By writing sin3A \sin 3A\,sin3A as sin(2A+A)\sin(2A + A)sin(2A+A), show that sin3A=3sinA−4sin3A\sin 3A = 3\sin A - 4\sin^3 Asin3A=3sinA−4sin3A
Solve, for 0≤A≤π0 \leq A \leq \pi0≤A≤π, the equation,
3sinA−4sin3A=12 3\sin A - 4\sin^3 A = \frac{1}{\sqrt{2}} 3sinA−4sin3A=21Give your answers in terms of π\piπ.
317 exam-style questions on AQA A Level Maths 1.8 E: Trigonometry, covering 1.8.1 Trigonometric definitions, rules and radians, 1.8.2 Small angle approximations (A-level only), 1.8.3 Trigonometric functions and exact values, 1.8.4 Reciprocal and inverse trigonometric functions (A-level only), 1.8.5 Trigonometric identities, 1.8.6 Compound and double angle formulae (A-level only), 1.8.7 Solving trigonometric equations, 1.8.8 Proofs with trigonometric identities, and 1.8.9 Trigonometry in context. Each one has a worked solution and a mark scheme showing where the marks go.