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I know that

$$\sin^2(2^n x) = 4^n\sin^2(x)\left(\prod_{j=1}^n \cos(2^{n-j}x)\right)^2$$

for $n \in \mathbb{N}$.

I am trying to derive an expression for $\sin^2(2^n x)$ in terms of only sines and cosines that are functions of $x$ only (i.e. without multiple-angles).

How can I rewrite $$\prod_{j=1}^n \cos(2^{n-j}x) = \cos(x)\cos(2x)\cdots\cos(2^{n-1}x)$$ in terms of $\sin(x), \cos(x)$ only?

Thanks for any help.

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Cosines of multiple angles are polynomial in the cosine of the angle, these are the Chebyshev polynomials. Squares of sines in multiple angles are polynomial in the sine squared of the angle, these are the spread polynomials.

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