2}} \]
Figure (svg): The solution to Worked example a mixed composition shown as a ladder of expressions, one row per legal move
\[ \sin(\arccos x)=\sqrt{1-x^2} \]
Verify: test a value
Why: At x equal to one half, arccos gives pi over three whose sine is root three over two; the formula gives the root of one minus one quarter, which is root three over two. They agree, confirming the triangle reading.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 784-786
Error analysis
A student simplifies a composition.
Annotate
On: \( \arccos\left(\cos\tfrac{7\pi}{4}\right)=\tfrac{7\pi}{4} \)
Checking whether the inner angle lies in the relevant range before cancelling costs one glance and prevents the error entirely. Going the other way — the cosine of an arccosine — always cancels, since the arccosine's output is inside by construction.
Faded example
Draw the triangle for theta with tangent x.
Fill in the blanks
\sin(\arctan x)=\frac21}}}}, \text___ x \text___ ___
Why: With opposite x and adjacent 1 the tangent is x as required, and the hypotenuse is the root of one plus x squared. Reading the sine as opposite over hypotenuse gives the formula, and the triangle method works for any mixed composition.
Prediction
Consider the sine of the arcsine of 0.3.
Predict first
What is the value?
Correct: 0.3, since this direction always cancels.
Why: The arcsine's output is inside the restricted domain by construction, so applying the sine to it returns the original input exactly. This direction is safe for any legal input; it is the other order that requires checking whether the inner angle is in range.
Explain it to yourself
One order always cancels and the other does not.
Discussion prompt
Explain why the two directions behave differently.
Hint: Where does the intermediate value live in each case?
Answer:
Applying the inverse first produces an angle that is inside the restricted domain — that is what the range of the inverse means. So the outer function is being applied exactly where the restriction says it behaves invertibly, and the round trip closes.
Applying the original first can start with any angle at all, including ones far outside the restriction. The original's output carries no memory of which angle produced it.
So the inverse can only send it back to the one representative it is allowed to return. The asymmetry is the restriction showing itself: information about which of the many equivalent angles you began with is genuinely lost, and no rule can recover it.
Section
Section 5
Concept
Solving a trigonometric equation starts with an inverse to get one angle, then uses symmetry to find the others in a turn and periodicity to extend to all of them.
The last point matters in applications. A triangle's angles are between zero and a straight angle, so a solution the algebra produces may be geometrically impossible and must be discarded — the context, not the algebra, decides.
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 769-786
Picture it
For a positive input both inverses land in the first quadrant.
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
The second solution to an equation is in whichever other quadrant gives the same sign, which is what the symmetry step finds after the inverse has given the first.
Worked example
One inverse and one reflection.
\[ \text{Solve } \cos\theta=-\tfrac{\sqrt{3}}{2} \text{ for } 0\le\theta<2\pi. \]
Apply the inverse
Why: Gives the second-quadrant angle.
\[ 5 \pi / 6 \]
Identify the other quadrant
Why: The cosine is negative in the third too.
Reflect across the horizontal axis
Why: Subtract from two pi.
\[ 2 \pi - 5 \pi / 6 \]
State both
Why: Two solutions in one turn.
\[ 5 \pi / 6\text{ and } 7 \pi / 6 \]
Figure (svg): The unit circle with the arcsine's output region on the right half and the arccosine's on the top half shaded differently
\[ \theta=\tfrac{5\pi}{6},\;\tfrac{7\pi}{6} \]
Verify: check the second solution
Why: Seven pi over six has reference angle pi over six and lies in the third quadrant where the cosine is negative, so its cosine is negative root three over two — matching. Both solutions are in the required interval, and there are exactly two, as expected for a cosine equation over one turn.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 770-775
Prediction
The sine of an angle is 0.4, with the angle between zero and a full turn.
Predict first
How many solutions are there?
Correct: Two.
Why: The sine is positive in the first and second quadrants, giving one solution in each, and a full turn contains each quadrant once. The inverse returns only the first-quadrant one, so the second has to be produced by subtracting from a straight angle.
Worked example
Here the physical range decides.
\[ \text{A ramp rises } 2 \text{ ft over a run of } 10 \text{ ft. Find its angle.} \]
Identify the ratio
Why: Opposite over adjacent.
Write the equation
Why: The tangent of the angle is the ratio.
\[ \tan A = 0.2 \]
Apply the inverse
Why: Gives one angle.
\[ \arctan(0.2) \]
Evaluate
Why: About 11.3 degrees.
\[ 11.3 ^\circ \]
Figure (svg): The solution to Worked example a triangle application shown as a ladder of expressions, one row per legal move
\[ A\approx 11.3^\circ \]
Verify: check plausibility
Why: The tangent of 11.3 degrees is about 0.200, matching the ratio. And an acute angle is the only sensible answer for a ramp, so the second mathematical solution — 11.3 degrees plus a straight angle — is discarded on physical grounds rather than algebraic ones.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 776-786
Trap
\[ \sin A=0.6 \;\Longrightarrow\; A=\arcsin(0.6)\approx 36.9^\circ \text{ only} \]
Take the calculator's answer as the sole solution
Why: The obtuse possibility is not considered.
A valid obtuse triangle is missed entirely.
A triangle's angle can be obtuse, and the supplement of 36.9 degrees is 143.1 degrees, whose sine is also 0.6.
Both are in the legal range for a triangle angle, so both must be checked against the rest of the triangle — the remaining angles must still sum correctly.
The inverse gives one angle; the context decides how many are admissible. Discarding without checking is as much an error as keeping an impossible one.
Faded example
The arctangent gives one angle; the tangent is also positive in the third quadrant.
Fill in the blanks
\theta_2 = \theta_1 + pi, \textpi___
Why: For the tangent the second solution in a turn is a half turn away rather than a reflection, because its period is pi. That is a genuine difference from the sine and cosine, whose second solutions come from reflecting rather than translating.
Sorting
The context is an angle in a triangle.
Sort into buckets
Sort each candidate.
Explain it
The inverse gives one angle but equations usually have more.
Discussion prompt
Explain to a classmate how to get from one angle to all of them.
Hint: What produces the second, and what produces the rest?
Answer:
The inverse gives the reference solution, the one inside its range. That is the starting point and never the whole answer.
The second solution in one turn comes from symmetry: reflect to the other quadrant where the function has the same sign — subtract from pi for the sine, from two pi for the cosine, add pi for the tangent.
Then add full turns to both to reach every solution. A good explanation stresses that the inverse is deliberately single-valued and that recovering the rest is your job, not the function's — which is why 'the calculator gave one answer' is never a reason to stop.
Comparison
Fill the blanks from memory. The ranges are what everything else depends on.
Comparison matrix
| arcsin | arccos | arctan | |
|---|---|---|---|
| domain | -1 to 1 | -1 to 1 | all reals |
| range | -pi/2 to pi/2 | 0 to pi | open, -pi/2 to pi/2 |
| output for a negative input | negative, fourth quadrant | second quadrant | negative, fourth quadrant |
| asymptotes | none | none | horizontal at plus and minus pi/2 |
The third row is where the arccosine parts company with the other two, and it is where errors cluster — negative inputs are the only case in which the three ranges visibly differ.
Pattern
Five steps, and the third is the one that distinguishes the three functions.
Step 5 catches every version of the commonest error, since an answer outside the stated range is wrong even when the trigonometric value is right.
OpenStax Algebra and Trigonometry 2e, §8.3 Inverse Trigonometric Functions §8.3
Check
The range decides the quadrant.
Check your understanding
What is the arccosine of negative one half?
Answer: A
Why: The reference angle is pi over three, and a negative cosine puts the angle in the second or third quadrant. The arccosine's range of zero to pi allows only the second, giving pi minus pi over three.
Check
Check whether the inner angle is in range.
Check your understanding
What does arcsin(sin(2pi/3)) equal?
Answer: A
Why: Two pi over three is outside the arcsine's range, so the composition does not cancel. Its sine is root three over two, and the angle in the range with that sine is pi over three.
Check
The domain is the original's range.
Check your understanding
Which expression is undefined?
Answer: A
Why: No angle has a sine of 1.4, since the sine never exceeds 1 in size. The arcsine's domain is exactly the sine's range, from negative one to one, so any input outside that is illegal.
Real world
Anything that converts a slope or a ratio back into an angle uses an inverse.
Discussion prompt
Software drawing a line from one point to another needs the line's angle. Why is the arctangent alone not enough?
Hint: How many directions share a slope?
Answer:
The arctangent of the slope gives an angle, but its range is only the right half of the circle — it can never return an angle pointing leftward.
Two opposite directions share the same slope, so the arctangent cannot distinguish 'up and to the right' from 'down and to the left'. Both give the same answer.
Which is why graphics libraries provide a two-argument arctangent that takes the two coordinates separately rather than their ratio. Keeping both signs identifies the quadrant, so it returns a full-circle angle. The restriction that makes the inverse a function is precisely what that second argument works around, and it is one of the most-used functions in graphics and navigation code.
Commit first
State your confidence along with your answer.
Predict first
Why do the inverse trigonometric functions have restricted ranges?
Correct: Because the originals are not one-to-one, so a single output must be chosen.
Why: Infinitely many angles share each trigonometric value, and a function must return exactly one. The restriction is what makes the inverse well defined. The particular intervals chosen are conventional, but the need for some restriction is not.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate asks why their calculator gives 30 degrees when they wanted 150. What do you say?
Hint: Both have the same sine.
Answer:
Both angles have a sine of one half, so the calculator cannot tell which was wanted. It has to pick one, and it always picks the one in the inverse sine's range.
That range is the right half of the circle, so 30 degrees is inside it and 150 is not. The calculator is not wrong — it is doing the only thing a single-valued function can do.
To get 150, subtract 30 from 180. Finding the other solutions is the user's job, and a good explanation adds why: the inverse was deliberately built to return one angle, so the information about which one you wanted was never available to it.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where most errors live, since the arccosine's range differs from the other two exactly when the input is negative. The fourth is what makes the whole section usable in later work.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw a unit circle and shade the arcsine's output region and the arccosine's in different colours, labelling each with its interval. Beside it, write the arctangent's range and note why its endpoints are excluded. Underneath, write one composition that cancels and one that does not, with the reason for each.
If your two shaded regions overlap in the first quadrant and your two compositions differ only in which order the functions were applied, the section's two hardest points are both on the page.
Recap
Five things, and the first explains all the others.
| if you remember one thing | it should be this |
|---|---|
| about the ranges | arcsin right half, arccos top half, arctan open right half |
| about negative inputs | the arccosine goes to the second quadrant, never negative |
| about compositions | inverse-then-original always cancels; the other order may not |
| about solving | the inverse gives one angle, and finding the rest is your job |
Chapter 7 uses these constantly: every trigonometric equation ends with an inverse and a symmetry argument, and every identity verification has to respect the same domain restrictions established here.
OpenStax, Precalculus, §6.3 Inverse Trigonometric Functions §6.3, pp. 764-780 — everything on these slides traces back here
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