Express 2sinx−3cosx2 \sin x - 3 \cos x2sinx−3cosx in the form Rsin(x−α)R \sin (x - \alpha)Rsin(x−α), where R>0 R > 0\,R>0 and 0≤α≤π2\displaystyle 0 \leq \alpha \leq \frac{\pi}{2}0≤α≤2π
Hence find the greatest value of (2sinx−3cosx)2(2 \sin x - 3 \cos x)^2(2sinx−3cosx)2 and find, the smallest positive value of x x\,x for which this maximum occurs
Solve, for 0≤θ≤2π0 \leq \theta \leq 2\pi0≤θ≤2π,
2sinx−3cosx=1 2 \sin x - 3 \cos x = 1 2sinx−3cosx=1Give your answers to 3 decimal places.
137 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.