Lesson 7: Trigonometry

A-level trigonometry leaves right-angled triangles behind and enters the world of radians, identities, and inverse functions — the language of waves, oscillations, and rotations that physics and engineering run on. This lesson covers radian measure, the small-angle approximations, exact identities (compound angle, double angle, R-formulae), and solving trigonometric equations generally. The identities are not lists to memorise blindly but tools derived from a few core truths; learn the derivations and the formulas become unforgettable.

Radians — the natural angle

Degrees are arbitrary (why 360? Babylonian astronomy). Radians are natural: one radian is the angle subtending an arc equal to the radius, so a full circle is 2π radians. Conversion: π radians = 180°, so 1 rad ≈ 57.3°. Exact values: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2. A-level calculus with trig functions requires radians — the derivative of sin x is cos x only in radians. Using degrees in calculus is simply wrong.

Arc length s = rθ and sector area A = (1/2)r²θ (θ in radians) — cleaner than the degree versions because radians absorb the π. Worked example: a wheel of radius 0.4 m rolls through 5 radians: distance = rθ = 2 m. A sector of radius 12 cm with angle 1.2 rad has area (1/2)(144)(1.2) = 86.4 cm². Small-angle approximations: for small θ (radians), sin θ ≈ θ, tan θ ≈ θ, cos θ ≈ 1 − θ²/2 — proved geometrically or via series; essential for physics (pendulums) and for limits like lim(θ→0) (sin θ)/θ = 1.

Worked example: approximate sin(0.05) and cos(0.05): sin ≈ 0.05, cos ≈ 1 − 0.0025/2 = 0.99875 (calculator: 0.99875026... — six-figure accuracy!). But approximations fail for large angles: sin(1) ≈ 1 is poor (true 0.841). Always check 'small' means |θ| ≲ 0.3 rad for good accuracy.

Identities — compound and double angles

The compound-angle formulae derive from geometry (or Euler's formula in Further Maths): sin(A ± B) = sin A cos B ± cos A sin B; cos(A ± B) = cos A cos B ∓ sin A sin B; tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B). Setting B = A gives the double-angle formulae: sin 2A = 2 sin A cos A; cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A; tan 2A = 2tan A/(1 − tan²A). The three forms of cos 2A are each useful: 2cos²A − 1 for integrating cos², 1 − 2sin²A for integrating sin².

Worked example: express sin 75° exactly. 75° = 45° + 30°: sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2) = √2(√3 + 1)/4 = (√6 + √2)/4. Exact-value problems reward choosing the right split. Worked example: simplify (sin 2x)/(1 + cos 2x) = 2 sin x cos x/(1 + 2cos²x − 1) = 2 sin x cos x/(2cos²x) = tan x. Identity-proving is algebraic persistence: expand one side using compound formulae until it matches.

Worked example (exact value by subtraction). Find tan 15° exactly. Write 15° = 45° − 30°: tan(45° − 30°) = (tan 45° − tan 30°)/(1 + tan 45° tan 30°) = (1 − 1/√3)/(1 + 1/√3). Multiply top and bottom by √3: (√3 − 1)/(√3 + 1); rationalise by multiplying by (√3 − 1): (√3 − 1)²/2 = (4 − 2√3)/2 = 2 − √3 ≈ 0.268. Three techniques in one line — compound formula, surd arithmetic, rationalising — which is why exact-value questions are examiner favourites.

Learn derivations, not just results: cos(A + B) from the unit circle or from sin via cofunction; double-angle by setting B = A; the tan formula by dividing sin by cos. In the exam you are given the formulae (in the formula booklet) — the skill is choosing which to apply, not recalling them. Examiners test recognition: 'which identity simplifies cos⁴x − sin⁴x?' (difference of squares → cos²x − sin²x → cos 2x).

The R-formulae (harmonic form)

Any expression a sin x + b cos x can be written as R sin(x + α) or R cos(x − α), where R = √(a² + b²) and tan α = b/a (for the sin form). Proof: R sin(x + α) = R sin x cos α + R cos x sin α; matching coefficients: R cos α = a, R sin α = b; squaring and adding gives R² = a² + b²; dividing gives tan α = b/a. Worked example: 3 sin x + 4 cos x = 5 sin(x + α) with tan α = 4/3, α ≈ 53.1°. So 3 sin x + 4 cos x = 5 sin(x + 53.1°).

Why bother? The R-form reveals amplitude and phase instantly: the maximum of 3 sin x + 4 cos x is 5 (when sin(x + α) = 1), occurring at x = 90° − 53.1° = 36.9°. Without the R-form, maximising a sin x + b cos x needs calculus; with it, one line. Applications: alternating current V = V₀ sin(ωt + φ) is an R-form; sound wave superposition; tidal modelling. Physics runs on R-formulae.

Care with quadrants: for 3 sin x − 4 cos x = 5 sin(x − α) still (α ≈ 53.1°, since tan α = −4/3 gives α = −53.1°). For −3 sin x + 4 cos x, R = 5 but the angle α ≈ 126.9° — the signs of a and b determine the quadrant. Sketch a and b as a right triangle to place α correctly; blind arctan gives the wrong quadrant half the time.

Solving trigonometric equations

A-level equations need all solutions in a range, using exact values and CAST. Example: solve 2sin²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°. Factorise: (2 sin x + 1)(sin x − 1) = 0. sin x = 1 → x = 90°. sin x = −1/2 → x = 210°, 330° (CAST: sine negative in quadrants 3 and 4). Solutions: 90°, 210°, 330°. Giving only the principal value is the top exam error in this topic.

Equations with multiple angles: solve cos 2x = 1/2 for 0 ≤ x ≤ 2π. The argument 2x ranges over [0, 4π]: 2x = π/3, 5π/3, 7π/3, 11π/3, so x = π/6, 5π/6, 7π/6, 11π/6. The method: enlarge the range for the argument first, solve there, then divide — never divide the range before solving.

Quadratic-in-disguise and identity-requiring equations: solve 2cos²x + 3 sin x = 0 on [0°, 360°). Replace cos²x = 1 − sin²x: 2(1 − sin²x) + 3 sin x = 0 → 2sin²x − 3 sin x − 2 = 0 → (2 sin x + 1)(sin x − 2) = 0. sin x = 2 impossible; sin x = −1/2 → x = 210°, 330°. Strategy: reduce to a single trig function via identities, substitute (u = sin x), solve the algebra, then solve for x with CAST. Check solutions in the original equation — squaring steps can introduce extras.

Inverse trig and modelling

arcsin, arccos, arctan invert the trig functions on restricted domains: arcsin: [−1, 1] → [−π/2, π/2]; arccos: [−1, 1] → [0, π]; arctan: ℝ → (−π/2, π/2). So arcsin(1/2) = π/6 (not 5π/6 — the principal value). Composition: sin(arcsin(0.7)) = 0.7, but arcsin(sin(5π/6)) = arcsin(1/2) = π/6 ≠ 5π/6 — the inverse 'folds' back into the principal range. Examiners test this folding explicitly.

Differentiating inverse trig: d/dx(arcsin x) = 1/√(1 − x²), d/dx(arctan x) = 1/(1 + x²) — via implicit differentiation (if y = arcsin x then sin y = x, cos y · dy/dx = 1, dy/dx = 1/cos y = 1/√(1 − x²)). These derivatives integrate to give the standard forms ∫ 1/√(1 − x²) dx and ∫ 1/(1 + x²) dx — the circle of calculus closing.

Modelling with trig: Southampton tides follow h(t) = 2.5 sin(πt/6.2 + 1.1) + 3.2 (metres, t hours after midnight). Maximum depth 5.7 m when the sine is 1: πt/6.2 + 1.1 = π/2 → t = (π/2 − 1.1) × 6.2/π ≈ 0.93 h, about 00:56. When is the depth 4 m? Solve 2.5 sin(...) = 0.8 → sin = 0.32 → two solutions per cycle — high tide rising and falling give two times. Sinusoidal models describe tides, daylight hours, AC circuits, and seasonal demand — find amplitude, period, phase shift, vertical shift, then read off answers.

Secant, cosecant, cotangent

The reciprocal trig functions complete the family: sec x = 1/cos x, cosec x = 1/sin x, cot x = 1/tan x = cos x/sin x. They simplify identities: dividing sin²x + cos²x = 1 by cos²x gives tan²x + 1 = sec²x; dividing by sin²x gives 1 + cot²x = cosec²x. These three Pythagorean identities (in sin/cos, tan/sec, cot/cosec forms) are the complete set — learn all three, since examiners use whichever makes the algebra collapse.

Their calculus is clean: d/dx(tan x) = sec²x, d/dx(sec x) = sec x tan x, d/dx(cosec x) = −cosec x cot x, d/dx(cot x) = −cosec²x — each proved via the quotient rule in one line. Integrals reverse them: ∫ sec²x dx = tan x, and ∫ sec x tan x dx = sec x. The derivatives explain why tan appears in the integral of 1/(1 + x²)... via the substitution x = tan θ (Lesson 6): dx = sec²θ dθ, and 1 + tan²θ = sec²θ cancels beautifully.

Graphs: sec x has vertical asymptotes where cos x = 0 (x = π/2 + kπ), with U-shaped branches between — sketch by reciprocating the cosine graph (where cos is small, sec is huge; where cos = ±1, sec = ±1). Domain restrictions matter: equations like sec x = 2 become cos x = 1/2 (x ≠ π/2 + kπ automatically satisfied) — but always check candidate solutions against the original function's domain, since reciprocals exclude zeros of the original.

Common mistakes

• Degrees in calculus. d/dx(sin x) = cos x only in radians. Any differentiation or integration of trig functions in degrees is wrong by a factor of π/180.
• Missing solutions in the range. sin x = 1/2 has two solutions in [0°, 360°), not one. Use CAST or the symmetry of the graph for every equation, and enlarge the range first for multiple angles like 2x or x/2.
• Wrong form of cos 2x. Pick the version matching the rest of the equation: 2cos²x − 1 when cos² appears, 1 − 2sin²x when sin² appears. Choosing the other doubles the work and invites errors.
• Blind arctan in the R-formula. tan α = b/a gives α in the wrong quadrant when a is negative. Determine the quadrant from the signs of a and b (or use the cos form) — and verify by expanding your answer.
• Treating arcsin(sin x) as x. The inverse folds into [−π/2, π/2]: arcsin(sin(5π/6)) = π/6, not 5π/6.

Exam Experience

Questions in the style of the A-level pure paper. Attempt them before reading the mark scheme.

1. Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π, giving your answers in radians. [4]

2. Prove the identity (sin x)/(1 + cos x) + (1 + cos x)/(sin x) ≡ 2/(sin x), for sin x ≠ 0 and cos x ≠ −1. [3]

3. Express 5 sin x + 12 cos x in the form R sin(x + α), where R > 0 and 0 < α < π/2. Hence state the maximum value of 5 sin x + 12 cos x. [3]

Mark scheme.

• 1. Factorise: (2cos x + 1)(cos x − 1) = 0 (M1) for reducing to a single function and factorising. cos x = 1 → x = 0, 2π (A1). cos x = −1/2 → x = 2π/3, 4π/3 (A1) — both second/third-quadrant solutions required, (A1) for all four solutions with none extra (endpoints 0 and 2π both count in a closed range).
• 2. Combine over the common denominator sin x(1 + cos x): numerator = sin²x + (1 + cos x)² (M1). = sin²x + 1 + 2cos x + cos²x = 2 + 2cos x = 2(1 + cos x) (M1) for using sin²x + cos²x = 1. Hence the expression = 2(1 + cos x)/(sin x(1 + cos x)) = 2/(sin x) (A1) — identity proved; the cancelled factor (1 + cos x) is non-zero by the stated conditions.
• 3. R = √(25 + 144) = 13 (M1). Matching coefficients: R cos α = 5, R sin α = 12 → tan α = 12/5 → α ≈ 1.18 rad (67.4°) (A1). So 5 sin x + 12 cos x = 13 sin(x + 1.18); maximum value 13 (A1) (attained when the sine equals 1).