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.)
317 exam-style questions on OCR A Level Maths 1.5 Trigonometry, covering 1.5.1 Definitions for all arguments, 1.5.2 Sine and cosine rules, 1.5.3 Area of a triangle, 1.5.4 Radian measure (A-level only), 1.5.5 Small angle approximations (A-level only), 1.5.6 Graphs of basic trigonometric functions, 1.5.7 Exact values in radians (A-level only), 1.5.8 Reciprocal and inverse trigonometric ratios (A-level only), 1.5.9 Graphs of reciprocal and inverse functions (A-level only), 1.5.10 Trigonometric identities, 1.5.11 Further trigonometric identities (A-level only), 1.5.12 Double angle and compound angle formulae (A-level only), 1.5.13 Geometrical proofs of formulae (A-level only), 1.5.14 Harmonic form Rcos / Rsin (A-level only), 1.5.15 Trigonometric equations, 1.5.16 Proof involving trigonometric functions (A-level only), and 1.5.17 Trigonometric functions in context. Each one has a worked solution and a mark scheme showing where the marks go.