Answer to Question #91337 in Algebra for Ra

Question

Use DeMoivre's theorem to show that cos3θ=4cos³θ-3cosθ

Answer 1

De Moivre's theorem states that for any real number θ and integer n it holds that

$\big(\cos(\theta) + i \sin(\theta)\big)^n = \cos(n\theta) + i\sin(n\theta).$

Let n = 3, then

$\big(\cos(\theta) + i \sin(\theta)\big)^3 = \cos(3\theta) + i\sin(3\theta).$

Then use the Binomial theorem to expand the terms in the brackets:

$\cos ^3(\theta )+3 i \sin (\theta ) \cos ^2(\theta )-3 \sin ^2(\theta ) \cos (\theta )-i \sin ^3(\theta ) = \cos(3\theta) + i\sin(3\theta).$

Thus, equating the real and imaginary parts, we get

$\cos ^3(\theta )-3 \sin ^2(\theta ) \cos (\theta ) = \cos(3\theta),$ $3 \sin (\theta ) \cos ^2(\theta )-\sin ^3(\theta ) = \sin(3\theta).$

Then

$\cos(3\theta) = \cos ^3(\theta )-3 \big(1-\cos^2(\theta )\big) \cos (\theta ) = 4\cos ^3(\theta )-3\cos (\theta ).$