By writing sin3θ \sin 3\theta\,sin3θ as sin(2θ+θ)\sin(2\theta + \theta)sin(2θ+θ), show that
sin3θ≡3sinθ−4sin3θ\sin 3\theta \equiv 3\sin\theta - 4\sin^3\thetasin3θ≡3sinθ−4sin3θ
Solve, for 0°≤θ≤180°0° \leq \theta \leq 180°0°≤θ≤180°, the equation
3sinθ−4sin3θ=0.43\sin\theta - 4\sin^3\theta = 0.43sinθ−4sin3θ=0.4
Give your answers to one decimal place.
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.