Please help me in verifying this identity!
1+cos(2theta)+cos(4theta)+cos(6theta)=4cos(theta)c os(2theta)cos(3theta)
I'm stuck and can't seem to figure it out! Please show full work/steps so I can understand the process.. Thanks!

It's a bit of a bear, but one technique is to convert each of the \cos(n \theta) factors into its most basic expression in terms of \cos(\theta). For example:

cos(2x) = 2 cos^2x -1
cos(3x) = cos(x+2x) = cosxcos2x - sinxsin2x = cosx(2cos^2x-1)-sinx(2sinxcosx) = 2 cos^3x-cosx -2(1-cos^2x)cosx = 4 cos^3x-3cosx.

cos(4x) = cos(2(2x)) = 2 cos^2(2x)-1, etc...
cos(6x) = cos(2(3x)) =2 cos^2(3x)-1, etc...

Then set the left hand side against the right hand side and it works out.