# Derive $\cos(3\theta)=(4\cos\theta)^3 − 3\cos\theta$

I'm having trouble with the following derivation:

Q: We can use Euler's Theorem ($$e^{i\theta} = \cos\theta + i\sin\theta$$), 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 $$(e^{i\theta})^2$$, we can evaluate it two different ways. First, we can multiply exponents, obtaining $$e^{i \, 2\theta}$$ and then applying Euler's formula to get $$\cos(2\theta) + i \sin(2\theta)$$, or we can apply Euler's formula to the inside, obtaining $$(\cos\theta + i \sin\theta)^2$$, which we then evaluate via the binomial theorem: \begin{align} \cos(2\theta) + i \sin(2\theta) &= (\cos\theta + i \sin\theta)^2 \\ &= (\cos\theta)^2 + 2 i \cos\theta \sin\theta + i^2 (\sin\theta)^2 \\ &=(\cos\theta)^2 + 2 i \cos\theta \sin\theta − (\sin\theta)^2 \end{align} Equating real and imaginary parts gives us \begin{align} \cos(2\theta) &= (\cos\theta)^2 − (\sin\theta)^2 \\ \sin(2\theta) &= 2 \cos\theta \sin\theta \end{align} We can then rewrite the first of these identities, using $$1=(\sin\theta)^2+(\cos\theta)^2$$ to get $$(\cos\theta)^2=1−(\sin\theta)^2$$, whence the familiar

$$\cos(2\theta)=1−2(\sin\theta)^2$$

Use this same approach to show $$\cos(3\theta)=(4\cos\theta)^3−3\cos\theta$$.

A: This is my work so far: \begin{align} e^{i \, 3\theta} &= \cos(3\theta) + i \sin(3\theta) = (\cos\theta)^3 + 3 i (\cos\theta)^2 \sin\theta - 3\cos\theta (\sin\theta)^2 - i(\sin\theta)^3 \\ \cos(3\theta) &= (\cos\theta)^3 - 3(\sin\theta)^2 \cos\theta \\ \sin(3\theta) &= 3\sin\theta(\cos\theta)^2 - (\sin\theta)^3 \end{align} But now I'm unsure how to get $$\cos(3\theta) = (4 \cos\theta)^3 − 3\cos\theta$$ from what I've derived.

• cos(3θ) = (cosθ)^3 - 3 (sinθ)^2 cosθ Hint: $\sin^2 \theta = 1 - \cos^2 \theta$
– dxiv
Commented Feb 3, 2017 at 2:14

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

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

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

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

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

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

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}