Question

Prove :-\[{\frac{2\cos2^n \theta}{2\cos \theta +1}=(2\cos \theta -1)(2\cos 2 \theta-1)(2\cos 2^2 \theta-1)....(2\cos 2^{n-1} \theta-1)}\]

Answer 1

@Michele_Laino sir

Answer 2

It's easier to prove the equivalent statement
\[{2\cos(2^n \theta)
\\ =(2\cos \theta +1) (2\cos \theta -1)(2\cos 2 \theta-1)(2\cos 2^2 \theta-1)....(2\cos 2^{n-1} \theta-1)} \]We will be using the identity \( (2 \cos u + 1)(2 \cos u - 1) = ( 2 \cos2u + 1 ) \)

-Proof-\[ (2\cos \theta +1) (2\cos \theta -1)(2\cos 2 \theta-1)(2\cos 2^2 \theta -1)...(2\cos 2^{n-1} \theta-1)
\\ =(2\cos 2\theta +1) (2\cos 2 \theta-1)(2\cos 2^2 \theta-1)...(2\cos 2^{n-1} \theta-1)
\\ =(2\cos 2^2\theta +1) (2\cos 2^2 \theta-1)...(2\cos 2^{n-1} \theta-1)
\\ =(2\cos 2^3\theta +1)...(2\cos 2^{n-1} \theta-1)
\\ =... = (2\cos 2^{n-1} \theta +1) (2\cos 2^{n-1} \theta-1) = (2\cos 2^{n} \theta-1)
\] If you want to prove this rigorously, you can use induction

Answer 3

Than a lot for ur help :)