Your calculator, asked to solve sin θ = 0.5, says θ = 0.5236 (that's π/6) — and stops. But look at the unit circle: height 1/2 is reached at π/6 AND at 5π/6. And again at π/6 + 2π. And at 5π/6 + 2π. And infinitely many more.
The calculator isn't wrong — it's an inverse function, so it must return exactly one output (the treaty from Lesson 20). Finding the rest of the solutions is your job, and it's a job the exam checks on every single form. The routine: one angle from the inverse, its partner from symmetry, then ± multiples of the period.
Worked skeleton for sin θ = 1/2 on [0, 2π): θ₁ = π/6; θ₂ = π − π/6 = 5π/6; extension adds nothing new inside the window. {π/6, 5π/6}.
General solution (no interval given): θ = π/6 + 2kπ or θ = 5π/6 + 2kπ, k any integer.
For sin(2θ) = 1 on [0, 2π): let u = 2θ, with u running over [0, 4π) — the substitution stretches the window.
sin u = 1 → u = π/2 or u = π/2 + 2π = 5π/2 (both inside [0, 4π))
θ = u/2 = π/4 or 5π/4
The most-lost point in this topic: forgetting the enlarged u-window and missing half the solutions. A b-multiplied argument produces b times as many solutions per window.
Treat sin θ as a variable: 2sin²θ − sin θ − 1 = 0 factors as (2sin θ + 1)(sin θ − 1) = 0 → sin θ = −1/2 or sin θ = 1 → run the master routine on each. Isolate before inverting; never divide both sides by a trig factor you could be setting to zero (factor instead — dividing by cos θ throws away the cos θ = 0 solutions).
sec θ = 1/cos θ csc θ = 1/sin θ cot θ = cos θ/sin θ (= 1/tan θ where both defined)
Values come free from the originals: sec(π/3) = 1/(1/2) = 2; csc(π/6) = 2; cot(π/4) = 1.
Structure (all inherited from the parent's zeros):
| domain excludes | vertical asymptotes | range | period | |
|---|---|---|---|---|
| sec | cos θ = 0 → θ = π/2 + kπ | at those θ | (−∞, −1] ∪ [1, ∞) | 2π |
| csc | sin θ = 0 → θ = kπ | at those θ | (−∞, −1] ∪ [1, ∞) | 2π |
| cot | sin θ = 0 → θ = kπ | at those θ | all reals | π |
Graph intuition: sec's graph is a chain of U's and ∩'s sitting on the peaks and hanging from the troughs of cos — wherever cos is small, sec is huge. The gap (−1, 1) is unreachable for sec and csc since |cos|, |sin| ≤ 1 forces their reciprocals' magnitudes ≥ 1.
[GRAPH: y = cos θ (light) with y = sec θ (bold) on [0, 2π]: sec's upward U touching cos's maximum at (0, 1) and (2π, 1), downward ∩ hanging below the minimum at (π, −1), vertical asymptotes dashed at π/2 and 3π/2. Shaded horizontal band −1 < y < 1 labeled "sec never enters".]
Problem: Solve 2cos θ = −√3 on [0, 2π).
Solution: cos θ = −√3/2. Reference angle: π/6. Cosine negative → QII and QIII: θ = π − π/6 = 5π/6 and θ = π + π/6 = 7π/6.
Interpretation: For cosine, the two solutions sit symmetric about the x-axis (5π/6 and 7π/6 = 2π − 5π/6 ✓). Quadrant placement first, arithmetic second.
Problem: Solve sin(3θ) = √2/2 on [0, 2π).
Solution: Let u = 3θ ∈ [0, 6π). sin u = √2/2 → base solutions u = π/4, 3π/4, then + 2π and + 4π copies:
u = π/4, 3π/4, 9π/4, 11π/4, 17π/4, 19π/4 (six values, all < 6π)
θ = u/3 = π/12, π/4, 3π/4, 11π/12, 17π/12, 19π/12
Interpretation: b = 3 → 3× the window → 6 solutions instead of 2. Count before you finish: a sin(bθ) = k equation (with |k| < 1) has 2b solutions per full period window.
Problem: Solve cos θ = 0.3 on [0, 2π), to three decimal places.
Solution: θ₁ = arccos(0.3) ≈ 1.266 (QI). Cosine positive also in QIV: θ₂ = 2π − 1.266 ≈ 5.017. Solutions: {1.266, 5.017}.
Interpretation: The calculator supplies exactly one angle; the unit circle's symmetry supplies its twin. Writing only 1.266 is the single most common calc-active error in Unit 3.
Problem: Evaluate sec(5π/6), and explain why sec has a vertical asymptote at π/2 but csc does not.
Solution: cos(5π/6) = −√3/2 → sec(5π/6) = −2/√3 = −2√3/3. At θ = π/2: cos = 0, so sec = 1/0 — undefined, with |sec| → ∞ nearby: vertical asymptote. But sin(π/2) = 1, so csc(π/2) = 1 — perfectly defined (it's the minimum of csc's upper branch). Each reciprocal blows up only at its own parent's zeros.
Interpretation: "Asymptotes live at the parent's zeros" answers every domain question about sec/csc/cot without memorization.
1. (B). Reference π/6; sine positive in QI and QII: π/6, 5π/6. (C) pairs the wrong quadrants.
2. (A). Cosine reaches −1 only at the far-left point (−1, 0): θ = π.
3. (B). Tangent's period is π: π/3 + kπ captures everything. (A) misses half the solutions (e.g., 4π/3).
4. (A). sin θ = √3/2: QI and QII: π/3, 2π/3. (B) solves sin = 1/2; (C) puts the partner in QIV (that's cosine's symmetry).
5. (C). 1/cos(π/3) = 1/(1/2) = 2. (B) is sec(π/6); (A) inverts the wrong way.
6. (A). 1/sin(π/6) = 1/(1/2) = 2. (D) is csc(π/3).
7. (D). cos/sin at π/4 = (√2/2)/(√2/2) = 1. (A) is cot(π/2); (B) is cot(0).
8. (B). Parent zero rule: cos = 0 at π/2 + kπ. (A) is csc/cot's exclusion set.
9. (C). u = 2θ ∈ [0, 4π): u = π/2, 5π/2 → θ = π/4, 5π/4. (A) forgets the stretched window; (D) uses sine's π − partner on θ instead of u.
10. (D). θ₁ = arccos 0.3 ≈ 1.266; twin 2π − θ₁ ≈ 5.017. (B) uses π + θ₁ (that's where cos = −0.3); (A) stops at one.
11. (C). Reciprocal of a value in [−1, 1]{0} has magnitude ≥ 1: (−∞, −1] ∪ [1, ∞). (A) is cos's range.
12. (FRQ-style, 6 points) (i) [3 pts] (2cos θ − 1)(cos θ + 1) = 0 [1 pt] → cos θ = 1/2: θ = π/3, 5π/3 [1 pt]; cos θ = −1: θ = π [1 pt]. Set: {π/3, π, 5π/3}. (ii) [2 pts] csc θ = 2 → sin θ = 1/2 → θ = π/6, 5π/6. (iii) [1 pt] csc θ = 1/2 requires sin θ = 2, impossible since |sin θ| ≤ 1 — csc's range excludes (−1, 1). No solutions.
12. (FRQ-style) 🚫 Solve each completely: (i) 2cos²θ + cos θ − 1 = 0 on [0, 2π) (factor first). (ii) csc θ = 2 on [0, 2π). (iii) Explain why csc θ = 1/2 has no solutions.
(a) Solve 2sin²θ − sin θ − 1 = 0 on the interval [0, 2π). (Factor; then solve each resulting basic equation.)
(b) Find the general solution of cos(2θ) = √2/2.
(c) The function g(θ) = sec θ. (i) State the domain restriction of g on [0, 2π). (ii) Solve sec θ = −2 on [0, 2π).
(a) [2 pts] (2sin θ + 1)(sin θ − 1) = 0 [1 pt factoring] → sin θ = −1/2: θ = 7π/6, 11π/6 (QIII, QIV); sin θ = 1: θ = π/2. Solutions: {π/2, 7π/6, 11π/6} [1 pt all three, no extras].
(b) [2 pts] Let u = 2θ: cos u = √2/2 → u = π/4 + 2kπ or u = −π/4 + 2kπ [1 pt] → dividing by 2: θ = π/8 + kπ or θ = −π/8 + kπ, k any integer [1 pt]. (Halving the 2π-period extension makes it a kπ extension — the compressed function repeats twice as often.)
(c) [2 pts] (i) [1 pt] sec θ undefined where cos θ = 0: θ ≠ π/2, 3π/2. (ii) [1 pt] sec θ = −2 → cos θ = −1/2 → θ = 2π/3, 4π/3.
1. (B). Reference π/6; sine positive in QI and QII: π/6, 5π/6. (C) pairs the wrong quadrants.
2. (A). Cosine reaches −1 only at the far-left point (−1, 0): θ = π.
3. (B). Tangent's period is π: π/3 + kπ captures everything. (A) misses half the solutions (e.g., 4π/3).
4. (A). sin θ = √3/2: QI and QII: π/3, 2π/3. (B) solves sin = 1/2; (C) puts the partner in QIV (that's cosine's symmetry).
5. (C). 1/cos(π/3) = 1/(1/2) = 2. (B) is sec(π/6); (A) inverts the wrong way.
6. (A). 1/sin(π/6) = 1/(1/2) = 2. (D) is csc(π/3).
7. (D). cos/sin at π/4 = (√2/2)/(√2/2) = 1. (A) is cot(π/2); (B) is cot(0).
8. (B). Parent zero rule: cos = 0 at π/2 + kπ. (A) is csc/cot's exclusion set.
9. (C). u = 2θ ∈ [0, 4π): u = π/2, 5π/2 → θ = π/4, 5π/4. (A) forgets the stretched window; (D) uses sine's π − partner on θ instead of u.
10. (D). θ₁ = arccos 0.3 ≈ 1.266; twin 2π − θ₁ ≈ 5.017. (B) uses π + θ₁ (that's where cos = −0.3); (A) stops at one.
11. (C). Reciprocal of a value in [−1, 1]{0} has magnitude ≥ 1: (−∞, −1] ∪ [1, ∞). (A) is cos's range.
12. (FRQ-style, 6 points) (i) [3 pts] (2cos θ − 1)(cos θ + 1) = 0 [1 pt] → cos θ = 1/2: θ = π/3, 5π/3 [1 pt]; cos θ = −1: θ = π [1 pt]. Set: {π/3, π, 5π/3}. (ii) [2 pts] csc θ = 2 → sin θ = 1/2 → θ = π/6, 5π/6. (iii) [1 pt] csc θ = 1/2 requires sin θ = 2, impossible since |sin θ| ≤ 1 — csc's range excludes (−1, 1). No solutions.
🎯 Exam tip: After solving any trig equation, count your answers against the formula "2 solutions per period for sin/cos (1 for tangent), times b for multiplied angles, per 2π of window." If the count is off, you've either missed a partner or leaked one in from outside the interval.