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Watch the free lessonDe Moivre's theorem, the whole lesson, no account neededFree A worked example from the lesson on De Moivre's theorem: expressing cos 3θ in powers of cos θ.

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Complex numbers

Not needed for Year 12Hard

Multiple angles · Year 13 · 4 marks

  1. Use de Moivre's theorem to show that cos⁡3θ\cos 3\theta=4cos3⁡θ{}= 4\cos^{3}\theta−3cos⁡θ{}- 3\cos \theta.
    (4)
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  1. (cos⁡θ+isin⁡θ)3(\cos\theta + \mathrm{i}\sin\theta)^3=cos⁡3θ{}= \cos 3\theta+isin⁡3θ{}+ \mathrm{i}\sin 3\theta
  2. Binomial expansion, real part.Re:cos3⁡θ\text{Re}: \cos^3\theta−3cos⁡θsin2⁡θ{}- 3\cos\theta\sin^2\theta
  3. Use sin2⁡θ\sin^{2}\theta=1−cos2⁡θ{}= 1 - \cos^{2}\theta.=4cos3⁡θ−3cos⁡θ= 4\cos^3\theta - 3\cos\theta

Total4

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