<s>[INST] Find $f(\frac{2\pi k}{2^n \pm1})$ given $f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$. We are required to find $f(\frac{2\pi k}{2^n \pm1})$ given
$f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$
Let's start by removing the product notation
$\Rightarrow f(x)=(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)$
Now multiplying and dividing by $2\cos x+1$,
$\Rightarrow f(x)=\frac{(2\cos x+1)(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
We know, $a^2-b^2=(a+b)(a-b)$
$\Rightarrow f(x)=\frac{(4\cos^2x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
Now, $\cos2x=2\cos^2x-1$
$\Rightarrow f(x)=\frac{(2\cos2x+1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
$\Rightarrow f(x)=\frac{(2\cos 2^2x+1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$
We can now repeatedly apply the difference of squares formula and our function reduces to
$\Rightarrow f(x)=\frac{4\cos^2 2^{n-1}x-1}{2\cos x+1}$
$\Rightarrow f(x)=\frac{2\cos 2^nx+1}{2\cos x+1}$
How do I proceed from here on out?
Edit- We just need to know that $\cos(2k)=\cos(2^nk)$. Thus the denominator and the numerator cancel out and we get
$f(\frac{2\pi k}{2^n \pm1})=1$ and that's the answer.
 [/INST] In fact you have solved roughly all of it .
You just need to know that:

*

*$\frac{2^n2\pi k}{2^n +1} +\frac{2\pi k}{2^n+1}=2\pi k$


*$\frac{2^n2\pi k}{2^n -1} -\frac{2\pi k}{2^n-1}=2\pi k$
So in the end you get the answer 1.
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