We know that, #cos 2x=2cos^2x-1=1-2sin^2x#
From these, we get, #sin^2x=(1-cos2x)/2,...........(1)# and, #cos^2x=(1+cos2x)/2.................(2)#
Now, #cos^4x=(cos^2x)^2={(1+cos2x)/2}^2=1/4{1+2cos2x+cos^2(2x)}#
Here, to convert #cos^2(2x)# into multiple angle, we will reuse #(2),# and, put #2x# in place of #x,# to get,
#cos^2(2x)=(1+cos2(2x))/2=(1+cos4x)/2.#
Therefore, #cos^4x=1/4{1+2cos2x+(1+cos4x)/2}=1/8(3+4cos2x+cos4x).........................(3)#
Finally, we have, #cot^2x=cos^2x/sin^2x=(1+cos2x)/(1-cos2x).....(4)#
Using, #(1),(3) and (4),# we have,
#f(theta)=1/2(1-cos2theta)+6{(1+cos2theta)/(1-cos2theta)}-(3+4cos2theta+cos4theta)#
We can still go further and express #f(theta)# as a rational function, as follows :
#f(theta)={(1-cos2theta)^2+12(1+cos2theta)-2(1-cos2theta)(3+4cos2theta+cos4theta)}/{2(1-cos2theta)#
Where, #Nr.=1-2cos2theta+cos^2(2theta)+12+12cos2theta-2(3+4cos2theta+cos4theta-3cos2theta-4cos^2(2theta)-cos4thetacos2theta)#
#=13+10cos2theta+cos^2(2theta)-6-2cos2theta-2cos4theta+8cos^2(2theta)+2cos4thetacos2theta#
#=7+8cos2theta-2cos4theta+9cos^2(2theta)+cos(4theta+2theta)+cos(4theta-2theta)#
#=7+8cos2theta-2cos4theta+9{(1+cos4theta)/2}+cos6theta+cos2theta#
#=1/2(23+18cos2theta+5cos4theta+2cos6theta)#
#:. f(theta)=(23+18cos2theta+5cos4theta+2cos6theta)/(4(1-cos2theta))#
