<s>[INST] Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?
 [/INST] It might be tedious, but really there's no reason you shouldn't get to the answer.
Taking (for reduced finger strain) $c=\cos x$ and $s=\sin x$, and noting that $s^2 = 1-c^2$
$$\begin{align} \cos 6x &= \Re((c+is)^6) \\
&= \Re(c^6+6ic^5s-15c^4s^2-20ic^3s^3 +15c^2s^4+6ics^5-s^6)\\
&= c^6-15c^4s^2+15c^2s^4-s^6\\
&= c^6-15c^4(1-c^2)+15c^2(1-2c^2+c^4)-(1-3c^2+3c^4-c^6)\\
&= c^6(1+15+15+1)+c^4(-15-30-3) + c^2(15+3) -1\\
&= 32c^6-48c^4+18c^2-1\\
&= 32\cos^6 x-48\cos^4 x+18\cos^2 x-1\\
\end{align}$$
as required.
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