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.)
Practise AQA A Level Maths 1.8 E: Trigonometry with exam-style questions for A Level Maths. 239 questions covering 1.8.1 Trigonometric definitions, rules and radians, 1.8.2 Small angle approximations (A-level only), 1.8.3 Trigonometric functions and exact values, 1.8.4 Reciprocal and inverse trigonometric functions (A-level only), 1.8.5 Trigonometric identities, 1.8.6 Compound and double angle formulae (A-level only), 1.8.7 Solving trigonometric equations, 1.8.8 Proofs with trigonometric identities, and 1.8.9 Trigonometry in context, matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
