Lesson 5: Trigonometric Ratios and Equations

Trigonometry is the mathematics of triangles, circles and cycles (AA Topic 3), and it splits cleanly into two halves: geometry with degrees and radians, and algebra with equations and identities. IB works in radians for everything beyond basic triangle geometry — set your GDC to radian mode and check it before every paper; the examiners' reports return to this error every single session. A radian is the angle subtended when the arc length equals the radius: 2π radians = 360°, so 1 radian ≈ 57.3°.

Radian measure: arcs and sectors

With θ in radians, two formulas run everything: arc length s = rθ and sector area A = ½r²θ. They are one-liners in radians and unusable in degrees — that is precisely why IB insists on radians. A segment (sector minus triangle) is the classic composite: A_segment = ½r²θ − ½r² sin θ.

Worked Example 1 — arc and segment. A circle has radius 8 cm. A sector has angle 1.2 rad. Find (a) the arc length, (b) the area of the segment cut off by the chord joining the arc's endpoints.

Solution. (a) s = rθ = 8 × 1.2 = 9.6 cm. (b) Sector area = ½ × 64 × 1.2 = 38.4 cm². Triangle area = ½r² sin θ = ½ × 64 × sin(1.2) = 32 × 0.9320 ≈ 29.8 cm². Segment = 38.4 − 29.8 = 8.57 cm² (3 s.f.). Method marks are earned by computing sector and triangle separately and visibly; a single unexplained line forfeits them if the final value slips.

The sine and cosine rules

For a triangle with sides a, b, c opposite angles A, B, C: sine rule a/sin A = b/sin B = c/sin C (use it when you know a side-angle pair) and cosine rule a² = b² + c² − 2bc cos A (use it with two sides and the included angle, or all three sides). Area = ½ab sin C. The sine rule hides the ambiguous case: given two sides and a non-included angle, two different triangles may fit — check whether the supplement of your angle also works before finishing.

Worked Example 2 — a surveying problem (past-paper style). Two points P and Q lie 400 m apart on a straight coast road. A lighthouse L is seen from P at a bearing such that angle QPL = 52°, and from Q with angle PQL = 71°. Find the distance from P to the lighthouse.

Solution. The third angle at L is 180 − 52 − 71 = 57°. The side PQ = 400 m is opposite the 57° angle; PL is opposite 71°. Sine rule: PL / sin 71° = 400 / sin 57°, so PL = 400 × sin 71° / sin 57° = 400 × 0.9455 / 0.8387 ≈ 451 m (3 s.f.). Diagrams: IB awards a method mark for a correctly labelled sketch even when the arithmetic later slips — draw one, always.

Trigonometric equations

Solving sin x = k or cos x = k over a specified interval is a core Paper 1 skill. Method: find the principal value (calculator or known angles), generate the family from the symmetry of the unit circle (sin: θ and π − θ; cos: ±θ; tan: θ + π periodicity), then keep only solutions inside the interval. Known exact values must be automatic: sin/cos/tan of 0, π/6, π/4, π/3, π/2 and their reflections.

Worked Example 3 — solving over an interval. Solve 2 sin x − 1 = 0 for 0 ≤ x ≤ 2π, and 3 tan x = −√3 for 0 ≤ x < 2π.

Solution. sin x = 1/2: principal value π/6; sine is also positive in the second quadrant: π − π/6 = 5π/6. Solutions: π/6, 5π/6. For tan x = −√3/3 = −1/√3: reference angle π/6 where tan is positive; tangent is negative in quadrants II and IV: x = π − π/6 = 5π/6 and x = 2π − π/6 = 11π/6. Solutions: 5π/6, 11π/6. On Paper 2 you may find these from a GDC graph, but Paper 1 demands the quadrant reasoning — and note that "solve" defaults to radians throughout AA unless degrees are explicitly requested.

Graphs of trig functions (bridge to Lesson 6)

y = a sin(bx) + d has amplitude |a|, period 2π/b, vertical shift d — the vocabulary of every modelling question: tides, temperature, daylight, Ferris wheels. Fitting means reading amplitude (half the range), period (from the cycle length), and midline (the average of max and min) off a graph or a story. Lesson 6 builds the unit-circle machinery beneath these graphs; this lesson's job is fluency with triangles and equations.

Common mistakes

• Degree mode on the GDC — the most expensive slip in the course. Check the mode line before Papers 2 and 3.
• Arc/sector formulas in degrees — s = rθ demands radians.
• Ambiguous case ignored: reporting one triangle when two exist (compare the given sides: if the side opposite the given angle is shorter than the other given side, suspect two solutions).
• Solutions outside the interval left in the answer, or the family generated but the interval never applied.
• Premature rounding of angles mid-question; keep at least 5 figures in working, 3 s.f. at the end — and angles in IB are usually expected to 3 s.f. too, not to the nearest degree, unless stated.
• sin⁻¹ used as 1/sin — inverse function versus reciprocal; write csc if you mean the reciprocal.

IB Exam Experience

Question 1 [2 marks]. A sector of a circle of radius 5 cm has area 30 cm². Find the angle of the sector, in radians. *Marking notes:* M1 for 30 = ½ × 25 × θ; A1 for θ = 2.4 rad.

Question 2 [3 marks]. In triangle ABC, AB = 7 cm, BC = 9 cm and angle BAC = 40°. Find angle BCA, giving all possible values. *Marking notes:* M1 for the sine rule 9/sin 40° = 7/sin C → sin C = 7 sin 40°/9 ≈ 0.500; A1 for C ≈ 30.0°; A1 for the ambiguous second value C ≈ 150° (valid since 150 + 40 < 180). The second value is the entire point of the question.

Question 3 [4 marks]. Solve 2 cos² x − cos x − 1 = 0 for 0 ≤ x ≤ 2π. *Marking notes:* M1 for treating it as a quadratic in cos x and factorising: (2 cos x + 1)(cos x − 1) = 0; A1 for cos x = 1 and cos x = −1/2; A1 for x = 0 and 2π (accept either or both per the interval's closed ends — markschemes accept x = 0, 2π, 2π/3, 4π/3); A1 for x = 2π/3, 4π/3 from the second factor. Four solutions expected; missing the endpoint duplication is the classic loss.

Extended problem (Paper 2 style) [10 marks]. The depth of water at a harbour entrance is modelled by d(t) = 2.4 cos(πt/6) + 3.8 metres, t hours after midnight. (a) Find the maximum and minimum depths. [2] (b) Find the depth at 09:00. [2] (c) A ship needs at least 3.0 m of water. Find the total time between midnight and the next midnight during which the ship can enter. [6] *Marking notes:* (a) A1 A1: max 6.2 m, min 1.4 m (amplitude 2.4 about midline 3.8). (b) A2: d(9) = 2.4 cos(3π/2) + 3.8 = 3.8 m. (c) M1 for solving 2.4 cos(πt/6) + 3.8 = 3.0, i.e. cos(πt/6) = −1/3; M1 for the first solution πt/6 = arccos(−1/3) ≈ 1.911 → t ≈ 3.65 h; M1 for symmetry/period 12 h to find the second crossing t ≈ 12 − 3.65 = 8.35 h (and the pattern repeating in the second 12 hours); A1 for the unsafe window 3.65–8.35 (4.70 h) per cycle; A1 for two cycles per day; A1 for total safe time 24 − 2 × 4.70 ≈ 14.6 hours. Examiner's note: the conclusion must be an interval of *safe* time — answering with the unsafe window, or with raw solutions only, loses the final marks.

Exploration (IA) idea: record sunrise times or temperature through a day (or a year) for your city, fit a sinusoid, and evaluate how well a pure cosine captures reality — asymmetric real data motivates a discussion of model limits, exactly what IA Criterion E rewards.

Going deeper: the area of a triangle, three ways

IB papers reward flexibility: the same area falls to ½ab sin C (two sides + included angle), to the sine rule plus ½ab sin C, or to coordinates. Choosing the route that matches the *given* data is the skill.

Extra worked example. A triangular plot has sides 13 m, 14 m and 15 m. Find its area. Route: the cosine rule supplies an angle first: cos C (opposite the 15 side) = (13² + 14² − 15²)/(2 × 13 × 14) = (169 + 196 − 225)/364 = 140/364 = 5/13. Then sin C = 12/13 (positive, since C is a triangle angle), and area = ½ × 13 × 14 × 12/13 = 84 m². (Heron's formula agrees: s = 21, area = √(21 × 8 × 7 × 6) = √7056 = 84 ✓.) The 5-12-13 triple hiding inside the working is typical of designed questions — when cos comes out as a famous fraction, expect sin to be its partner.

Extra exam practice [4 marks]. Two ships leave a port at the same time. Ship A sails 30 km on bearing 040°; ship B sails 40 km on bearing 130°. Find the distance between the ships. *Marking notes:* M1 for the angle at the port between the tracks: 130 − 40 = 90° (bearings differ by 90° — the diagram earns this); M1 for the cosine rule (or Pythagoras, spotting the right angle): d² = 30² + 40² − 2(30)(40)cos 90° = 2500; A1: d = 50 km; A1 for units and a labelled sketch (the method mark's insurance).