<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] the easier way to do this is to derive the chebyshev recurrence relation for $$a_n = \cos nt, x = \cos t$$ by the sum to product formula. that is, $$a_{n+2} +a_n = \cos (nt+2t) + \cos(nt) = 2\cos t \cos(nt) = 2xa_n, a_0 = 1, a_1 = x. $$
now we can compute $$a_2 = 2x^2 - 1, a_3 = 2x(2x^2-1) - x=4x^3 - 3x,\\a_4=8x^4-6x^2-(2x^2-1) = 8x^4-8x^2 + 1\\a_5 = 16x^5-16x^3+2x-(4x^3-3x) = 16x^5-20x^3+5x\\a_6 = 32x^6-40x^4+10x^2-(8x^4-8x^2+1)=32x^6-48x^4+18x^2-1. $$
</s>