Let's prove this identity by induction.
A. For #n=1# we have to check that
#(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-1#
Indeed, using identity #cos(2theta)=2cos^2(theta)-1#, we see that
#2cos(2theta)+1 = 2(2cos^2(theta)-1)+1 = 4cos^2(theta)-1 =#
#=(2cos(theta)-1)*(2cos(theta)+1)#
from which follows that
#(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-1#
So, for #n=1# our identity holds true.
B. Assume that the identity is true for #n#
So, we assume that
#(2cos(2^ntheta)+1)/(2cos(theta)+1) = Pi _(j in [0,n-1])[2cos(2^jtheta)-1]#
(symbol #Pi# is used for product)
C. Using assumption B above, let's prove the identity for #n+1#
We have to prove that from assumption B follows
#(2cos(2^(n+1)theta)+1)/(2cos(theta)+1) = Pi _(j in [0,n])[2cos(2^jtheta)-1]#
(notice that the right boundary for an index of multiplication is #n# now).
PROOF
Using an identity #cos(2x)=2cos^2(x)-1# for #x=2^ntheta#,
#2cos(2^(n+1)theta)+1 = 2cos(2*(2^n*theta))+1 =#
#= 2[2cos^2(2^ntheta)-1]+1 =#
#=4cos^2(2^ntheta)-1 =#
#=[2cos(2^ntheta)-1]*[2cos(2^ntheta)+1] #
Divide beginning and ending expressions by #[2cos(theta)+1]#, getting
#[2cos(2^(n+1)theta)+1]/[2cos(theta)+1] = #
# = [2cos(2^ntheta)-1]*[2cos(2^ntheta)+1] /[2cos(theta)+1] #
Now we use assumption B getting
#[2cos(2^(n+1)theta)+1]/[2cos(theta)+1] = #
# = [2cos(2^ntheta)-1]*Pi _(j in [0,n-1])[2cos(2^jtheta)-1] = #
# = Pi _(j in [0,n])[2cos(2^jtheta)-1] #
(notice the range of an index now is extended to #n#).
The last formula is exactly the same for #n+1# as original is for #n#. That completes the proof by induction that our formula is true for any #n#.