consider `f_n(theta)=(2costheta-1)(2cos2theta-1)(2cos(2^2 theta)-1).....(2cos(2^n theta)-1)` the value of `f_n(theta)` is equal to

The value (2cos theta-1)(2cos2 theta-1)(2cos4 theta-1)(2cos8 theta-1)=

For a positive integer n, let f_(n)(theta)=(2cos theta+1)(2cos theta-1)(2cos2 theta-1)(2cos(2^(2))theta-1)...(2cos(2^(n-1))theta-1) Which one of the following hold(s) good?

Range of f(theta)=cos^2theta(cos^2theta+1)+2 sin^2theta is

Prove that, (2 cos theta - 1)(2 cos 2 theta-1)(2 cos2^(2) theta-1)...(2 cos 2^(n-1) theta-1) = (2 cos 2^(n) theta + 1)/(2 cos theta + 1)

Range of f(theta)=cos^2theta(cos^2theta+1)+2sin^2theta is

Range of f(theta)=cos^2theta(cos^2theta+1)+2sin^2theta is

Range of f(theta)=cos^2theta(cos^2theta+1)+2sin^2theta is

Prove that: (2cos2^ntheta+1)/(2costheta+1)=(2costheta-1)(2cos2theta-1)(2cos2^2theta-1) ...(2cos2^(n-1)theta-1)

Prove that: (2cos2^ntheta+1)/(2costheta+1)=(2costheta-1)(2cos2theta-1)(2cos2^2theta-1) ...(2cos2^(n-1)theta-1)

Show that (2costheta-1) (2cos2theta-1) (2cos2^2theta-1)……(2cos2^(n-1)theta-1) = (2cos2^ntheta+1)/(2costheta+1)