Answer to Question #86945 in Algebra for sonali mansingh

Question

Use De Moivre’s theorem to show that cos3θ = 4cos θ − 3cosθ
3
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Answer 1

De Moivre's theorem: $z^n = \cos(n\phi)+ i\sin(n\phi).$

So $z^3 = \cos(3\theta) + i\sin(3\theta)$

In the second part of derivation, we need to bring to third power this expression $(\cos(\theta)+i\sin(\theta))$

$(\cos(\theta) + i\sin(\theta))^3 = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta)+3i\cos^2(\theta)\sin(\theta) - i\sin^3(\theta) = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) + i(3\cos^2(\theta)\sin(\theta) - i\sin^3(\theta)$ Now we can equate real parts of first and second expressions:

$\cos(3\theta) = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) = \cos^3(\theta) - 3\cos(\theta)(1 - \cos^2(\theta)) = \cos^3(\theta) - 3\cos(\theta) + 3\cos^3(\theta) = 4\cos^3(\theta) - 3\cos(\theta)$