Derive cos(3θ)=(4cosθ)^3−3cosθ

Question

I'm having trouble with the following derivation:

Q: We can use Euler's Theorem (e^iθ=cosθ+isinθ), where e is the base of the natural logarithms, and i = sqrt(-1), together with the binomial theorem as above, to derive a number of trigonometric identities. E.g., if we consider (e^iθ)^2, we can evaluate it two different ways. First, we can multiply exponents, obtaining e^(i⋅2θ) and then applying Euler's formula to get cos(2θ)+isin(2θ), or we can apply Euler's formula to the inside, obtaining (cosθ+isinθ)^2, which we then evaluate via the binomial theorem:

cos(2θ)+isin(2θ)=(cosθ+isinθ)^2 =(cosθ)^2+2icosθsinθ+i^2(sinθ)^2 =(cosθ)^2+2icosθsinθ−(sinθ)^2

Equating real and imaginary parts gives us

cos(2θ)=(cosθ)^2−(sinθ)^2 sin(2θ)=2cosθsinθ

We can then rewrite the first of these identities, using 1=(sinθ)^2+(cosθ)^2 to get (cosθ)^2=1−(sinθ)^2 , whence the familiar

cos(2θ)=1−2(sinθ)^2

Use this same approach to show cos(3θ)=(4cosθ)^3−3cosθ.

A: This is my work so far:

e^(i⋅3θ) = cos(3θ) + isin(3θ) = (cosθ)^3 + 3i(cosθ)^2sinθ - 3cosθ(sinθ)^2 - i(sinθ)^3

cos(3θ) = (cosθ)^3 - 3(sinθ)^2cosθ

sin(3θ) = 3sinθ(cosθ)^2 - (sinθ)^3

But now I'm unsure how to get cos(3θ)=(4cosθ)^3−3cosθ from what I've derived.

Answer 1

Continue from your answer,

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

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

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

$= \cos^3 \theta - 3\cos \theta + 3\cos^3\theta$

$= 4\cos^3 \theta - 3\cos \theta$

Similarly you can change $\cos^2\theta = 1 - \sin^2\theta$ to get the formula of $\sin(3\theta)$.

Answer 2

I believe your claim is actually incorrect. You should have the identity \begin{align} \cos 3\theta = 4\cos^3\theta -3\cos\theta \end{align} not \begin{align} \cos 3\theta = (4\cos\theta)^3 -3\cos\theta. \end{align} Observe \begin{align} e^{i3\theta} =&\ (\cos\theta + i\sin\theta)^3 = \cos^3\theta+3(i\sin\theta)\cos^2\theta+3(i\sin\theta)^2\cos\theta +(i\sin\theta)^3\\ =&\ \cos^3\theta -3\sin^2\theta \cos\theta+i(3\sin\theta\cos^2\theta-\sin^2\theta). \end{align} Hence taking the real part yields \begin{align} \cos 3\theta =&\ \cos^3\theta -3\sin^2\theta \cos \theta\\ =&\ \cos^3\theta - 3(1-\cos^2\theta)\cos \theta\\ =&\ 4\cos^3\theta -3\cos\theta. \end{align}