Question

Use de Moivre’s Formula to find integers A, B, C such that
cos 4θ = Acos^4 θ +Bcos^2 θ +C.

Answer 1

\[\cos (4\theta) = \mathbb{Real} ~ e^{i 4\theta} = \mathbb{Real} ~ (\cos \theta + i \sin \theta)^4 \]

Answer 2

expand it using binomial thm, take the real part and compare coefficients

Answer 3

What do you mean by take the real part?

Answer 4

take the terms that dont have "\(i\)" in them

Answer 5

for example : Real part of \(a + ib\) is \(a\)

Answer 6

\[\cos (4\theta) = \mathbb{Real} ~ e^{i 4\theta} = \mathbb{Real} ~ (\cos \theta + i \sin \theta)^4 \\ =\mathbb{Real} ~ \cos^4 \theta + 4 \cos^3 \theta (i \sin \theta ) + 6 \cos^2 \theta (i \sin \theta)^2 + 4 \cos \theta (i \sin \theta )^3 + (i \sin \theta)^4 \\ = \mathbb{Real} ~ \cos^4 \theta + 4i \cos^3 \theta\sin \theta - 6 \cos^2 \theta sin^2 \theta - 4i \cos \theta \sin^3 \theta + \sin^4 \theta \]

Answer 7

take the real part of above ^

Answer 8

Thank you :)

Answer 9

np :) u will need to change everything into cosines and compare

Answer 10

\[\cos (4\theta) = \mathbb{Real} ~ e^{i 4\theta} = \mathbb{Real} ~ (\cos \theta + i \sin \theta)^4 \\ =\mathbb{Real} ~ \cos^4 \theta + 4 \cos^3 \theta (i \sin \theta ) + 6 \cos^2 \theta (i \sin \theta)^2 + 4 \cos \theta (i \sin \theta )^3 + (i \sin \theta)^4 \\ = \mathbb{Real} ~ \cos^4 \theta + 4i \cos^3 \theta\sin \theta - 6 \cos^2 \theta \sin^2 \theta - 4i \cos \theta \sin^3 \theta + \sin^4 \theta \\ = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta \]

Answer 11

so

\[ \cos 4\theta = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta = A\cos^4 \theta +B\cos^2 \theta +C \]

Answer 12

rewrite sin^2 as 1-cos^2 and get it into the right hand side form, then compare