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?

This can be done with a few trig identities. We start with the double angle formula for cosine:
$$\cos(6x) = \cos(2\cdot 3x) = 2\cos^2(3x) - 1$$
Now we need to find an expression for $\cos(3x)$ before we proceed:
$$\cos(3x) = \cos(2x + x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$$
$$= (2\cos^2(x) - 1) \cos(x) - 2\sin^2(x)\cos(x)$$
$$=2\cos^3(x) - \cos(x) - 2(1-\cos^2(x))\cos(x)$$
$$=2\cos^3(x) - 3\cos(x) + 2\cos^3(x)$$
$$=4\cos^3(x) - 3\cos(x).$$
Thus,
$$\cos(6x) = 2(4\cos^3(x) - 3\cos(x))^2 - 1$$
$$=32 \cos^6(x) - 48 \cos^4(x) + 18 \cos^2(x) - 1.$$
This was accomplished using the following:
Cosine Sum Identity: $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$
Pythagorean Identity: $\sin^2(x) + \cos^2(x) = 1$
Cosine Double Angle Identity (or Cosine Sum Identity): $\cos(2x) = 2\cos^2(x) - 1$
Sine Double Angle Identity: $\sin(2x) = 2\sin(x)\cos(x)$
