prove that 2 Cos 2^n A + 1 / 2cos A+1 = (2 cos A -1 ) (2cos 2A -1) (2cos^2A -1)...................(2cos 2 ^2n-1 - 1)from SL...

We'll multiply both sides by 2cosA + 1 and we'll
get:

2 cos`2^(n)` A + 1 = (2cos A + 1)(2cos A - 1)(2cos 2A
- 1)...(2cos `2^(n-1)` A - 1)

We notice that the product of
the first two factors fro the right returns the difference of two
squares:

(2cos A + 1)(2cos A - 1) = 4`cos^(2)` A - 1 =
2`cos^(2)` A + 2`cos^(2)` A - 1

But 2`cos^(2)` A - 1 = cos
2A

(2cos A + 1)(2cos A - 1) = 2`cos^(2)` A + cos
2A

We'll add and subtract 1 to the
right:

(2cos A + 1)(2cos A - 1) = 2`cos^(2)` A - 1 + 1 +
cos 2A

(2cos A + 1)(2cos A - 1) = cos 2A + 1 + cos
2A

(2cos A + 1)(2cos A - 1) = 2cos 2A +
1

Therefore, instead of the product (2cos A + 1)(2cos A -
1) , we'll put the result 2cos 2A + 1.

2 cos `2^(n)` A + 1
= (2cos 2A + 1)(2cos 2A - 1)...(2cos`2^(n-1)` A - 1)

We
notice that the product of the first two factors fro the right returns the difference of
two squares:

(2cos 2A + 1)(2cos 2A - 1) = 4`cos^(2)` 2A - 1
= 2`cos^(2)` 2A + cos 4A

(2cos 2A + 1)(2cos 2A - 1) = 2cos
4A + 1 = 2cos `2^(2)` A + 1

This result will be multiplied
by (2cos `2^(2)` A - 1 ):

(2cos `2^(2)` A - 1)(2cos`2^(2)`
A + 1) = 2cos`2^(3)` A + 1

Each result will be multiplied
by the conjugate factor till we'll reach to the result 2cos`2^(n-1)` A +
1.

(2cos`2^(n-1)` A + 1)(2cos`2^(n-1)` A - 1) = 2cos
`2^(n)` A + 1

We notice that the result we've came up to
the right side represents the same expression with the one from the left
side.

Therefore, the identity is
verified!