Show that
1tanθ−tanθ≡cos2θsinθcosθ \frac{1}{\tan \theta} - \tan \theta \equiv \frac{\cos 2\theta}{\sin \theta \cos \theta} tanθ1−tanθ≡sinθcosθcos2θfor θ≠nπ2\theta \neq \frac{n\pi}{2}θ=2nπ where n∈Zn \in \mathbb{Z}n∈Z.
Solve, for 0∘≤x<90∘0^\circ \le x < 90^\circ0∘≤x<90∘, the equation
5sin2(2x−15∘)=2 5 \sin^2(2x - 15^\circ) = 2 5sin2(2x−15∘)=2giving your answers in degrees to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)
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.