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1.8 E: Trigonometry

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Question 21

The displacement ddd of a mechanical oscillator is modeled by the equation

d=0.9cos⁡(1.5t)−4.0sin⁡(1.5t) d = 0.9 \cos(1.5t) - 4.0 \sin(1.5t) d=0.9cos(1.5t)−4.0sin(1.5t)

where ddd is measured in millimetres and ttt is time in seconds. To analyze the peak amplitude of the oscillation, the expression is rewritten in the form Rcos⁡(1.5t+α)R \cos(1.5t + \alpha)Rcos(1.5t+α), where R>0R > 0R>0.

Find the value of RRR.

Circle the correct answer from the options below:

3.13.13.1 \qquad 4.054.054.05 \qquad 4.14.14.1 \qquad 4.94.94.9

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Markscheme

1.8 E: Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.8 E: Trigonometry

317 exam-style questions on AQA A Level Maths 1.8 E: Trigonometry, 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. Each one has a worked solution and a mark scheme showing where the marks go.

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