Trigonometry carries about 40 of the 150 marks in Paper 2, and the question types repeat every year. Cover each answer below, try the question on paper, then compare your working line by line.
Question 1: Reduction formulas
Simplify sin(180° − x) · cos(90° + x) / tan(360° − x). Use sin(180° − x) = sin x, cos(90° + x) = −sin x and tan(360° − x) = −tan x. That gives (sin x)(−sin x) / (−tan x) = sin²x / tan x. Answer: sin x cos x.
Question 2: General solution
Find the general solution of 2 sin x = 1. Then sin x = ½, so the reference angle is 30°. Sine is positive in quadrants 1 and 2. Answer: x = 30° + k·360° or x = 150° + k·360°, k ∈ ℤ. Common slip: leaving out k ∈ ℤ or the second solution.
Question 3: Prove an identity
Prove that (1 − cos²x) / (sin x cos x) = tan x. Start with the left side: 1 − cos²x = sin²x, so the fraction becomes sin²x / (sin x cos x) = sin x / cos x = tan x. Work on one side only, and never move terms across the equals sign.
Question 4: Double angle
If sin A = 3/5 and A is acute, find sin 2A without a calculator. Draw the triangle: cos A = 4/5. Then sin 2A = 2 sin A cos A = 2(3/5)(4/5). Answer: 24/25.
Question 5: Compound angle
Find cos 75° in surd form. Write 75° = 45° + 30°, so cos 75° = cos 45° cos 30° − sin 45° sin 30° = (√2/2)(√3/2) − (√2/2)(½). Answer: (√6 − √2)/4.
Question 6: Cosine rule
In △ABC, a = 5, b = 7 and C = 60°. Find c. Use c² = a² + b² − 2ab cos C = 25 + 49 − 2(5)(7)(½) = 39. Answer: c = √39 ≈ 6.24.
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