#f(theta)=2cos^2(2 theta) + sin^2(4theta)-5tan^4theta =(2 cos(2theta)^2 cos(theta)^4 + sin(4theta)^2 cos(theta)^4 - 5 cdot sin(theta)^4)/cos(theta)^4 #
#f(theta)# is an even function such that
#int_{-pi}^{pi}f(theta)cos((2k+1)theta)d theta = 0# for #k = 0,1,2,3,...#
Defining #g(theta)=(sum_{k=0}^6a_{2k}cos(2ktheta))/cos(theta)^4#
Choosing #theta_k = (k pi)/12# for #k=0,1,2,cdots,6# and solving the equations
#f(theta_k)-g(theta_k)# for #k = 1,2,cdots,6# and solving for #a_{2k}#
we have
#(a_0= -5/4, a_2=7/2, a_4=-3/32,a_6=1/8, a_8=-1/8,a_{10}= -1/8,a_{12} =-1/32)#
giving
#cos(theta)^4g(theta)=-5/4 + 7/2 cos(2theta) - 3/32 cos(4theta) + 1/8 cos(6theta)-1/8 cos(8theta) - 1/8 cos(10theta) - 1/32cos(12theta)#
Using the same procedure we determine
#cos(theta)^4=3/8 + 1/2 Cos(2theta) + 1/8 Cos(4theta)#
