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1.4 Trigonometry

1.4 Trigonometry

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Question 29
a.

Show that the equation

3cos⁡θ−2=5sin⁡θtan⁡θ 3\cos \theta - 2 = 5 \sin \theta \tan \theta 3cosθ−2=5sinθtanθ

can be written in the form

8cos⁡2θ−2cos⁡θ−5=0 8\cos^2 \theta - 2\cos \theta - 5 = 0 8cos2θ−2cosθ−5=0
[3]
b.

Hence solve, for 0≤x<π0 \le x < \pi0≤x<π,

3cos⁡2x−2=5sin⁡2xtan⁡2x 3\cos 2x - 2 = 5 \sin 2x \tan 2x 3cos2x−2=5sin2xtan2x

giving your answers, where appropriate, to 2 decimal places.

[4]
Markscheme

1.4 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.4 Trigonometry

84 exam-style questions on CCEA A Level Maths 1.4 Trigonometry, covering 1.4.1 Trigonometry, 1.4.2 Trigonometry, 1.4.3 Trigonometry, 1.4.4 Trigonometry, 1.4.5 Trigonometry, 1.4.6 Trigonometry, and 1.4 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.

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