Mathematics · 9. Trigonometry and trigonometric equations
Unit circle and the six canonical trigonometric equations
Master trigonometric identities, addition, double-angle, power-reduction, and the six canonical trigonometric equation methods.
Driving question
How does the unit circle representation enable complete control over periodic solution sets in trigonometric equations?
Curriculum coverage
Vietnam Mathematics 10–11; Cambridge IGCSE / AS & A-Level: Trigonometry.
Model scope and limitations
Trigonometric equations yield periodic families with period 2π or π; always specify k in Z and verify domain restrictions for tan(x) and cot(x).
Learning objectives
- Apply sin²(x) + cos²(x) = 1, addition, double-angle and power-reduction formulas.
- Solve the 6 canonical forms: basic, quadratic/power-reduction, linear sin-cos, homogeneous quadratic, symmetric, and factorisation/amplitude bounding.
- Represent and verify solutions on the unit circle against domain conditions to eliminate extraneous roots.
Prerequisite knowledge
- Angles, radians and right-triangle trigonometric ratios.
- Algebraic quadratic equations and substitution bounds.
Core knowledge
Canonical trigonometric equations & 6 special root positions
sin(u) = sin(v) ⇔ u = v + 2kπ or u = π − v + 2kπ; cos(u) = cos(v) ⇔ u = ±v + 2kπ; tan(u) = tan(v) ⇔ u = v + kπ. Six special positions yielding a single periodic family: sin(x) = 0 ⇔ x = kπ; sin(x) = ±1 ⇔ x = ±π/2 + 2kπ; cos(x) = 0 ⇔ x = π/2 + kπ; cos(x) = 1 ⇔ x = 2kπ; cos(x) = −1 ⇔ x = π + 2kπ.
Quadratic trigonometric equations & Power reduction
Form a·sin²(x) + b·sin(x) + c = 0 or using power-reduction identities. Substitute t = sin(x) or cos(x) with strict bound −1 ≤ t ≤ 1. For tan(x) or cot(x), t in R.
Linear combination: a·sin(x) + b·cos(x) = c
Condition for real roots is a² + b² ≥ c². Divide both sides by √(a² + b²) to rewrite as sin(x + α) = c / √(a² + b²), where cos(α) = a / √(a² + b²) and sin(α) = b / √(a² + b²). Used for finding ranges of trigonometric fractions.
Homogeneous quadratic equations in sin and cos
Form a·sin²(x) + b·sin(x)·cos(x) + c·cos²(x) = d. Step 1: Check if cos(x) = 0 is a solution. Step 2: For cos(x) ≠ 0, divide by cos²(x) and use 1/cos²(x) = 1 + tan²(x) to obtain a quadratic in tan(x).
Symmetric and skew-symmetric equations
Form a·(sin(x) ± cos(x)) + b·sin(x)·cos(x) + c = 0. Substitute t = sin(x) ± cos(x) = √2·sin(x ± π/4) with |t| ≤ √2, yielding sin(x)·cos(x) in terms of t² and solving the resulting quadratic.
Product-to-sum factorisation & Amplitude bounding
Apply sum-to-product identities cos u ± cos v and sin u ± sin v to factorise into A(x)·B(x) = 0. For extreme-value equations, use amplitude bounds |sin u| ≤ 1 and |cos v| ≤ 1 to establish equality conditions.
Worked example
Solve √3·sin(x) − cos(x) = √2. Here a = √3, b = −1, c = √2. Check condition: a² + b² = 4 ≥ 2 (valid). Divide by √(a² + b²) = 2: (√3/2)·sin(x) − (1/2)·cos(x) = √2/2 ⇔ sin(x − π/6) = sin(π/4). Roots: x = 5π/12 + 2kπ or x = 11π/12 + 2kπ (k in Z).
Misconceptions to avoid
- Forgetting a² + b² ≥ c² causes attempts to solve inconsistent equations that have no real roots.
- Forgetting |t| ≤ √2 for t = sin(x) ± cos(x) leads to accepting extraneous roots outside [−√2, √2].
- Omitting the cos(x) = 0 case before dividing a homogeneous equation by cos²(x).
Virtual experiment procedure: Trigonometric circle and trigonometric equations
- Identify equation type (basic, quadratic/power-reduction, linear sin-cos, homogeneous, symmetric, or product).
- State domain restrictions (for tan, cot, fractions) and bounds on auxiliary variables.
- Transform into basic equations, write solutions with periods 2kπ or kπ (k in Z), and verify domain conditions.
Safety and cautions
- Always state k in Z; plot solutions on the unit circle to check for overlapping roots or forbidden points.
Evidence to collect
- A structured solution showing solvability condition checks, trigonometric identities used, variable bounds, and verified root sets.
Diagnostic check
1. The necessary and sufficient condition for a·sin(x) + b·cos(x) = c to have real roots is?
Show answer and explanation
Correct answer: a² + b² ≥ c²
Dividing by √(a² + b²) gives sin(x + α) = c / √(a² + b²). Because |sin| ≤ 1, we require |c| / √(a² + b²) ≤ 1 ⇔ a² + b² ≥ c².
2. When substituting t = sin(x) + cos(x), what is the valid range of t?
Show answer and explanation
Correct answer: −√2 ≤ t ≤ √2
Since t = sin(x) + cos(x) = √2·sin(x + π/4) and −1 ≤ sin ≤ 1, we have −√2 ≤ t ≤ √2.
3. What is the solution set of sin(x) = 1/2?
Show answer and explanation
Correct answer: x = π/6 + 2kπ or x = 5π/6 + 2kπ (k ∈ Z)
sin(x) = sin(α) ⇔ x = α + 2kπ or x = π − α + 2kπ. For α = π/6, π − π/6 = 5π/6.
Application practice
1. How many real solutions does 3·sin(x) + 4·cos(x) = 6 have?
Show answer and explanation
Correct answer: 0 solutions
Check condition: a² + b² = 3² + 4² = 25 < 6² = 36, hence the equation has no real solutions.
2. What are the solutions of cos(2x) = 0?
Show answer and explanation
Correct answer: x = π/4 + k·π/2 (k ∈ Z)
cos(2x) = 0 ⇔ 2x = π/2 + kπ ⇔ x = π/4 + k·π/2 (k ∈ Z).
3. How many solutions does 2·sin²(x) + sin(x) − 1 = 0 have in [0, π]?
Show answer and explanation
Correct answer: 2 solutions (x = π/6 and x = 5π/6)
Setting t = sin(x) yields 2t² + t − 1 = 0 ⇔ t = 1/2 or t = −1. On [0, π], sin(x) ≥ 0 excludes t = −1; t = 1/2 gives exactly x = π/6 and x = 5π/6.
Academic references
- Cambridge IGCSE Mathematics 0580 — official syllabus and resources
- Cambridge AS & A Level Mathematics 9709 — official syllabus
- OpenStax Precalculus 2e — functions, geometry and probability
This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.
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