Mathematics · OCR H240 · Topic 1.05

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 coming soon All OCR Mathematics topics

What the specification says

Every sub-topic OCR 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. 1.05a-dSin, cos and tan, the sine and cosine rules, and radiansY13
  2. 1.05eSmall angle approximationsY13
  3. 1.05f-gGraphs and exact values of the trigonometric functionsY13
  4. 1.05h-iInverse and reciprocal trigonometric ratiosY13
  5. 1.05j-kTrigonometric identitiesY13
  6. 1.05l-nFurther trigonometric identitiesY13
  7. 1.05oTrigonometric equationsY13
  8. 1.05pProof involving trigonometric functionsY13
  9. 1.05qTrigonometric 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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