Why the trigonometric functions need restricted domains before they can be inverted, the ranges of inverse sine, cosine and tangent, evaluating inverse expressions exactly, solving a trigonometric equation in a stated quadrant, and finding angles in applied right triangles.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 13 — Trigonometric Ratios and Functions
Evaluate Inverse Trigonometric Functions
Objectives
Five outcomes. The question turned around.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 875-879 — the lesson these objectives are drawn from
Warm-up
So far every question gave an angle and asked for a ratio.
Discussion prompt
Turn it round. Which angle has a sine of 0.5? Try to name every one.
Hint: Where does the horizontal line y = 0.5 cross the unit circle?
Answer:
\[ \tfrac{\pi}{6}, \; \tfrac{5\pi}{6}, \; \tfrac{13\pi}{6}, \; \tfrac{17\pi}{6}, \; -\tfrac{7\pi}{6}, \; \dots \]
There are infinitely many, two in every full turn. So which angle has a sine of 0.5 is not a question with an answer.
To make it one, the sine function's domain is cut down to a stretch where no value repeats. Only then does an inverse exist, and the choice of stretch is what the inverse sine's range records.
Concept
The trigonometric functions repeat, so they fail the one-to-one test and have no inverses on their full domains. Restricting each to a stretch that hits every value exactly once creates the inverse sine, inverse cosine and inverse tangent, each of which returns exactly one angle.
inverse sine — For a between negative 1 and 1, the inverse sine of a is the unique angle theta with sine equal to a and theta between negative pi over 2 and pi over 2, that is, between negative 90 and 90 degrees.
\[ \theta = \sin^{-1}a \;\Longleftrightarrow\; \sin\theta = a \text{ and } -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2} \]
The restriction is a decision, not a discovery. Any stretch hitting every value once would work, and the standard ones were chosen for convenience.
Figure (svg): Two columns comparing the direct and inverse questions
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 874-875
Section
Section 1
Concept
A function has an inverse only if each output comes from exactly one input. Sine returns to every value twice per turn and forever afterwards, so on its full domain it has no inverse at all.
\[ \sin\tfrac{\pi}{6} = \sin\tfrac{5\pi}{6} = \sin\tfrac{13\pi}{6} = \tfrac{1}{2} \]
Cosine has the same problem, and so does tangent, which repeats every half turn rather than every full one.
Figure (svg): Many different angles all having the same sine
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 874-875 — Investigating Inverse Trigonometric Functions
Picture it
The horizontal line at height one half.
Figure (svg): Many different angles all having the same sine
The line crosses the circle twice, and every extra revolution adds two more crossings. No single angle can be called the answer.
Worked example
The activity on page 874.
\[ \text{Explain why } f(\theta) = \sin\theta \text{ has no inverse on } -\pi \le \theta \le \pi. \]
Test two inputs
Why: Pi over 6 and 5 pi over 6.
\[ \text{both give } \frac{1}{2} \]
State the failure
Why: One output, two inputs.
Say what that costs
Why: An inverse would have to choose.
Check the graph
Why: A horizontal line meets it twice.
Figure (svg): Many different angles all having the same sine
\[ \sin\tfrac{\pi}{6} = \sin\tfrac{5\pi}{6} \]
Verify: check cosine on the same domain
Why: Cosine of pi over 3 and cosine of negative pi over 3 are both one half, so cosine fails on that domain too — for a different reason. Sine repeats by reflecting across the line theta equals pi over 2; cosine repeats by reflecting across the vertical axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 874-874
Fill the middle
The list on page 875.
Fill in the blanks
\sin\tfrac5___ = \tfrac______ \text___ \sin\tfrac___\pi}___ = \tfrac______
Why: Five pi over 6, which is 150 degrees. It has reference angle pi over 6 and lies in quadrant two, where sine is still positive.
Worked example
The Draw Conclusions questions on page 874.
\[ \text{Give a restricted domain on which sine has an inverse, and one for cosine.} \]
Look for a stretch with no repeats
Why: Sine rises steadily from -1 to 1.
\[ -\frac{\pi}{2}\text{ to } \frac{\pi}{2} \]
Confirm it covers every value
Why: The endpoints give -1 and 1.
Repeat for cosine
Why: Cosine falls steadily from 1 to -1.
\[ 0\text{ to } \pi \]
Note these are not the only choices
Why: Any stretch with no repeats works.
\[ \frac{\pi}{2}\text{ to } 3 \pi / 2\text{ also works} \]
Figure (svg): The solution to Worked example choosing a restricted domain shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}; \qquad 0\le\theta\le\pi \]
Verify: check each stretch covers all values once
Why: Across negative pi over 2 to pi over 2 the sine climbs from negative 1 to 1 without ever turning back, so every value between appears exactly once. Cosine does the same in reverse across 0 to pi. Those two properties, covering everything and repeating nothing, are precisely what an inverse needs.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 874-874
Trap
\[ \sin\theta = \tfrac{1}{2} \]
Apply the inverse sine and stop
Why: The inverse is treated as if it undid the sine completely.
\[ \theta = \sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6} \text{, the only solution} \]
Five pi over 6 also has sine one half, and so do infinitely many others. The inverse returns ONE of them by design.
\[ \theta = \tfrac{\pi}{6} \text{ from the inverse} \]
Treat the inverse's output as a starting point
Why: Then use reference angles to find the others.
\[ \theta = \tfrac{\pi}{6}+2\pi k \; \text{ or } \; \tfrac{5\pi}{6}+2\pi k \]
The restriction is what makes the inverse a function at all; it does not make the other solutions disappear.
Sorting
Ask whether any value repeats.
Sort into buckets
Sort each function and domain.
The three domains in the left bucket are exactly the standard restrictions, which is no accident: they were chosen as the simplest stretches with this property.
Matching
Every value is reached more than once.
Match the pairs
Why: The last is the exception that proves the rule: sine reaches its maximum of 1 at only one angle per turn, because that is where it turns around. Every value strictly between the extremes is reached twice.
Prediction
Commit before reasoning.
Predict first
Sine is inverted on negative pi over 2 to pi over 2. Could a different domain have been chosen?
Correct: Yes — pi over 2 to 3 pi over 2 also hits every value exactly once and would work equally well.
\[ \sin\!\left[\tfrac{\pi}{2},\tfrac{3\pi}{2}\right] \text{ also covers } [-1,1] \text{ once} \]
Why: Across that stretch sine falls steadily from 1 to negative 1, covering everything once, so an inverse could have been built on it. The standard choice is a convention favoured because it is centred at zero and keeps the inverse of a positive value positive. Knowing the restriction is a choice rather than a law explains why calculators must agree on one, and why answers outside the chosen range are wrong even when their sine is right.
Section
Section 2
Concept
Inverse sine returns an angle between negative pi over 2 and pi over 2; inverse cosine returns one between 0 and pi; inverse tangent returns one strictly between negative pi over 2 and pi over 2.
\[ \sin^{-1}: \!\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]; \; \cos^{-1}: [0,\pi]; \; \tan^{-1}: \!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) \]
Sine and cosine accept only inputs from negative 1 to 1, since no angle has a sine or cosine outside that. Tangent accepts any real number.
Figure (svg): The output range of each of the three inverse functions
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 875-875 — Inverse Trigonometric Functions
Picture it
Where each inverse function's answers live.
Figure (svg): The output range of each of the three inverse functions
Cosine takes the upper half of the circle and the other two take the right half, with tangent's endpoints excluded because tangent is undefined there.
Worked example
The boxed definitions on page 875.
\[ \text{State the domain and range of each inverse trigonometric function.} \]
Inverse sine
Why: Inputs from -1 to 1.
\[ \text{outputs } -\frac{\pi}{2}\text{ to } \frac{\pi}{2} \]
Inverse cosine
Why: Inputs from -1 to 1.
\[ \text{outputs } 0\text{ to } \pi \]
Inverse tangent
Why: Any real input.
\[ \text{outputs strictly between } -\frac{\pi}{2}\text{ and } \frac{\pi}{2} \]
Note the excluded endpoints
Why: Tangent is undefined at those angles.
Figure (svg): The output range of each of the three inverse functions
\[ \!\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], \; [0,\pi], \; \!\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right) \]
Verify: explain why tangent's input is unrestricted
Why: Tangent is y over x and grows without bound as x approaches zero, so it really does take every real value. Sine and cosine have r underneath and r is the largest of the three quantities, which is precisely why they are trapped between negative 1 and 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 875-875
Matching
Which half-circle each one uses.
Match the pairs
Why: Sine and tangent share the same interval but tangent excludes the endpoints, because tangent is undefined at plus and minus pi over 2 and so never reaches those angles.
Worked example
The boxed definitions on page 875.
\[ \text{Why does inverse cosine return angles in } [0,\pi] \text{ rather than } \!\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]? \]
Test cosine on the sine range
Why: Cosine of pi over 3 and of negative pi over 3.
\[ \text{both equal } \frac{1}{2} \]
Note the repeat
Why: Cosine is symmetric about zero.
Try the upper half instead
Why: Cosine falls from 1 at 0 to -1 at pi.
Conclude
Why: That stretch works.
\[ \text{range is } 0\text{ to } \pi \]
Figure (svg): The solution to Worked example why cosine gets a different range shown as a ladder of expressions, one row per algebraic move
\[ \cos\tfrac{\pi}{3} = \cos\!\left(-\tfrac{\pi}{3}\right) = \tfrac{1}{2} \]
Verify: check the consequence for negative inputs
Why: Since the range is 0 to pi, the inverse cosine of a negative number lands in quadrant two — for instance the inverse cosine of negative one half is 2 pi over 3, or 120 degrees. The inverse SINE of a negative number lands in quadrant four instead. The two behave differently on negative inputs for exactly this reason.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 875-875
Error analysis
A student evaluates the inverse sine of one half.
Annotate
On: \( \sin^{-1}\tfrac{1}{2} = \tfrac{5\pi}{6} \)
Having the right sine is not enough. An inverse value must also lie in the range, and a value failing that test is simply wrong.
Sorting
Positive and negative inputs behave differently.
Sort into buckets
Sort each inverse value by the quadrant of its answer.
This is the practical difference between the two ranges: a negative input sends inverse cosine into quadrant two but sends inverse sine into quadrant four.
Comparison
Fill the blanks. Same input, different answers.
Comparison matrix
| Question | Inverse sine | Inverse cosine |
|---|---|---|
| Range | -pi/2 to pi/2 | 0 to pi |
| Answer for input 1/2 | pi/6 | pi/3 |
| Answer for input -1/2 | -pi/6 | 2 pi/3 |
| Quadrant for a negative input | IV | II |
For positive inputs both give acute angles, though different ones. For negative inputs they part company completely, and mixing them up produces answers that are off by a full quadrant.
Prediction
Commit before reasoning.
Predict first
Inverse sine rejects an input of 2, but inverse tangent accepts it. Why?
Correct: Because sine is trapped between -1 and 1 while tangent takes every real value.
\[ |y| \le r \;\Longrightarrow\; |\sin\theta| \le 1 \]
Why: Sine is y over r, and since r is the distance from the origin it is at least as large as the size of y, forcing the ratio into the band from negative 1 to 1. Tangent is y over x, where x can be as small as you like, so the ratio runs off to infinity. Asking for an angle with sine 2 is asking for a leg longer than the hypotenuse; asking for one with tangent 2 is perfectly reasonable and gives about 63.4 degrees.
Section
Section 3
Concept
To evaluate an inverse expression, ask which angle inside that function's range produces the given value. If the value lies outside the function's domain, the expression is undefined.
\[ \cos^{-1}\frac{\sqrt3}{2} = \tfrac{\pi}{6} = 30^\circ \]
The special-angle table of Lesson 13.1 supplies most of these directly, read from right to left.
Figure (svg): Three inverse evaluations, one of them undefined
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876 — Evaluate inverse trigonometric functions
Picture it
Example 1, all three parts.
Figure (svg): Three inverse evaluations, one of them undefined
The middle one is undefined because no angle has a sine of 2. Checking the input against the domain is the first thing to do.
Worked example
Example 1, all three parts.
\[ \text{Evaluate } \cos^{-1}\tfrac{\sqrt3}{2}, \; \sin^{-1}2, \; \tan^{-1}(-\sqrt3) \text{ in radians and degrees.} \]
First: check the input and range
Why: Root 3 over 2 is in range; answers lie in 0 to pi.
First: recall the special value
Why: Cosine of 30 degrees.
\[ \frac{\pi}{6},\text{ or } 30 ^\circ \]
Second: check the input
Why: Sine never exceeds 1.
Third: use tangent's range
Why: Between -90 and 90 with tangent -root 3.
\[ -\frac{\pi}{3},\text{ or } -60 ^\circ \]
Figure (svg): Three inverse evaluations, one of them undefined
\[ \tfrac{\pi}{6}; \quad \text{undefined}; \quad -\tfrac{\pi}{3} \]
Verify: check the third lands in the right range
Why: Negative pi over 3 is negative 60 degrees, safely inside tangent's open range. The angle 2 pi over 3 also has tangent negative root 3, but it lies outside the range and so is not the inverse tangent's value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876
Fill the middle
Example 1a.
Fill in the blanks
\cos\tfrac6___ = \tfrac______ \;\Longrightarrow\; \cos^___\tfrac______ = \tfrac______}
Why: Pi over 6, which is 30 degrees. The special-angle table read from right to left answers most inverse questions instantly.
Worked example
Guided Practice 1 to 4.
\[ \text{Evaluate } \sin^{-1}\tfrac{\sqrt2}{2}, \; \cos^{-1}\tfrac{1}{2}, \; \tan^{-1}(-1), \; \sin^{-1}\!\left(-\tfrac{1}{2}\right). \]
First: which angle has sine root 2 over 2
Why: In the range -90 to 90.
\[ \frac{\pi}{4},\text{ or } 45 ^\circ \]
Second: which angle has cosine one half
Why: In the range 0 to 180.
\[ \frac{\pi}{3},\text{ or } 60 ^\circ \]
Third: which angle has tangent -1
Why: In the range -90 to 90.
\[ -\frac{\pi}{4},\text{ or } -45 ^\circ \]
Fourth: which angle has sine -1/2
Why: In the range -90 to 90.
\[ -\frac{\pi}{6},\text{ or } -30 ^\circ \]
Figure (svg): The solution to Worked example four more shown as a ladder of expressions, one row per algebraic move
\[ 45^\circ; \; 60^\circ; \; -45^\circ; \; -30^\circ \]
Verify: compare the second with an inverse sine
Why: The inverse cosine of one half is 60 degrees, but the inverse SINE of one half is 30. They differ because the two functions ask about different sides of the triangle, and the two angles are complementary — which is true of every such pair.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876
Trap
\[ \sin^{-1}2 \]
Enter it and report whatever appears
Why: The expression looks like the others.
\[ \approx 1.57 \text{ or an error} \quad \text{(neither is an answer)} \]
No angle has a sine of 2, because sine never leaves the band from negative 1 to 1. The expression is undefined.
\[ 2 > 1 \;\Longrightarrow\; 2 \notin [-1,1] \]
Check the input against the domain first
Why: Before any evaluation.
\[ \sin^{-1}2 \text{ is undefined} \]
For inverse tangent no such check is needed, since every real number is an acceptable input there.
Fill the middle
Example 1c.
Fill in the blanks
\tan^60(-\sqrt3) = -___^\circ
Why: Negative 60 degrees. Tangent of 60 is root 3, and the negative sign sends the answer into quadrant four, which tangent's range allows.
Matching
Which angle in the range gives this?
Match the pairs
Why: Two of the four are negative, and both come from inverse sine or inverse tangent. An inverse cosine can never be negative, since its range starts at zero.
Prediction
Commit before reasoning.
Predict first
Is there any input for which the inverse cosine returns a negative angle?
Correct: No — the range is 0 to pi, so every value it returns is at least zero.
\[ \cos^{-1}\!\left(-\tfrac{1}{2}\right) = 120^\circ, \text{ not } -60^\circ \]
Why: A negative input pushes the answer past pi over 2 into quadrant two rather than below zero: the inverse cosine of negative one half is 120 degrees, not negative 60. This is the sharpest practical difference between the two inverses, and it is worth checking against every answer you produce — a negative inverse cosine is always a mistake.
Section
Section 4
Concept
A calculator returns only the angle inside the inverse function's range. When a problem asks for an angle in a different quadrant, take that value's size as the reference angle and rebuild the answer in the quadrant wanted.
\[ \sin^{-1}\!\left(-\tfrac{5}{8}\right) \approx -38.7^\circ \;\Longrightarrow\; \theta \approx 180^\circ+38.7^\circ \]
Checking the answer by taking its sine, cosine or tangent confirms both the size and the sign in one step.
Figure (svg): A calculator's inverse value moved into the required quadrant
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876 — Solve a trigonometric equation
Picture it
Example 2: sine of theta is negative five eighths, with theta between 180 and 270 degrees.
Figure (svg): A calculator's inverse value moved into the required quadrant
The calculator's negative 38.7 degrees has the right sine but the wrong quadrant. Its size, 38.7, is the reference angle for the answer that is wanted.
Worked example
Example 2.
\[ \text{Solve } \sin\theta = -\tfrac{5}{8} \text{ for } 180^\circ < \theta < 270^\circ. \]
Use the inverse sine
Why: What the calculator returns.
\[ \text{about } -38.7 ^\circ \]
Note the quadrant
Why: Negative 38.7 lies in quadrant four.
Take the reference angle
Why: The size of that answer.
\[ 38.7 ^\circ \]
Rebuild in quadrant three
Why: One eighty plus 38.7.
\[ \text{about } 218.7 ^\circ \]
Figure (svg): A calculator's inverse value moved into the required quadrant
\[ \theta \approx 218.7^\circ \]
Verify: take the sine of the answer
Why: Sine of 218.7 degrees is about negative 0.625, which is exactly negative five eighths. Checking the answer forward is the only way to be sure both the reference angle and the quadrant were handled right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876
Fill the middle
Example 2, Step 2.
Fill in the blanks
\theta' = 38.7^\circ, \; \text218.7 \;\Longrightarrow\; \theta = 180^\circ+38.7^\circ = ___^\circ
Why: Two hundred eighteen point seven degrees. Quadrant three angles are 180 plus their reference angle.
Worked example
Guided Practice 5, 6 and 9.
\[ \text{Solve } \cos\theta = 0.4 \text{ on } (270^\circ,360^\circ); \; \tan\theta = 2.1 \text{ on } (180^\circ,270^\circ); \; \sin\theta = 0.62 \text{ on } (90^\circ,180^\circ). \]
First: the inverse cosine
Why: About 66.4 degrees, in quadrant one.
\[ \text{reference angle } 66.4 \]
First: rebuild in quadrant four
Why: Three sixty minus 66.4.
\[ \text{about } 293.6 ^\circ \]
Second: inverse tangent and quadrant three
Why: About 64.5; add 180.
\[ \text{about } 244.5 ^\circ \]
Third: inverse sine and quadrant two
Why: About 38.3; subtract from 180.
\[ \text{about } 141.7 ^\circ \]
Figure (svg): The solution to Worked example three more equations shown as a ladder of expressions, one row per algebraic move
\[ 293.6^\circ; \; 244.5^\circ; \; 141.7^\circ \]
Verify: check each sign is consistent
Why: Cosine is positive in quadrant four and the input 0.4 was positive; tangent is positive in quadrant three and 2.1 was positive; sine is positive in quadrant two and 0.62 was positive. Every input's sign matches its quadrant, which is a necessary condition for the problem to have had a solution at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 876-876
Error analysis
A student solves sin theta = -5/8 for theta between 180 and 270 degrees.
Annotate
On: \( \theta = \sin^{-1}\!\left(-\tfrac{5}{8}\right) \approx -38.7^\circ \)
A calculator can only return the one angle in its range. Reading the required interval and rebuilding from the reference angle is the student's job, not the calculator's.
Fill the middle
Guided Practice 5.
Fill in the blanks
\cos^293.6(0.4) \approx 66.4^\circ, \; \text___ \;\Longrightarrow\; \theta = 360^\circ-66.4^\circ = ___^\circ
Why: Two hundred ninety-three point six degrees. Quadrant four angles are 360 minus their reference angle.
Matching
Given a reference angle theta prime.
Match the pairs
Why: These are exactly the reference-angle formulas of Lesson 13.3 solved for theta instead of for theta prime. Going forward you subtract to find the reference angle; coming back you add or subtract to rebuild the original.
Prediction
Commit before reasoning.
Predict first
Could the equation sin theta = 0.62 have a solution with theta between 180 and 270 degrees?
Correct: No — sine is negative throughout quadrant three, so it never equals a positive 0.62 there.
\[ \text{QIII: } y<0 \;\Longrightarrow\; \sin\theta<0 \]
Why: In quadrant three the y-coordinate is negative and r is positive, so the sine is negative at every angle. A problem asking for a positive sine there has no solution at all, and getting an answer anyway means a sign was mishandled somewhere. Checking that the given value's sign agrees with the required quadrant before starting saves the whole calculation.
Section
Section 5
Concept
In an applied right triangle, choose the ratio built from the two sides you were given, then apply the matching inverse function. Two legs call for tangent; a leg and the hypotenuse call for sine or cosine.
\[ \tan\theta = \tfrac{8}{20} \;\Longrightarrow\; \theta = \tan^{-1}0.4 \approx 21.8^\circ \]
Answers in applications are almost always wanted in degrees, so the calculator's mode matters as much as the arithmetic.
Figure (svg): A ramp's angle found from its height and horizontal length
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 877-877 — Write and solve a trigonometric equation
Picture it
Example 4: 8 feet high and 20 feet long horizontally.
Figure (svg): A ramp's angle found from its height and horizontal length
Both legs are given, so tangent is the ratio and the inverse tangent of 0.4 gives about 22 degrees.
Worked example
Example 4.
\[ \text{A ramp is } 8 \text{ feet high and } 20 \text{ feet long horizontally. Find its angle.} \]
Draw and label
Why: Height opposite, horizontal length adjacent.
Choose the ratio
Why: Opposite over adjacent is tangent.
\[ \tan \theta = \frac{8}{20} \]
Simplify
Why: Eight over 20.
\[ 0.4 \]
Apply the inverse
Why: Inverse tangent of 0.4.
\[ \text{about } 21.8 ^\circ \]
Figure (svg): A ramp's angle found from its height and horizontal length
\[ \theta = \tan^{-1}0.4 \approx 21.8^\circ \]
Verify: sanity-check against a known angle
Why: A ratio of 1 would give 45 degrees, and 0.4 is well under 1, so the angle must be well under 45 — as 21.8 is. Comparing the ratio with 1 places any inverse tangent answer immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 877-877
Fill the middle
Example 4.
Fill in the blanks
\tan\theta = \tfrac21.8___ = 0.4 \;\Longrightarrow\; \theta \approx ___^\circ
Why: About 21.8 degrees, usually rounded to 22. A tangent under 1 always gives an angle under 45 degrees.
Worked example
Example 3, and Guided Practice 14.
\[ \text{Find } \theta \text{ when the adjacent leg is } 5 \text{ and the hypotenuse } 14. \text{ Then find the angle of a } 26 \text{ foot ramp } 10 \text{ feet high.} \]
First: choose the ratio
Why: Adjacent over hypotenuse is cosine.
\[ \cos \theta = \frac{5}{14} \]
First: apply the inverse
Why: Inverse cosine of about 0.357.
\[ \text{about } 69.1 ^\circ \]
Second: choose the ratio
Why: The 26 feet is the ramp's LENGTH, so it is the hypotenuse.
\[ \sin \theta = \frac{10}{26} \]
Second: apply the inverse
Why: Inverse sine of about 0.385.
\[ \text{about } 22.6 ^\circ \]
Figure (svg): The solution to Worked example a leg and the hypotenuse shown as a ladder of expressions, one row per algebraic move
\[ \cos^{-1}\tfrac{5}{14} \approx 69.1^\circ; \qquad \sin^{-1}\tfrac{10}{26} \approx 22.6^\circ \]
Verify: check which side the 26 feet is
Why: Driving 26 feet ON the ramp means travelling along its sloped surface, which is the hypotenuse, not the horizontal length. Had 26 been the horizontal run the answer would have been the inverse tangent of 10 over 26, about 21.0 degrees — close enough that only careful reading tells them apart.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 877-877
Trap
\[ \theta = \tan^{-1}0.4 \]
Read whatever the display shows
Why: The mode is never checked.
\[ \theta \approx 0.38 \quad \text{(radians, not degrees)} \]
A ramp of 0.38 degrees would be almost flat. The answer is right but is in radians, and the question wanted degrees.
\[ \text{set degree mode, then } \tan^{-1}0.4 \]
Confirm the mode before evaluating
Why: Check that the inverse tangent of 1 returns 45.
\[ \theta \approx 21.8^\circ \]
Zero point 38 radians is 21.8 degrees, so the two agree — but only one of them answers the question asked.
Sorting
Look at which two sides you have.
Sort into buckets
Sort each situation by the inverse function it needs.
The last two items are the same ramp described differently: ramp LENGTH is the hypotenuse while horizontal length is a leg, and that one word changes which function you need.
Ranking
Smallest first.
Put in order
Why: The values are about 21.8, 22.6, 45, 69.1 and 90 degrees. The two ramp angles are close because their ratios are close, and the last is exactly 90 since the cosine of a right angle is zero.
Prediction
Commit before reasoning.
Predict first
The inverse tangent of 0.4 returns 0.38 in one mode and 21.8 in another. Is one of them wrong?
Correct: No — both are the same angle, one in radians and one in degrees, and 0.38 radians is 21.8 degrees.
\[ 0.38\!\left(\frac{180}{\pi}\right) \approx 21.8 \]
Why: Multiplying 0.38 by 180 over pi gives about 21.8, so the two displays agree completely. Neither is wrong mathematically; what is wrong is reporting one when the question asked for the other. The danger is that 0.38 looks like a plausible number rather than an obvious error, which is why checking that the inverse tangent of 1 returns 45 before starting is worth the two keystrokes.
Comparison
Fill the blanks. Same idea, three ranges.
Comparison matrix
| Question | Inverse sine | Inverse cosine | Inverse tangent |
|---|---|---|---|
| Accepts | -1 to 1 | -1 to 1 | any real number |
| Returns | -pi/2 to pi/2 | 0 to pi | -pi/2 to pi/2, open |
| Negative input gives | quadrant IV | quadrant II | quadrant IV |
| Can return 0? | yes | yes | yes |
Sine and tangent share a range and cosine takes a different one, which is why a negative input sends inverse cosine somewhere the other two never go.
Pattern
Range first, quadrant last.
A calculator returns only the one angle in the range. Every other solution has to be built from it.
OpenStax Algebra and Trigonometry 2e, §8.3 Inverse Trigonometric Functions §8.3
Check
The answer must lie in the range.
Check your understanding
What is the inverse sine of 1/2?
Answer: A
Why: 30 degrees has sine 1/2 and lies in the range from -90 to 90 degrees.
Check
Rebuild in the quadrant asked for.
Check your understanding
Solve sin theta = -5/8 for theta between 180 and 270 degrees.
Answer: A
Why: The inverse sine gives -38.7, whose size is the reference angle; in quadrant three, theta = 180 + 38.7.
Check
Two legs means tangent.
Check your understanding
A ramp is 8 feet high and 20 feet long horizontally. What is its angle?
Answer: A
Why: tan theta = 8/20 = 0.4, so theta is the inverse tangent of 0.4.
Real world
A wheelchair ramp must rise 30 inches. Building codes require the angle of the ramp to be at most 4.8 degrees.
Discussion prompt
What is the shortest horizontal run the ramp can have, and what angle does a 300 inch run give?
Hint: Work the direct problem for the first part and the inverse problem for the second.
Answer:
\[ \tan 4.8^\circ = \frac{30}{\text{run}} \;\Longrightarrow\; \text{run} = \frac{30}{\tan 4.8^\circ} \approx 357 \text{ inches} \]
\[ \text{with a } 300 \text{ inch run: } \theta = \tan^{-1}\frac{30}{300} = \tan^{-1}0.1 \approx 5.7^\circ \]
The ramp needs at least about 357 inches, roughly 30 feet, and a 300 inch run gives about 5.7 degrees, which fails the code.
This is the direct and the inverse question asked about the same triangle, and each needs its own tool. Given the angle, tangent finds the run; given the run, inverse tangent finds the angle. Notice too how unforgiving the geometry is: cutting the run by only 16 percent pushed the angle from a legal 4.8 degrees to an illegal 5.7, which is why accessible ramps take up so much more space than people expect.
Commit first
Answer, then rate your confidence honestly.
Predict first
The sine of 150 degrees is 1/2. Is the inverse sine of 1/2 therefore 150 degrees?
Correct: No — the inverse sine must return an angle from -90 to 90, so the answer is 30 degrees.
\[ \sin 150^\circ = \tfrac{1}{2} \text{ but } 150^\circ \notin \!\left[-90^\circ,90^\circ\right] \]
Why: Having the right sine is necessary but not sufficient: an inverse value must also lie in the range, and 150 degrees does not. This is the price of forcing a repeating function to have an inverse — the inverse can return only one of the infinitely many angles, and which one is fixed by the range. Both 30 and 150 solve the EQUATION sine theta equals one half; only 30 is the value of the inverse sine.
Explain it
They know that squaring and square-rooting undo each other.
Discussion prompt
In four sentences or fewer, explain why the inverse sine has to pick just one angle.
Hint: Compare it with the square root of 9.
Answer:
Both 3 and negative 3 square to 9, but the square root symbol gives only 3, because a function must have one output. The same problem is much worse for sine: infinitely many angles share each sine value.
So the inverse sine picks the one between negative 90 and 90 degrees and always returns that. The other angles still exist; the inverse just is not the tool for finding them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For ranges, remember cosine takes the upper half of the circle and the other two take the right half. For domains, only inverse tangent accepts anything outside -1 to 1. For rebuilding, take the size of the calculator's answer as the reference angle and use 180 minus, 180 plus or 360 minus. For applications, name the two sides you have and pick the ratio containing both.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an inverse trigonometry page. Top left: pick one value and list four angles whose sine equals it, then write one sentence on why that makes an inverse impossible without a restriction. Top right: draw three small circles and shade each inverse function's range on it, marking the domains beside them and noting which endpoints are excluded and why. Middle: evaluate six inverse expressions exactly, including one that is undefined, giving each answer in both radians and degrees. Bottom left: solve three trigonometric equations in stated quadrants, showing the calculator's answer, the reference angle and the rebuild as three separate lines. Bottom right: draw an applied right triangle of your own invention, name the two sides you are given, and solve for the angle in degrees.
If any inverse cosine on your page came out negative, it is wrong: that function's range starts at zero and never goes below it.
Recap
Five things, and the question finally running both ways.
| If you see | Then |
|---|---|
| An inverse sine or cosine | Check the input lies between -1 and 1 |
| An inverse tangent | Any real input is acceptable |
| A negative input to inverse cosine | The answer lands in quadrant two |
| A negative input to inverse sine | The answer lands in quadrant four |
| A quadrant specified in the question | Rebuild from the reference angle |
| An applied triangle | Two legs means tangent; a leg and the hypotenuse means sine or cosine |
Lesson 13.5 leaves right triangles behind altogether: the law of sines solves triangles with no right angle at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.4 Evaluate Inverse Trigonometric Functions §13.4, pp. 875-879 — everything on these slides traces back here
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