By writing sin3θ \sin 3\theta\,sin3θ as sin(2θ+θ)\sin (2\theta + \theta)sin(2θ+θ) show that 3sinθ−4sin3θ=sin3θ3\sin\theta - 4\sin^3 \theta = \sin 3\theta3sinθ−4sin3θ=sin3θ
Verify, for 0≤θ≤1800 \leq \theta \leq 1800≤θ≤180, that the solutions of the equation,
3sinθ−4sin3θ=0.5 3\sin\theta - 4\sin^3 \theta = 0.5 3sinθ−4sin3θ=0.5are
θ=10.0∘, 50.0∘, 130.0∘, 170.0∘ \theta = 10.0^\circ,\ 50.0^\circ,\ 130.0^\circ,\ 170.0^\circ θ=10.0∘, 50.0∘, 130.0∘, 170.0∘Give your answers to 1 decimal place.
348 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.