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1.5 Trigonometry

1.5 Trigonometry

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Question 164

A deep-space communication laser's alignment angle ψ \psi\,ψ is modeled by the equation

6sin⁡ψcos⁡ψcos⁡ψ+sin⁡ψ=(5+sec⁡2ψ)(cos⁡ψ−sin⁡ψ) \frac{6 \sin \psi \cos \psi}{\cos \psi + \sin \psi} = (5 + \sec 2\psi)(\cos \psi - \sin \psi) cosψ+sinψ6sinψcosψ​=(5+sec2ψ)(cosψ−sinψ)
a.

Show that this equation can be simplified to the form

6sin⁡2ψ−10cos⁡2ψ=2 6 \sin 2\psi - 10 \cos 2\psi = 2 6sin2ψ−10cos2ψ=2
[6]
b.

For an observation window 0<t<π0 < t < \pi0<t<π, determine the values of the signal time t t\,t satisfying

6sin⁡tcos⁡tcos⁡t+sin⁡t=(5+sec⁡2t)(cos⁡t−sin⁡t) \frac{6 \sin t \cos t}{\cos t + \sin t} = (5 + \sec 2t)(\cos t - \sin t) cost+sint6sintcost​=(5+sec2t)(cost−sint)

giving your answers to 3 significant figures.

[4]
Markscheme

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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