The double angle formulas are given as;
sin(2theta) = 2sin(theta) cos(theta)sin(2θ)=2sin(θ)cos(θ)
cos(2theta) = cos^2(theta)-sin^2(theta) = 2cos^2(theta)-1 = 1-2sin^2(theta)cos(2θ)=cos2(θ)−sin2(θ)=2cos2(θ)−1=1−2sin2(θ)
tan(2theta) = (2tan(theta))/(1-tan^2(theta))tan(2θ)=2tan(θ)1−tan2(θ)
Notice that the double angle formulas reduce the term inside the trig function by half. If we apply the appropriate double angle formula to our function, we get;
cot(4theta)-cos(2theta)cot(4θ)−cos(2θ)
=1/tan(4theta) - (cos^2(theta)-sin^2(theta))=1tan(4θ)−(cos2(θ)−sin2(θ))
=(1-tan^2(2theta))/(2tan(2theta))-cos^2(theta)+sin^2(theta)=1−tan2(2θ)2tan(2θ)−cos2(θ)+sin2(θ)
Use the double angle formula again on the tantan terms to get everything in terms of thetaθ.
(1-((2tan(theta))/(1-tan^2(theta)))^2)/(2((2tan(theta))/(1-tan^2(theta))))-cos^2(theta)+sin^2(theta)1−(2tan(θ)1−tan2(θ))22(2tan(θ)1−tan2(θ))−cos2(θ)+sin2(θ)
After some simplification, we get;
(1-tan^2(theta))/(4tan(theta)) - tan(theta)/(1-tan^2(theta))-cos^2(theta) + sin^2(theta)1−tan2(θ)4tan(θ)−tan(θ)1−tan2(θ)−cos2(θ)+sin2(θ)
