Question

Prove that :-(2cos A/2 +1) ( 2 cos A/2 -1) (2cos A -1) = cos2A+1​

Answer 1

Step-by-step explanation:

refer to the attchment

Answer 2

Answer:

hey mate here you go with the answer !!!!

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

2 cos2^(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) = 4cos^(2) A - 1 = 2cos^(2) A + 2cos^(2) A - 1

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

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

We'll add and subtract 1 to the right:

(2cos A + 1)(2cos A - 1) = 2cos^(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)...(2cos2^(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) = 4cos^(2) 2A - 1 = 2cos^(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)(2cos2^(2) A + 1) = 2cos2^(3) A + 1

Each result will be multiplied by the conjugate factor till we'll reach to the result 2cos2^(n-1) A + 1.

(2cos2^(n-1) A + 1)(2cos2^(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!