Let's prove this identity by induction.
A. For n=1n=1 we have to check that
(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-12cos(2θ)+12cos(θ)+1=2cos(θ)−1
Indeed, using identity cos(2theta)=2cos^2(theta)-1cos(2θ)=2cos2(θ)−1, we see that
2cos(2theta)+1 = 2(2cos^2(theta)-1)+1 = 4cos^2(theta)-1 =2cos(2θ)+1=2(2cos2(θ)−1)+1=4cos2(θ)−1=
=(2cos(theta)-1)*(2cos(theta)+1)=(2cos(θ)−1)⋅(2cos(θ)+1)
from which follows that
(2cos(2theta)+1)/(2cos(theta)+1) = 2cos(theta)-12cos(2θ)+12cos(θ)+1=2cos(θ)−1
So, for n=1n=1 our identity holds true.
B. Assume that the identity is true for nn
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.
