de Moivre's theorem

Please help me solve this /sci/.


If z = cos x + i sin x, show that
    z^n + z^-n = 2 cos nx, and
    z^n - z^-n = 2i sin nx.

Hence deduce that
    cos^6 x + sin^6 x = (3cos 4x +5)/8

z = cos x + i sin x = e^ix
z^n + z^-n = e^inx + e^-inx = cos nx + i sin nx + cos -nx + i sin -nx
= 2 cos nx

Similarly for the second part

too lazy for last part

:(

is the second part [3cos(4x+5)]/8 or [3cos(4x)+5]/8?

cos^6(x)
=[(z+1/z)/2]^6
=(z^6 + 6z^4 + 15z^2 + 20 + 15z^-2 + 6z^-4 + z-6) / 2^6
=[(z^6 + z^-6) + 6(z^4 + z^-4) + 15(z^2 + z^-2) + 20] /64
= (cos(6x) + 6cos(4x) + 15cos(2x) + 10)/32

sin^6(x)
= [(z-1/z)/2i]^6
=(z^6 - 6z^4 + 15z^2 - 20 + 15z^-2 - 6z^-4 + z-6) / 2i^6
=[(z^6 + z^-6) - 6(z^4 + z^-4) + 15(z^2 + z^-2) - 20] /-64
= (-cos(6x) + 6cos(4x) - 15cos(2x) + 10)/32

sin^6(x) + cos^6(x)
= (cos(6x) + 6cos(4x) + 15cos(2x) + 10)/32 + (-cos(6x) + 6cos(4x) - 15cos(2x) + 10)/32
= (12cos(4x) + 20)/32
= [3cos(4x)+5)/8