Mathematics · Pearson Edexcel 9MA0 · Topic 5

Trigonometry

Sine and cosine rules, radians, trigonometric graphs, identities, the addition and double angle formulae, and solving trigonometric equations.

  • 9 sub-topics
  • Taught in Year 13
  • 159 practice questions
  • 9 animated lessons
Practice questions (159) Lessons Worksheets All Edexcel Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 5.1Trigonometry, radians, arc length and sector areaY13
  2. 5.2Small angle approximationsY13
  3. 5.3Graphs, symmetries and periodicity of sin, cos and tanY13
  4. 5.4Reciprocal and inverse trigonometric functionsY13
  5. 5.5Trigonometric identitiesY13
  6. 5.6Double angle and compound angle formulaeY13
  7. 5.7Solve trigonometric equations in a given intervalY13
  8. 5.8Proofs involving trigonometric identitiesY13
  9. 5.9Trigonometric functions in contextY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 159 questions in this topic, free, no account needed.

Trigonometry

Easy

The sine rule · Year 12 · 3 marks

In triangle ABCABC, A=40∘A = 40^\circ, B=65∘B = 65^\circ and a=8a = 8 cm.
  1. Find bb.
    (3)
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  1. Sine rule: each side over the sine of the opposite angle.bsin⁡65∘\dfrac{b}{\sin 65^\circ}=8sin⁡40∘{}= \dfrac{8}{\sin 40^\circ}
  2. bb=8sin⁡65∘sin⁡40∘{}= \dfrac{8\sin 65^\circ}{\sin 40^\circ}
  3. =11.3= 11.3 cm

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Trigonometry

Not needed for Year 12Medium

A quadratic in cosine · Year 13 · 5 marks

  1. Solve 2sin2⁡x=3cos⁡x2\sin^2 x = 3\cos x for 0∘≤x<360∘0^\circ \le x < 360^\circ.
    (5)
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  1. Use sin2⁡x=1−cos2⁡x\sin^2 x = 1 - \cos^2 x so there is only one trig function.2(1−cos2⁡x)=3cos⁡x2(1 - \cos^2 x) = 3\cos x
  2. Rearrange into a quadratic in cos⁡x\cos x.2cos2⁡x+3cos⁡x−2=02\cos^2 x + 3\cos x - 2 = 0
  3. Factorise as you would 2c2+3c−22c^2 + 3c - 2.(2cos⁡x−1)(cos⁡x+2)=0(2\cos x - 1)(\cos x + 2) = 0
  4. Cosine is always between −1 and 1.cos⁡x=12\cos x = \tfrac{1}{2} (cos⁡x=−2\cos x = -2 has no solutions)
  5. Cosine is positive in the first and fourth quadrants.x=60∘, 300∘x = 60^\circ,\ 300^\circ

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Trigonometry

Not needed for Year 12Hard

Solving with the R form · Year 13 · 5 marks

  1. Solve 3sin⁡x+4cos⁡x=23\sin x + 4\cos x = 2 for 0∘≤x<360∘0^\circ \le x < 360^\circ.
    (5)
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  1. Use R=5R = 5, α=53.13∘\alpha = 53.13^\circ.5sin⁡(x+53.13∘)=25\sin(x + 53.13^\circ) = 2
  2. sin⁡(x+53.13∘)=0.4\sin(x + 53.13^\circ) = 0.4
  3. The range for x+53.13∘x + 53.13^\circ is 53.13∘53.13^\circ to 413.13∘413.13^\circ; 23.58∘23.58^\circ is too small.x+53.13∘=156.42∘x + 53.13^\circ = 156.42^\circ or 383.58∘383.58^\circ
  4. x=103.3∘x = 103.3^\circ or x=330.4∘x = 330.4^\circ

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