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7.3PureStretch

The Double-Angle Formulae

Setting B = A in the addition formulae gives the double-angle formulae. The cosine version has three equivalent forms — choose the one that suits the problem.

28 min Trigonometry and Modelling
Edexcel A level Maths: 7.3 The Double Angle FormulaeCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Derive the double-angle formulae
  • Know the three forms of cos 2A
  • Use them to simplify and solve
  • Apply them to halve powers of sin/cos
1

The formulae

Putting B=AB=A into the addition formulae:

sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
Sine double angle.
cos⁡2A=cos⁡2A−sin⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A
Three forms of cosine.
tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}
Tangent double angle.
2

Choosing the cos form

The forms 2cos⁡2A−12\cos^2 A - 1 and 1−2sin⁡2A1 - 2\sin^2 A are ideal for rewriting cos⁡2A\cos^2 A or sin⁡2A\sin^2 A — useful in integration too.

Example 1Show cos⁡2A=12(1+cos⁡2A)\cos^2 A = \tfrac12(1 + \cos 2A).
1From cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1.
2Rearrange: cos⁡2A=12(1+cos⁡2A)\cos^2 A = \tfrac12(1 + \cos 2A).
Answer
cos⁡2A=12(1+cos⁡2A)\cos^2 A = \tfrac12(1 + \cos 2A)

Tip — Pick the cos 2A form whose “leftover” function matches what you want to keep.

Formula recap

sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
Sine.
cos⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A
Cosine (two power-reducing forms).
tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A = \tfrac{2\tan A}{1 - \tan^2 A}
Tangent.

Common mistakes to avoid

Writing sin 2A = 2 sin A.
sin 2A = 2 sin A cos A.
Forgetting cos 2A has three forms.
cos²−sin², 2cos²−1, and 1−2sin² are all valid.

Key takeaways

  • sin 2A = 2 sinA cosA.
  • cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A.
  • tan 2A = 2tanA/(1 − tan²A).

Test yourself

Answer a few questions, then review anything you missed.

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Answer each question before opening its answer. This self-check does not record a quiz score.

  1. sin⁡2A\sin 2A equals:

    • 2sin⁡Acos⁡A2\sin A\cos A
    • 2sin⁡A2\sin A
    • sin⁡2A\sin^2 A
    • cos⁡2A−sin⁡2A\cos^2 A - \sin^2 A
    Show answer and explanation

    Answer: 2sin⁡Acos⁡A2\sin A\cos A

    From sin(A+A).

  2. Which is NOT a form of cos⁡2A\cos 2A?

    • 2sin⁡Acos⁡A2\sin A\cos A
    • 2cos⁡2A−12\cos^2 A - 1
    • 1−2sin⁡2A1 - 2\sin^2 A
    • cos⁡2A−sin⁡2A\cos^2 A - \sin^2 A
    Show answer and explanation

    Answer: 2sin⁡Acos⁡A2\sin A\cos A

    That is sin 2A.

  3. cos⁡2A\cos^2 A in terms of cos⁡2A\cos 2A is:

    • 12(1+cos⁡2A)\tfrac12(1 + \cos 2A)
    • 12(1−cos⁡2A)\tfrac12(1 - \cos 2A)
    • 1+cos⁡2A1 + \cos 2A
    • cos⁡2A\cos 2A
    Show answer and explanation

    Answer: 12(1+cos⁡2A)\tfrac12(1 + \cos 2A)

    From 2cos²−1.

  4. tan⁡2A\tan 2A equals:

    • 2tan⁡A1−tan⁡2A\tfrac{2\tan A}{1 - \tan^2 A}
    • 2tan⁡A2\tan A
    • tan⁡A1−tan⁡A\tfrac{\tan A}{1 - \tan A}
    • tan⁡2A\tan^2 A
    Show answer and explanation

    Answer: 2tan⁡A1−tan⁡2A\tfrac{2\tan A}{1 - \tan^2 A}

    From tan(A+A).

  5. sin⁡2A\sin^2 A in terms of cos⁡2A\cos 2A is:

    • 12(1−cos⁡2A)\tfrac12(1 - \cos 2A)
    • 12(1+cos⁡2A)\tfrac12(1 + \cos 2A)
    • 1−cos⁡2A1 - \cos 2A
    • cos⁡2A\cos 2A
    Show answer and explanation

    Answer: 12(1−cos⁡2A)\tfrac12(1 - \cos 2A)

    From 1−2sin².

Next step

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