Monday, November 10, 2014

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


But 2 A - 1 = cos
2A


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


We'll add and subtract 1 to the
right:


(2cos A + 1)(2cos A - 1) = 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 A + 1
= (2cos 2A + 1)(2cos 2A - 1)...(2cos 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 2A - 1
= 2 2A + cos 4A


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


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


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


Each result will be multiplied
by the conjugate factor till we'll reach to the result 2cos A +
1.


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

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