Sorry, another one.
If you could just explain how to split up the cos4x and sin4x, that would be helpful as well!
Thank you:
cot 2x = (1+cos4x)/sin4x
Use the identities $\displaystyle \cos{2x} = 2\cos^2{x} - 1$ and $\displaystyle \sin{2x} = 2\sin{x}\cos{x}$.
$\displaystyle \frac{1 + \cos{4x}}{\sin{4x}} = \frac{1 + \cos{2\times 2x}}{\sin{2\times 2x}}$
$\displaystyle = \frac{1 + 2\cos^2{2x} - 1}{2\sin{2x}\cos{2x}}$
$\displaystyle = \frac{2\cos^2{2x}}{2\sin{2x}\cos{2x}}$
$\displaystyle = \frac{\cos{2x}}{\sin{2x}}$
$\displaystyle = \cot{2x}$.