How do you simplify #cos4theta# to trigonometric functions of a unit #theta#?

Here are a couple of methods...

Method 1

Using:

#cos 2 theta = cos^2 theta - sin^2 theta#

#cos^2 theta + sin^2 theta = 1#

So:

#cos 2 theta = cos^2 theta - sin^2 theta = cos^2 theta - (1 - cos^2 theta) = 2cos^2 theta - 1#

So:

#cos 4 theta = 2cos^2 2 theta - 1#

#color(white)(cos 4 theta) = 2(2cos^2 theta - 1)^2 - 1#

#color(white)(cos 4 theta) = 2(4cos^4 theta - 4cos^2 theta + 1) - 1#

#color(white)(cos 4 theta) = 8cos^4 theta - 8cos^2 theta + 1#

Method 2

Using de Moivre's theorem:

#(cos theta + i sin theta)^n = cos n theta + i sin n theta#

and

#cos^2 theta + sin ^2 theta = 1#

So:

#cos 4 theta + i sin 4 theta#

#= (cos theta + i sin theta)^4#

#= cos^4 theta + 4i cos^3 theta sin theta - 6 cos^2 theta sin^2 theta - 4i cos theta sin^3 theta + sin^4 theta#

Equating real parts:

#cos 4 theta = cos^4 theta - 6 cos^2 theta sin^2 theta + sin^4 theta#

#color(white)(cos 4 theta) = cos^4 theta - 6 cos^2 theta (1-cos^2 theta) + (1-cos^2 theta)^2#

#color(white)(cos 4 theta) = cos^4 theta - 6 cos^2 theta + 6 cos^4 theta + 1-2cos^2 theta+cos^4 theta#

#color(white)(cos 4 theta) = 8cos^4 theta - 8 cos^2 theta + 1#