Yesterday's functions answered "the angle is θ — what's the height?" Today inverts the question: "the height is 0.5 — what was the angle?"
There's a catch. Infinitely many angles have sine 0.5 (π/6, 5π/6, 13π/6, …). A function may give only ONE answer. So mathematicians made a treaty: arcsine always answers from [−π/2, π/2] — the right half of the unit circle. Arccosine answers from [0, π] — the top half. Learn the treaty; the exam tests the treaty.
tan θ = sin θ/cos θ = the slope of the terminal ray (Lesson 17). The graph:
θ = π/2 + kπ. Near them the slope blows up: lim θ→(π/2)⁻ tan θ = +∞, lim θ→(π/2)⁺ tan θ = −∞.Transformations follow the usual template, with one twist: y = tan(bx) has period π/|b| (π, not 2π, on top).
[GRAPH: y = tan θ over (−3π/2, 3π/2): three identical increasing branches separated by dashed vertical asymptotes at θ = ±π/2 and ±3π/2; zeros at −π, 0, π marked; each branch sweeps from −∞ up to +∞. Inflection points at the zeros noted.]
Sine, cosine, tangent all fail the horizontal line test — so each is restricted to one well-chosen piece before inverting (Lesson 14's playbook):
| function | restricted domain | inverse | domain of inverse | range of inverse |
|---|---|---|---|---|
| sin θ | [−π/2, π/2] | arcsin x (sin⁻¹) | [−1, 1] | [−π/2, π/2] (right half-circle: QIV & QI) |
| cos θ | [0, π] | arccos x (cos⁻¹) | [−1, 1] | [0, π] (top half: QI & QII) |
| tan θ | (−π/2, π/2) | arctan x (tan⁻¹) | all reals | (−π/2, π/2) (open — asymptote angles excluded) |
Evaluation protocol: "arcsin(1/2) = the angle in [−π/2, π/2] whose sine is 1/2" → π/6. For negative inputs:
Negative inputs to arcsin/arctan give negative angles; negative inputs to arccos give obtuse angles. That asymmetry is heavily tested.
sin(arcsin x) = x always (for x in [−1, 1]) — inverse-inside compositions are safe.arcsin(sin θ) = θ only when θ is already in [−π/2, π/2]. Otherwise the arcsin returns the treaty-approved angle with the same sine:arcsin(sin(5π/6)) = arcsin(1/2) = π/6 (NOT 5π/6)
Protocol: evaluate the inside numerically, then apply the outside function honestly.
Problem: Evaluate tan(3π/4), and describe the behavior of tan θ as θ → (π/2)⁻.
Solution: 3π/4 is QII, reference π/4: tan = sin/cos = (√2/2)/(−√2/2) = −1. As θ → (π/2)⁻ the terminal ray approaches vertical and its slope grows without bound: lim θ→(π/2)⁻ tan θ = +∞.
Interpretation: Slopes: at 3π/4 the ray is the line y = −x. Vertical rays have no slope — hence the asymptote.
Problem: Evaluate arcsin(−1/2), arccos(−1/2), arctan(−1).
Solution: - arcsin(−1/2) = −π/6 (range [−π/2, π/2] → negative angle) - arccos(−1/2) = π − π/3 = 2π/3 (range [0, π] → obtuse) - arctan(−1) = −π/4
Interpretation: Same input sign, different treatment — arccos alone sends negatives to QII. If your three answers to a problem like this are all negative (or you wrote −π/3 for the arccos), the treaty was violated.
Problem: Evaluate: (a) arcsin(sin(5π/6)) (b) arccos(cos(−π/3)) (c) sin(arccos(3/5)).
Solution: (a) sin(5π/6) = 1/2 → arcsin(1/2) = π/6. (b) cos(−π/3) = 1/2 → arccos(1/2) = π/3 (not −π/3: arccos never outputs negatives). (c) Let θ = arccos(3/5): cos θ = 3/5 with θ in [0, π] — and since 3/5 > 0, θ is acute. Pythagorean triple: sin θ = 4/5 (positive, QI).
Interpretation: (c)'s pattern — "sine of an arccosine" — is a right-triangle question: adjacent 3, hypotenuse 5, opposite 4. Draw the triangle; confirm the sign from the inverse's range.
Problem: For g(x) = 3 tan(2x): period, asymptotes, and the range.
Solution: Period π/2 (π divided by b = 2). Asymptotes where 2x = π/2 + kπ → x = π/4 + kπ/2. Range: all real numbers — the 3 stretches vertically but tangent was already unbounded.
Interpretation: Two tangent-specific traps in one problem: the period formula's π numerator, and the fact that no vertical stretch changes an unbounded range (asking for tangent's "amplitude" is a category error).
1. (B). Opposite rays share a slope: period π. (A) is sin/cos's period.
2. (C). cos θ = 0 at π/2 + kπ. (A) lists the zeros instead.
3. (D). QII, reference π/4, tan negative in QII: −1. (A) drops the sign.
4. (A). The angle in [−π/2, π/2] with sine 1/2: π/6. (B) has the right sine but violates the range.
5. (B). π − arccos(√2/2) = π − π/4 = 3π/4. (A) is the arcsin-style error; arccos is never negative.
6. (C). tan(π/4) = 1 → π/4. (B) treats arctan(1) as the number 1.
7. (A). Treaty range: [−π/2, π/2], endpoints included (arcsin(±1) = ±π/2). (C) wrongly opens the interval — that's arctan's range.
8. (D). [0, π], top half of the circle. (A) belongs to arcsin.
9. (B). sin(5π/6) = 1/2; arcsin(1/2) = π/6. (A) forgets the range restriction.
10. (A). π/|b| = π/2. (C) uses the sinusoid formula 2π/b.
11. (C). Each tangent branch rises from −∞ to +∞: always increasing (with an inflection at 0, but never a turnaround).
12. (FRQ-style, 6 points) (i) [2 pts] arctan's range is (−π/2, π/2); since 2 > 0, θ is in (0, π/2): Quadrant I. (ii) [2 pts] Slope 2 → triangle legs: opposite 2, adjacent 1, hypotenuse √5. sin θ = 2/√5 = 2√5/5, cos θ = 1/√5 = √5/5 (both positive, QI ✓). (iii) [2 pts] tan(θ + π) = tan θ = 2 (period π), so arctan(tan(θ + π)) = arctan(2) = θ. The angle θ + π lies in QIII, outside arctan's range (−π/2, π/2), so arctan returns the in-range angle with the same tangent — θ itself.
12. (FRQ-style) 🚫 Let θ = arctan(2) — an angle whose terminal ray has slope 2. (i) State which quadrant θ lies in and why (use arctan's range). (ii) Find sin θ and cos θ exactly (draw the right triangle). (iii) Evaluate arctan(tan(θ + π)) and explain why the answer is θ and not θ + π.
Consider f(θ) = tan θ on the interval (−π/2, 3π/2), and the inverse function arctan.
(a) (i) State all vertical asymptotes of f in the given interval and the zeros of f there. (ii) Write one-sided limit statements for f at θ = π/2.
(b) (i) Explain why tan θ, restricted to (−π/2, π/2), has an inverse function. (ii) State the domain and range of arctan and explain both via the swap.
(c) (i) Evaluate lim x→∞ arctan x. (ii) Explain what feature of the tangent function this horizontal asymptote of arctan corresponds to.
(a) [2 pts]
(i) [1 pt] Asymptotes at θ = π/2 (cos = 0 there); zeros at θ = 0 and θ = π. (θ = 3π/2 is an endpoint, excluded.)
(ii) [1 pt] lim θ→(π/2)⁻ tan θ = +∞ and lim θ→(π/2)⁺ tan θ = −∞.
(b) [2 pts] (i) [1 pt] On (−π/2, π/2), tan is strictly increasing (its branch runs from −∞ to +∞ without turning), so it is one-to-one — each output comes from exactly one input — and therefore invertible. (ii) [1 pt] arctan has domain ℝ (= the range of the restricted tangent, which sweeps all real slopes) and range (−π/2, π/2) (= the restricted domain), by the inverse swap.
(c) [2 pts]
(i) [1 pt] lim x→∞ arctan x = π/2 — the graph of arctan has horizontal asymptote y = π/2.
(ii) [1 pt] It mirrors tangent's vertical asymptote at θ = π/2: inputs to tan approaching π/2 produce unboundedly large outputs, so unboundedly large inputs to arctan produce angles approaching (never reaching) π/2. Inverting a function turns vertical asymptotes into horizontal ones.
1. (B). Opposite rays share a slope: period π. (A) is sin/cos's period.
2. (C). cos θ = 0 at π/2 + kπ. (A) lists the zeros instead.
3. (D). QII, reference π/4, tan negative in QII: −1. (A) drops the sign.
4. (A). The angle in [−π/2, π/2] with sine 1/2: π/6. (B) has the right sine but violates the range.
5. (B). π − arccos(√2/2) = π − π/4 = 3π/4. (A) is the arcsin-style error; arccos is never negative.
6. (C). tan(π/4) = 1 → π/4. (B) treats arctan(1) as the number 1.
7. (A). Treaty range: [−π/2, π/2], endpoints included (arcsin(±1) = ±π/2). (C) wrongly opens the interval — that's arctan's range.
8. (D). [0, π], top half of the circle. (A) belongs to arcsin.
9. (B). sin(5π/6) = 1/2; arcsin(1/2) = π/6. (A) forgets the range restriction.
10. (A). π/|b| = π/2. (C) uses the sinusoid formula 2π/b.
11. (C). Each tangent branch rises from −∞ to +∞: always increasing (with an inflection at 0, but never a turnaround).
12. (FRQ-style, 6 points) (i) [2 pts] arctan's range is (−π/2, π/2); since 2 > 0, θ is in (0, π/2): Quadrant I. (ii) [2 pts] Slope 2 → triangle legs: opposite 2, adjacent 1, hypotenuse √5. sin θ = 2/√5 = 2√5/5, cos θ = 1/√5 = √5/5 (both positive, QI ✓). (iii) [2 pts] tan(θ + π) = tan θ = 2 (period π), so arctan(tan(θ + π)) = arctan(2) = θ. The angle θ + π lies in QIII, outside arctan's range (−π/2, π/2), so arctan returns the in-range angle with the same tangent — θ itself.
🎯 Exam tip: Before answering any inverse-trig item, write the range triple in the margin: arcsin → [−π/2, π/2], arccos → [0, π], arctan → (−π/2, π/2). Ten seconds of setup immunizes you against every wrong-quadrant distractor on the page.