<s>[INST] Complex numbers and trig identities: $\cos(3\theta) + i \sin(3\theta)$ Using the equally rule $a + bi = c + di$ and trigonometric identities how do I make...
$$\cos^3(\theta) - 3\sin^2(\theta)\ \cos(\theta) + 3i\ \sin(\theta)\ \cos^2(\theta) - i\ \sin^3(\theta)=
\cos(3\theta) + i\ \sin(3\theta)$$
Apparently it's easy but I can't see what trig identities to substitute
PLEASE HELP!
 [/INST] $(\cos x+i\sin x)^3=(\cos 3x+i\sin 3x)$(by De Moivre's theorem)
But $(\cos x+i\sin x)^3=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i\cos^2 x\sin x-3\cos x\sin^2x=\cos^3x+i^3\sin^3x+3\cos x i \sin x(\cos x+i\sin x)=\cos^3 x-i\sin^3x+3i(1-\sin^2x) \sin x-3\cos x(1-\cos^2 x)=4\cos ^3x-3\cos x+i(3\sin x-4\sin^3x)$
Now by equating we have, $\cos 3x=4\cos ^3x-3\cos x$ and $\sin 3x=3\sin x-4\sin^3x$
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