The inverse functions put to work. Covers approximating the three inverses a calculator does not provide, finding the domain and range of a transformed inverse function by tracking key points, applications where a measured ratio must be converted into an angle, and the main point of the section: solving equations whose values are not special angles, using an inverse function to name the reference angle exactly as a square root names an irrational solution.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.6 The Inverse Trigonometric Functions, pp. 833-841
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. 833-841 — the pages these objectives are drawn from
Warm-up
You can solve sine of theta equals one half completely. This lesson asks the same question with a value that is not on the special-angle list, and the answer is more familiar than it looks.
Discussion prompt
Solve x squared equals 4, then solve x squared equals 7. What did you do differently the second time, and is the second answer less exact than the first?
Hint: Write both answers down and compare their form.
Answer:
\[ x^2 = 4 \;\Longrightarrow\; x = \pm 2 \qquad x^2 = 7 \;\Longrightarrow\; x = \pm\sqrt{7} \]
The second answer is exactly as exact as the first. Root seven is a number, precisely specified; it simply has no shorter name. Writing 2.6458 instead would make it less exact, not more.
That is the whole idea of this lesson. Sine of theta equals one half has friendly answers; sine of theta equals one third does not — and arcsine of one third is the exact answer, in the same way that root seven is.
Concept
Until now every equation in this course had special-angle answers. Most equations do not. An inverse function is the notation that lets an unfriendly answer be written down exactly, and the parallel with the square root is close enough to be worth leaning on.
So the method for solving does not change at all. What changes is that the reference angle is written as an arcsine or arctangent instead of being read off a table.
Figure (svg): A side by side comparison of solving x squared equals four with friendly answers against x squared equals seven needing a square root, and sine of theta equals one half against sine of theta equals one third needing an arcsine
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 838-838
Section
Section 1
Concept
Calculators provide only arcsine, arccosine and arctangent. The other three must be converted, and the conversion is straightforward for positive inputs and needs care for negative ones.
Always finish by checking that the value returned lies in the range of the function you asked for. That single test catches every branch error a calculator can commit.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 833-835
Picture it
The reference-angle route works for every one of the three missing inverses.
Figure (svg): The recommended route for solving a secant equation, converting to a cosine equation first so that no arcsecant convention has to be chosen
Better still, when the problem is an equation rather than an evaluation, converting to a cosine or sine at the outset avoids the whole issue.
Worked example
Example 10.6.5, part d. The same input, two genuinely different answers.
\[ \text{Approximate } \operatorname{arccsc}\left(-\frac{3}{2}\right) \text{ under each convention.} \]
Use the identity, trigonometry friendly
Why: There it holds for the whole domain.
\[ \arcsin(-\frac{2}{3}) = -0.7297 \]
Note the identity fails, calculus friendly
Why: There it needs a positive input.
Find the reference angle
Why: The reference angle has cosecant three halves, so it is arcsine of two thirds.
\[ \alpha = \arcsin(\frac{2}{3}) \]
Place it in quadrant three
Why: The calculus-friendly range puts a negative cosecant between pi and three pi over two.
\[ \pi + \alpha = 3.8713 \]
Figure (svg): The solution to Worked example a negative arccosecant, both conventions shown as a ladder of expressions, one row per legal move
\[ \text{trig: } -0.7297; \qquad \text{calculus: } \pi + \arcsin\left(\tfrac{2}{3}\right) \approx 3.8713 \]
Verify: check each against its own range
Why: Negative 0.7297 is between negative pi over two and zero, which is inside the trigonometry-friendly range. And 3.8713 is between pi, about 3.14, and three pi over two, about 4.71 — inside the calculus-friendly range. Each answer is right for its own convention and wrong for the other.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 835-835
Sorting
A detour is needed whenever the reciprocal identity's condition fails.
Sort into buckets
Sort each evaluation by whether the direct conversion works.
Worked example
Example 10.6.5, part c. Here there is only one convention, so only one answer.
\[ \text{Approximate } \operatorname{arccot}(-2) \text{ to four decimal places.} \]
Locate the answer from the range
Why: Arccotangent has range zero to pi, and a negative cotangent puts the angle in quadrant two.
\[ \frac{\pi}{2} < t < \pi \]
Find the reference angle
Why: Its cotangent is positive 2, so it is the arctangent of one half.
\[ \alpha = \arctan(\frac{1}{2}) \]
Place it in quadrant two
Why: The angle is pi minus its reference angle.
\[ t = \pi - \arctan(\frac{1}{2}) \]
Evaluate
Why: Pi minus about 0.4636.
\[ = 2.6779 \]
Figure (svg): The solution to Worked example a 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 a second route
Why: The arctangent of negative one half is about negative 0.4636, and adding pi gives about 2.6779 — the same. The two routes agree because the tangent has period pi, so adding pi moves between the two angles with the same tangent.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 834-834
Trap
\[ \operatorname{arccot}(-2) \approx \arctan(-0.5) \approx -0.4636 \]
Type the reciprocal into the arctangent button and report the result
Why: The calculator gives an answer without complaint, and the arithmetic inside it is correct.
But the arccotangent's range is zero to pi, so a negative answer is impossible. The calculator answered the arctangent question, which is a different one.
\[ \operatorname{arccot}(-2) = \pi + \arctan(-0.5) \approx 2.6779 \]
Adjust the branch to land in the correct range
Why: Adding pi moves to the angle with the same tangent in the other half of the circle.
A calculator has no idea which function you meant. It applies a fixed branch, and checking the answer against the intended function's range is the only defence. This is the same discipline as the atan2 problem in the previous lessons.
Fill the middle
Approximate the arccotangent of negative one third.
Fill in the blanks
\text3 t = \pi - \arctan(1.2490) \approx \pi - ___ \approx 1.8925
Why: The reference angle has cotangent one third, so its tangent is 3 and it is the arctangent of 3, about 1.2490. Subtracting from pi gives about 1.8925, which is between pi over two and pi as a quadrant two angle must be. Note that a small cotangent means a steep angle, which is why the answer is close to pi over two rather than close to pi.
Prediction
You compute the arcsecant of 1.001 on a calculator.
Predict first
Roughly what should you get?
Correct: Close to 0.
Why: The arcsecant of exactly 1 is zero, since the secant of zero is 1. An input just above 1 gives an angle just above zero — about 0.0447 radians here. Estimating first is worth doing because the intuition can run the wrong way: it is the very large inputs that give angles near pi over two, and the inputs near 1 that give angles near zero.
Edge cases
A calculator is asked for the arcsine of 1.0000001.
Discussion prompt
What happens, and why is that the right behaviour rather than a limitation?
Hint: Is the input in the domain?
Answer:
It returns an error, because 1.0000001 is outside the domain of the arcsine — no angle has a sine exceeding 1.
This is correct behaviour rather than a limitation. The expression is genuinely undefined, and returning a number would be worse than returning an error.
It is worth knowing because this error appears in practice for a specific reason: rounding. A computation that should give exactly 1 can produce 1.0000000001 through accumulated floating-point error, and feeding that into an arcsine crashes. Numerical code routinely clamps the value into the interval before calling the function, precisely to prevent this.
Section
Section 2
Concept
A transformed inverse function such as pi over two minus arccosine of x over five has a domain and a range that can both be read off, and each comes from a different part of the formula.
This is the same procedure used for transformations of any function; nothing about it is specific to trigonometry. What is specific is that the inverse functions have unusual domains and ranges to start from.
Figure (svg): The graph of the arccosine transformed into pi over two minus arccosine of x over five, showing how three key points move and how the domain and range change
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 836-836
Picture it
Three points in, three points out, and both the domain and the range fall out.
Figure (svg): The graph of the arccosine transformed into pi over two minus arccosine of x over five, showing how three key points move and how the domain and range change
Notice how the division by five stretched the domain by a factor of five while leaving the range alone, and the subtraction from pi over two reflected and shifted the range while leaving the domain alone.
Worked example
Example 10.6.5, part 2a. Argument first, then the outside.
\[ \text{Find the domain and range of } f(x) = \frac{\pi}{2} - \arccos\left(\frac{x}{5}\right). \]
Set the argument in the arccosine's domain
Why: The inside must be between negative one and one.
\[ -1 \le \frac{x}{5} \le 1 \]
Solve for x
Why: Multiply through by 5.
\[ \text{domain } [-5, 5] \]
Track the three key points
Why: Arccosine's points are negative one to pi, zero to pi over two, one to zero.
Apply the outside operations
Why: Negate and add pi over two, giving negative pi over two, zero and pi over two.
\[ \text{range } [-\frac{\pi}{2}, \frac{\pi}{2}] \]
Figure (svg): The solution to Worked example domain and range of a transformed arccosine shown as a ladder of expressions, one row per legal move
\[ \text{domain } [-5, 5], \qquad \text{range } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \]
Verify: recognise the result
Why: The range came out as the arcsine's range, and the function is in fact the arcsine of x over five — since pi over two minus arccosine of u is arcsine of u, by the cofunction identity. That is a satisfying confirmation, and it explains why the range came out so tidily.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 836-836
Matching
Take the function A times arcsin of the quantity Bx plus C, plus D.
Match the pairs
Why: B and C sit inside the function and therefore act on the input, changing where the argument satisfies the domain condition. A and D sit outside and act on the output, changing the range. This division is completely general and holds for any function, not just an inverse trigonometric one — which is why the same reasoning worked for sinusoids in Lesson 10.5a.
Worked example
Example 10.6.5, part 2b. Here the asymptotes are what to track.
\[ \text{Find the domain and range of } f(x) = 3\arctan(4x). \]
Set the argument in the arctangent's domain
Why: The arctangent accepts every real number, so 4x is never a problem.
Identify the parent's range
Why: Strictly between negative pi over two and pi over two.
\[ (-\frac{\pi}{2}, \frac{\pi}{2}) \]
Apply the outside operation
Why: The multiplication by 3 stretches vertically.
\[ \text{multiply by } 3 \]
State the new range
Why: Still open at both ends, since the asymptotes stay asymptotes.
\[ (-3 \pi / 2, 3 \pi / 2) \]
Figure (svg): The solution to Worked example a transformed arctangent shown as a ladder of expressions, one row per legal move
\[ \text{domain } (-\infty, \infty), \qquad \text{range } \left(-\tfrac{3\pi}{2}, \tfrac{3\pi}{2}\right) \]
Verify: check what the 4 did
Why: The 4 inside compressed the graph horizontally by a factor of four, which changes how quickly the curve approaches its asymptotes but not the domain or the range at all. Only the 3 outside affected the range — the standard division of labour between inside and outside.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 836-836
Error analysis
A student finds the domain of two arcsin of x, plus 1.
Annotate
On: \( -1 \le 2\arcsin(x) + 1 \le 1 \;\Longrightarrow\; \text{domain} = [-1, 0] \)
Inside affects the domain and outside affects the range, and no operation does both. Writing the argument down explicitly before imposing any condition prevents this entirely.
Faded example
Find the domain of arccos of the quantity two x minus one.
Fill in the blanks
-1 \le 2x - 1 \le 1 \;\Longrightarrow\; 0 \le 2x \le 2 \;\Longrightarrow\; \text[0, 1] = ___
Why: Adding one throughout gives zero to two, and halving gives zero to one. The domain is the closed interval from zero to one — narrower than the arccosine's own domain, because the coefficient 2 compressed it. The range is unaffected and remains zero to pi, since nothing outside the function was changed.
Prediction
Consider the function arccotangent of x over 2, plus pi.
Predict first
What is its range?
Correct: from pi to 2 pi.
Why: The arccotangent's range is zero to pi, open at both ends, and adding pi shifts the whole thing up by pi, giving pi to two pi. The division by 2 inside affects only the domain, which remains all real numbers since arccotangent accepts everything. This is the function the book graphs on a calculator, and it needs a piecewise definition to do so — not because of the range but because the calculator has no arccotangent.
Socratic
To graph arccotangent of x over two, plus pi, the book writes it as three cases involving arctangent.
Discussion prompt
Explain why one arctangent formula cannot cover the whole graph.
Hint: What happens at x equal to zero, and what happens for negative x?
Answer:
For positive x the identity gives arctangent of two over x, plus pi. For negative x that formula lands on the wrong branch and needs a further pi added, giving arctangent of two over x plus two pi.
At x equal to zero neither formula works at all, because two over x is undefined. Yet the original function is perfectly well defined there: the arccotangent of zero is pi over two, so the value is three pi over two.
So three cases are needed: one for negative x, one for zero, one for positive x. The piecewise definition is an artefact of the calculator's missing button, not of the function, which is smooth and unremarkable across the whole real line.
Section
Section 3
Concept
Almost every practical use of an inverse trigonometric function has the same shape: two lengths have been measured, their ratio is known, and the angle is wanted.
For an acute angle in a triangle the restriction never causes trouble, because every inverse function's range already contains the whole of quadrant one. The convention problems only bite when the angle can be obtuse or negative.
Figure (svg): The side view of a roof with a six over twelve pitch, forming a right triangle with a rise of six feet and a run of twelve feet, and the angle of inclination marked at the base
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 837-838
Picture it
A builder's pitch is a ratio. Converting it to an angle is a single inverse function call.
Figure (svg): The side view of a roof with a six over twelve pitch, forming a right triangle with a rise of six feet and a run of twelve feet, and the angle of inclination marked at the base
Notice that the answer is under 30 degrees even though the roof looks reasonably steep. Pitch ratios and angles do not track each other intuitively, which is one reason both notations survive.
Worked example
Example 10.6.6. A 6 over 12 pitch means a rise of 6 feet over a run of 12.
\[ \text{A roof has a } 6/12 \text{ pitch. Find its angle of inclination in degrees.} \]
Identify which two sides are known
Why: The rise is opposite the angle and the run is adjacent to it.
Choose the matching function
Why: Opposite over adjacent is the tangent.
\[ \tan \theta = \frac{6}{12} \]
Simplify the ratio
Why: Six twelfths reduces to one half.
\[ \tan \theta = \frac{1}{2} \]
Apply the inverse and convert to degrees
Why: The angle is acute, so the arctangent applies without difficulty.
\[ \theta = \arctan(\frac{1}{2}) = 26.56 ^\circ \]
Figure (svg): The solution to Worked example the angle of a roof shown as a ladder of expressions, one row per legal move
\[ \theta = \arctan\left(\frac{1}{2}\right) \approx 26.56^\circ \]
Verify: sanity-check against a landmark
Why: A 12 over 12 pitch would be 45 degrees, since the rise would equal the run. A 6 over 12 pitch is half as steep in ratio, and the answer is less than half of 45 — which is right, because the tangent is not linear in the angle.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 837-838
Matching
The pair of sides you measured decides the function.
Match the pairs
Why: Rise over run and opposite over adjacent are the same ratio, so both give an arctangent. A height over a sloping distance is opposite over hypotenuse, giving an arcsine. A horizontal distance over a straight-line distance is adjacent over hypotenuse, giving an arccosine. The word hypotenuse rarely appears in the problem, so the skill is recognising which measured length is the sloping one.
Worked example
The choice is dictated by which sides were measured.
\[ \text{A ramp } 8 \text{ m long rises } 1.2 \text{ m. Find its angle of inclination.} \]
Identify the sides
Why: The 8 metres is the ramp itself, so it is the hypotenuse; the 1.2 metres is the rise, opposite the angle.
Choose the matching function
Why: Opposite over hypotenuse is the sine.
\[ \sin \theta = \frac{1.2}{8} \]
Simplify
Why: One point two over eight is 0.15.
\[ \sin \theta = 0.15 \]
Apply the arcsine
Why: The angle is acute, so no ambiguity arises.
\[ \theta = \arcsin(0.15) = 8.63 ^\circ \]
Figure (svg): The solution to Worked example choosing the right inverse shown as a ladder of expressions, one row per legal move
\[ \theta = \arcsin(0.15) \approx 8.63^\circ \]
Verify: check against the accessibility standard
Why: A common maximum ramp gradient for wheelchair access is 1 in 12, which is a rise over run of about 0.0833 and an angle of about 4.76 degrees. This ramp is nearly twice as steep, so it would not meet that standard — a useful sanity check that the number is in a plausible range for a real ramp.
Trap
\[ \text{ramp } 8 \text{ m long rising } 1.2 \text{ m}: \quad \tan\theta = \frac{1.2}{8} \]
Treat the ramp's length as the run
Why: Rise over run is the familiar formula, and the two given numbers slot into it without complaint.
But the 8 metres is the sloping length, not the horizontal distance. The horizontal run is shorter, about 7.91 metres, so the tangent is slightly larger than the ratio used.
\[ \sin\theta = \frac{1.2}{8} \;\Longrightarrow\; \theta \approx 8.63^\circ \]
Match the function to the sides that were actually measured
Why: Opposite over hypotenuse is a sine, not a tangent.
For a shallow angle the two answers barely differ — the tangent version gives 8.53 degrees against the correct 8.63 — which is exactly what makes this error dangerous. It stays small until the angle is steep, and then it grows quickly. Reading the problem for which side is which costs nothing and removes the risk entirely.
Faded example
A ladder 5 m long has its foot 1.4 m from a wall. Find the angle it makes with the ground.
Fill in the blanks
cos~theta = \fracarccos___ = 0.28 \;\Longrightarrow\; ~theta = ___(0.28) \approx 73.7^\circ
Why: The 1.4 metres is the horizontal distance, adjacent to the angle at the ground, and the 5 metres is the ladder itself, the hypotenuse. Adjacent over hypotenuse is a cosine, so the angle is the arccosine of 0.28, about 73.7 degrees. That is a steep and sensible ladder angle — safety guidance typically recommends about 75 degrees.
Estimation
A path rises 1 metre for every 20 metres of horizontal distance.
Predict first
Roughly what angle is that?
Correct: About 3 degrees.
Why: The tangent is one twentieth, or 0.05, and the arctangent of 0.05 is about 0.05 radians — because for small angles the tangent is very close to the angle itself in radians. Converting, 0.05 radians is about 2.86 degrees. The small-angle approximation is genuinely useful here: for gradients under about 1 in 10, the angle in radians is close to the gradient, and multiplying by 57.3 converts it.
Real world
A phone's accelerometer reports the components of gravity along its three axes. When the phone is tilted about one axis only, the reported values are g times the sine and g times the cosine of the tilt angle.
Discussion prompt
Explain how the phone computes its tilt angle, and why using the ratio of the two components is better than using either one alone.
Hint: What would happen to a single-component method if the sensor's calibration drifted?
Answer:
\[ \theta = \arctan\left(\frac{a_x}{a_z}\right) \]
Taking the ratio of the two components cancels the factor of g entirely, so the computed angle does not depend on knowing the local value of gravity or on the sensor's overall gain being correctly calibrated.
Using one component alone would require dividing by g, so any drift in the sensor's scale factor would appear directly as an angle error. The ratio method is self-calibrating in that respect, which is why the two-argument arctangent appears in essentially every tilt-sensing routine.
It is the same principle as Lesson 10.2c's observation that the cosine and sine do not depend on which point of the terminal side you pick. A ratio sees direction and ignores scale, and that is exactly the property an orientation sensor needs.
Section
Section 4
Concept
Solving sine of theta equals one third uses exactly the procedure from Lesson 10.2b. The reference angle is arcsine of one third instead of pi over six, and everything downstream is identical.
\[ \sin(\theta) = \tfrac{1}{3} \;\Longrightarrow\; \theta = \arcsin\left(\tfrac{1}{3}\right) + 2\pi k \;\text{ or }\; \theta = \pi - \arcsin\left(\tfrac{1}{3}\right) + 2\pi k \]
The answer is exact. Replacing arcsine of one third with 0.3398 makes it approximate, and unless a decimal was asked for, that is a loss rather than a simplification.
Figure (svg): A unit circle showing the two solutions of sine theta equals one third, one in quadrant one at the arcsine and one in quadrant two at pi minus the arcsine, both sharing the same reference angle
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 838-839
Picture it
Put the two side by side and the only difference is what the reference angle is called.
Figure (svg): Two columns contrasting a special-value equation with one whose value is not special, showing that the structure of the answer is identical
This is why the section is placed where it is: the inverse functions are not a new topic so much as the notation that lets the old topic handle every case rather than a handful of special ones.
Worked example
Example 10.6.7, part 1. Two families, exactly as in Lesson 10.2b.
\[ \text{Solve } \sin(\theta) = \frac{1}{3} \text{ for all } \theta. \]
Find the reference angle
Why: The value is positive, and the arcsine returns an acute angle directly.
\[ \alpha = \arcsin(\frac{1}{3}) \]
Identify the quadrants
Why: A positive sine puts the crossings above the axis, in quadrants one and two.
Write one solution per quadrant
Why: Quadrant one is the reference angle itself; quadrant two is pi minus it.
Attach the coterminal families
Why: Each repeats every revolution.
\[ \text{plus } 2 \pi k\text{ each} \]
Figure (svg): The solution to Worked example a sine equation with an unfriendly value shown as a ladder of expressions, one row per legal move
\[ \theta = \arcsin\left(\tfrac{1}{3}\right) + 2\pi k \quad\text{or}\quad \theta = \pi - \arcsin\left(\tfrac{1}{3}\right) + 2\pi k \]
Verify: check the second family numerically
Why: Arcsine of one third is about 0.3398, so the second family starts at about 2.8018. The sine of 2.8018 is about 0.3333, which is one third. Both families genuinely solve the equation.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 838-839
Comparison
Fill the blanks from memory. The two columns are structurally identical.
Comparison matrix
| Step | sin theta = 1/2 | sin theta = 1/3 |
|---|---|---|
| Reference angle | pi over 6 | arcsin(1/3) |
| Quadrants | one and two | one and two |
| First family | pi/6 + 2 pi k | arcsin(1/3) + 2 pi k |
| Second family | pi - pi/6 + 2 pi k | pi - arcsin(1/3) + 2 pi k |
| Exactness | exact | equally exact |
Every row matches. The only thing that changed is that the reference angle acquired a name instead of a value, which is a notational difference rather than a mathematical one.
Worked example
Example 10.6.7, part 2. One family suffices here, as always for tangent.
\[ \text{Solve } \tan(t) = -2 \text{ for all real } t. \]
Apply the arctangent directly
Why: It accepts negative inputs and returns an angle in quadrant four.
\[ \beta = \arctan(-2) \]
Identify the quadrants
Why: A negative tangent puts the solutions in quadrants two and four.
Note the two are half a revolution apart
Why: That is always true for tangent, since its period is pi.
Combine into one family
Why: A single family stepping by pi captures both.
\[ t = \arctan(-2) + \pi k \]
Figure (svg): The solution to Worked example a tangent equation shown as a ladder of expressions, one row per legal move
\[ t = \arctan(-2) + \pi k, \quad k \in \mathbb{Z} \]
Verify: check two members
Why: Arctangent of negative 2 is about negative 1.1071, whose tangent is negative 2. Adding pi gives about 2.0344, and the tangent there is also negative 2 — since the tangent has period pi. The single family covers both quadrants.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 839-840
Error analysis
A student solves sine of theta equals one third and writes the answer.
Annotate
On: \( \theta = 0.3398 + 2\pi k \;\text{ or }\; \theta = 2.8018 + 2\pi k \)
Reaching for a decimal is often reflexive and it costs exactness for nothing. Give the exact form, and add a decimal only when it is asked for or when the answer has to be used in a further numerical computation.
Faded example
Solve cosine of theta equals 0.4 for all theta.
Fill in the blanks
\alpha = \arccos(0.4), \; \textIV-arccos(0.4) \;\Longrightarrow\; \theta = \arccos(0.4) + 2\pi k \;\text___\; \theta = ___ + 2\pi k
Why: A positive cosine puts the crossings to the right of the y-axis, in quadrants one and four. Quadrant one gives the arccosine directly; quadrant four gives its negative, since the two are reflections across the x-axis. Writing the second family as two pi minus the arccosine would be equally correct — the two descriptions list the same angles.
Prediction
You solve tangent of t equals 7 for all real t.
Predict first
How many families of solutions will there be?
Correct: One.
Why: A tangent equation always yields one family, because the two solution quadrants are opposite each other and therefore half a revolution apart. The answer is arctangent of 7 plus pi k. This is the same fact that made tangent families combinable back in Lesson 10.3a, and it holds regardless of whether the value is special — the value affects only what the reference angle is called.
Explain it to yourself
The book compares arcsine of one third with root seven.
Discussion prompt
Explain how far the analogy goes. Is there any respect in which the two are different?
Hint: Think about what each notation is doing and where each came from.
Answer:
The analogy is close. Both are exact names for numbers with no expression in simpler terms, both arose because a useful function was not one-to-one and had to be restricted before inverting, and in both cases the restriction is why only one of the two solutions comes back.
One genuine difference: root seven is an algebraic number, a solution of a polynomial equation with integer coefficients, whereas arcsine of one third is not — it is transcendental. That difference does not matter for solving equations, but it is why no amount of algebraic manipulation will ever produce a surd expression for it, whereas root seven already is one.
For the purposes of this course the analogy holds completely: write the exact form, approximate only when asked.
Section
Section 5
Concept
An equation in secant or cosecant would force a choice of arcsecant convention. Converting it into a cosine or sine equation first avoids that entirely, because arccosine and arcsine have universally agreed ranges.
\[ \sec(x) = -\tfrac{5}{3} \;\Longrightarrow\; \cos(x) = -\tfrac{3}{5} \]
The same advice applies to cotangent equations: converting to a tangent equation avoids nothing about conventions, since arccotangent's range is agreed, but it does let you use the single-family shortcut.
Figure (svg): The recommended route for solving a secant equation, converting to a cosine equation first so that no arcsecant convention has to be chosen
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 840-840
Picture it
One reciprocation and the ambiguity is gone.
Figure (svg): The recommended route for solving a secant equation, converting to a cosine equation first so that no arcsecant convention has to be chosen
This is worth doing even if you know which convention your course uses, because the resulting answer is one that any reader can interpret without being told.
Worked example
Example 10.6.7, part 3. Convert, then solve as usual.
\[ \text{Solve } \sec(x) = -\frac{5}{3} \text{ for all real } x. \]
Reciprocate both sides
Why: This converts to a cosine equation with an agreed inverse.
\[ \cos x = -\frac{3}{5} \]
Identify the quadrants
Why: A negative cosine puts the crossings to the left, in quadrants two and three.
Write the quadrant two solution
Why: The arccosine of a negative value already lands in quadrant two.
\[ \beta = \arccos(-\frac{3}{5}) \]
Write the quadrant three solution
Why: Its reflection across the x-axis, which is the negative.
\[ -\beta \]
Figure (svg): The solution to Worked example a secant equation shown as a ladder of expressions, one row per legal move
\[ x = \arccos\left(-\tfrac{3}{5}\right) + 2\pi k \quad\text{or}\quad x = -\arccos\left(-\tfrac{3}{5}\right) + 2\pi k \]
Verify: check a member of each family
Why: Arccosine of negative three fifths is about 2.2143, whose cosine is negative 0.6, so its secant is about negative 1.667, which is negative five thirds. And the negative of that angle has the same cosine, since cosine is even. Both families check out.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 840-840
Sorting
Convert whenever the inverse involved has no agreed convention.
Sort into buckets
Sort each equation by whether converting first is worth doing.
Worked example
Most applied problems want the solutions in one revolution rather than all of them.
\[ \text{Solve } \cos(\theta) = -0.7 \text{ for } \theta \text{ in } [0, 2\pi). \]
Find the quadrant two solution
Why: The arccosine of a negative value returns an angle in quadrant two.
\[ \arccos(-0.7) = 2.3462 \]
Find the quadrant three solution
Why: Reflecting across the x-axis gives the negative, which must be adjusted into the window.
\[ -2.3462\text{ is outside} \]
Adjust into the window
Why: Add two pi to bring it into the interval from zero to two pi.
\[ 2 \pi - 2.3462 = 3.9370 \]
State both
Why: The window of width two pi contains exactly one member of each family.
Figure (svg): The solution to Worked example a complete solution in a restricted window shown as a ladder of expressions, one row per legal move
\[ \theta = \arccos(-0.7) \approx 2.3462 \quad\text{or}\quad \theta = 2\pi - \arccos(-0.7) \approx 3.9370 \]
Verify: check both are in the window and have the right cosine
Why: Both lie between zero and two pi. The cosine of 2.3462 is about negative 0.7000 and the cosine of 3.9370 is also about negative 0.7000. And the two sum to two pi, which is the signature of a reflected pair across the x-axis.
Trap
\[ \sec(x) = -\tfrac{5}{3} \;\Longrightarrow\; x = \operatorname{arcsec}\left(-\tfrac{5}{3}\right) + 2\pi k \]
Apply the arcsecant directly, as one would apply the arccosine
Why: It is the natural inverse for the function in the equation, so reaching for it is instinctive.
But arcsecant of negative five thirds is about 2.2143 under one convention and about 4.0689 under the other. A reader has no way to know which family is meant, and the second family has been omitted entirely.
\[ \cos(x) = -\tfrac{3}{5} \;\Longrightarrow\; x = \pm\arccos\left(-\tfrac{3}{5}\right) + 2\pi k \]
Convert to a cosine equation first
Why: Arccosine has one agreed range, so the answer means the same thing to every reader.
Two things were fixed at once. The ambiguity is gone, and so is the omission — writing the answer in terms of arccosine makes both families visible, whereas the arcsecant form quietly returned only one of them.
Fill the middle
Solve cosecant of x equals 4 for all real x.
Fill in the blanks
\csc x = 4 \;\Longrightarrow\; \sin x = 1/4 \;\Longrightarrow\; x = \arcsin(pi - arcsin(1/4)) + 2\pi k \;\text___\; x = ___ + 2\pi k
Why: Reciprocating gives a sine of one quarter, a positive value, so the crossings are in quadrants one and two. Quadrant one gives the arcsine directly and quadrant two gives pi minus it. Solving via arccosecant would have needed a convention and would have produced only one of the two families.
Elimination
The equation is secant of x equals 3.
Eliminate the wrong options
Which answer is both complete and free of convention issues?
Survives elimination: B
Why: Reciprocating gives a cosine of one third, a positive value, so the solutions are in quadrants one and four — at plus and minus the arccosine of one third. Writing it with a plus-or-minus captures both families in one line, uses only the universally agreed arccosine, and requires no statement of convention.
Explain it
A classmate solves every equation with whichever inverse function matches the function in the equation, including arcsecant.
Discussion prompt
In four sentences or fewer, explain why converting first is a better habit, giving both reasons.
Hint: There are two separate problems with the arcsecant route.
Answer:
A usable answer: two things go wrong. First, arcsecant has no agreed range, so anyone reading your answer cannot tell which angle you meant unless you also state your convention — and most people will not read the small print. Second, applying an inverse function returns one angle, so writing the answer as an arcsecant plus a family quietly loses the second family of solutions.
Converting to a cosine equation fixes both at once: the arccosine means the same thing to everybody, and the plus-or-minus structure of the answer makes the second family visible. It costs one line and removes two problems.
Comparison
Fill the blanks from memory. The method never changed; its reach did.
Comparison matrix
| Lesson 10.2b | This lesson | |
|---|---|---|
| Values handled | special angles only | any value in the range |
| Reference angle | read off the table | named by an inverse function |
| Number of families | two, or one for tangent | two, or one for tangent |
| Exactness of the answer | exact | equally exact |
| Structure of the answer | families with 2 pi k | families with 2 pi k |
Only the first two rows differ, and the second is purely notational. The inverse functions widened the method's applicability without changing the method.
Pattern
Whether the value is special or not, and whichever function appears, the same five moves cover it.
For a calculator answer, check that the value returned lies in the range of the function you intended. A calculator applies a fixed branch and cannot know which one you wanted.
OpenStax Algebra and Trigonometry 2e, §8.3 Inverse Trigonometric Functions §8.3
Check
Domain of a transformed inverse. The argument decides it.
Check your understanding
What is the domain of f of x equals arcsin of the quantity x over 3?
Answer: B
Why: The argument x over three must lie in the arcsine's domain, the interval from negative one to one. Solving that inequality for x multiplies through by 3, giving the interval from negative three to three.
Check
An applied ratio. Match the function to the sides.
Check your understanding
A guy wire 30 m long is anchored 12 m from the base of a mast. What angle does it make with the ground?
Answer: B
Why: The 30 metres is the wire itself, the hypotenuse, and the 12 metres is the horizontal distance, adjacent to the ground angle. Adjacent over hypotenuse is a cosine, so the angle is the arccosine of 12 over 30, about 66.4 degrees.
Check
Solving with an unfriendly value. Give the exact form.
Check your understanding
Which describes all solutions of tangent of t equals negative 4?
Answer: B
Why: A tangent equation has solutions in two opposite quadrants, half a revolution apart, so the two families combine into one stepping by pi. The arctangent supplies the reference solution directly, since it accepts negative inputs.
Real world
A solar panel installer needs to set the tilt angle so that panels face the sun at local noon on the equinox. At latitude L the optimal tilt from horizontal equals the latitude. A client's roof already has a 5 over 12 pitch, and the site is at latitude 41.6 degrees north.
Discussion prompt
Find the roof's angle, determine the extra tilt needed from mounting brackets, and say why the answer must be computed as an angle rather than as a ratio.
Hint: Convert the pitch to an angle first, then subtract.
Answer:
\[ \theta_{\text{roof}} = \arctan\left(\frac{5}{12}\right) \approx 22.62^\circ \]
\[ \text{extra tilt} = 41.6^\circ - 22.62^\circ \approx 18.98^\circ \]
The brackets must add about 19 degrees of tilt. The answer has to be computed as an angle because angles add and ratios do not: the pitch ratios 5 over 12 and the equivalent of 19 degrees cannot simply be summed, since the tangent is not linear.
Checking this concretely: a 5 over 12 pitch is 22.62 degrees and a 12 over 12 pitch is 45 degrees, so doubling the ratio does not double the angle. Anyone who tried to work in ratios throughout would get a materially wrong bracket specification, and the error grows with the angle. Converting to angles is what makes the quantities additive, and that is a large part of why inverse trigonometric functions matter in practice.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is arcsin(1/3) an exact answer or an approximation?
Correct: An exact answer, like root seven.
\[ \sin\left(\arcsin\left(\tfrac{1}{3}\right)\right) = \tfrac{1}{3} \quad \text{exactly} \]
Why: Arcsine of one third denotes a specific real number, precisely determined: the unique angle between negative pi over two and pi over two whose sine is one third. It has no shorter name, exactly as root seven has none, but that is a fact about notation rather than about precision. Writing 0.3398 instead would be the approximation. The third option confuses the function with the value it returns at a particular input, which is the same confusion as calling root seven a function.
Explain it
They have solved sine of theta equals one half happily and are stuck on sine of theta equals one third, saying it has no answer.
Discussion prompt
In no more than five sentences, unstick them. Use an analogy they will already accept.
Hint: Ask them to solve x squared equals seven.
Answer:
A usable answer: ask them what x squared equals seven comes to. They will say root seven without hesitating, and they will not feel they have failed to answer — root seven is the answer, it just has no tidier name.
Sine of theta equals one third is the same situation. The angle exists, it is just not one of the handful that have names like pi over six, so we write it as arcsine of one third and that is the exact answer. The rest of the working is identical to what they already did: same two quadrants, same two families, same plus two pi k.
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: Calculator work is fixed by converting to a function that exists and then checking the answer lies in the intended range. Domain and range are fixed by remembering that the argument controls the domain and everything outside controls the range. Applied choices are fixed by identifying which two sides were measured before naming any function. Complete solutions are fixed by writing both families and keeping the exact form. 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 write the two analogies side by side: x squared equals four against x squared equals seven, and sine theta equals one half against sine theta equals one third, and write one sentence under them saying what the square root and the arcsine each do. Below, draw a unit circle showing both solutions of sine theta equals one third, marking the reference angle as arcsine of one third and the quadrant two solution as pi minus it, and write both families beside the picture. To the right, draw the roof-pitch triangle and work out the angle for a 4 over 12 pitch. Underneath that, write the rule for domain and range of a transformed inverse — argument controls domain, outside controls range — and apply it to two arccos of the quantity x over four, minus one. In the bottom left, solve secant of x equals negative 3 completely, converting first and showing why. In the bottom right, list the three inverses a calculator lacks and how to get each. Finally, circle the one place on the page where a convention has to be stated, and write what to do instead.
The circled item should be the arcsecant, and what to do instead is convert the equation to a cosine one before solving — which is exactly what the bottom-left worked example does.
Recap
Five things, and the fourth is what the whole section was building toward.
| If the question says | Your first move is |
|---|---|
| Approximate an arccotangent of a negative | Reference angle from the positive, then pi minus it |
| Find the domain | Put the argument inside the inverse's own domain |
| Find the range | Track key points through the outside operations |
| An applied angle is wanted | Identify which two sides were measured |
| Solve a secant equation | Reciprocate it into a cosine equation first |
That completes Section 10.6. Equations with any value at all can now be solved, but only the simplest ones — a single function equal to a constant. The last section of the chapter takes on the general case: equations with several terms, several functions, and inequalities as well.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.6 The Inverse Trigonometric Functions §10.6, pp. 833-841 — everything on these slides traces back here
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