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
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.