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1.5 Trigonometry

1.5 Trigonometry

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Question 158

In a precision laser alignment system, the deviation ratio DDD for a small angle of incidence θ\thetaθ (measured in radians, where θ≠0\theta \neq 0θ=0) is modeled by the function:

D(θ)=3θsin⁡(4θ)1−cos⁡(5θ) D(\theta) = \frac{3\theta \sin(4\theta)}{1 - \cos(5\theta)} D(θ)=1−cos(5θ)3θsin(4θ)​

Using small angle approximations, show that for small values of θ\thetaθ, D(θ)≈KD(\theta) \approx KD(θ)≈K where KKK is a constant to be determined.

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1.5 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.5 Trigonometry

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.

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