<s>[INST] Prove $\cos 3\theta = 4 \cos^3\theta − 3 \cos \theta$ $\cos 3θ = 4 \cos^3 θ − 3 \cos θ$
Here's my attempt. Is it correct? Thanks!
$\cos(3θ)$
$= \cos(2θ + θ)$
$= \cos(2θ)\cos(θ) - \sin(2θ)\sinθ$
$= (2\cos^2θ - 1)\cosθ - (2\sinθ\cosθ)\sinθ$
$= 2\cos^3θ - \cosθ - 2\sin^2θ\cosθ$
$= 2\cos^3θ - \cosθ - 2(1 - \cos^2θ)\cosθ$
$= 2\cos^3θ - \cosθ - (2\cosθ - 2\cos^3θ)$
$= 4\cos^3θ - 3\cosθ $
 [/INST] Almost good. You made a mistake on the second last line: the $2$ should have been distributed to both the $1$ and the $\cos^2\theta$
Instead of $2\cos^3\theta-\cos\theta-(2\cos\theta-\cos ^3\theta)$, it should be $2\cos^3\theta-\cos\theta-(2\cos\theta-2\cos^3\theta)$
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