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.
225 exam-style questions on WJEC A Level Maths 3.5 Trigonometry (A-level only), covering 3.5.1 Trigonometry (A-level only), 3.5.2 Trigonometry (A-level only), 3.5.3 Trigonometry (A-level only), 3.5.4 Trigonometry (A-level only), 3.5.5 Trigonometry (A-level only), 3.5.6 Trigonometry (A-level only), 3.5.7 Trigonometry (A-level only), and 3.5.8 Trigonometry (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.