Is there an alternative form/identity for $\prod_{k=1}^{m} \cos(2^kx)$?

Question

I am working with a family of integrals that involve a product of cosines of the form $\prod_{k=1}^{m} \cos(2^kx)$. Is there a formula or identity for simplifying $\prod_{k=1}^{m} \cos(2^kx)$ to a polynomial number of terms in $x$? Or $\cos x$, $\sin x$?

Answer 1

Check by induction that $$ \frac{\sin(2^{m+1}x)}{2^{m+1}\sin(x)} = \prod_{j=0}^m \cos(2^{j} x)$$ since $$ \sin(2^{m+1}x)=2^{}\cos(2^{m})\sin(2^{m}x)\\=2^{2}\cos(2^{m})\cos(2^{m-1}) \sin(2^{m-1}x)\\=2^{3}\cos(2^{m})\cos(2^{m-1}) \cos(2^{m-2})\sin(2^{m-2}x) =.......$$

Answer 2

$$\cos x\cdot \cos (2x)\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P$$

Multiply both sides by $\sin x$, so

$$\frac{1}{2}\sin 2x\cdot \cos (2x)\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P\sin x$$

$$\frac{1}{2^2}\sin 4x\cdot\cos(4x)\cdot\cos(8x)...\cos(2^mx)=P\sin x$$

$$\frac{1}{2^3}\sin 8x\cdot\cos(8x)...\cos(2^mx)=P\sin x$$

keep doing that and get,

$$\frac{1}{2^{m+1}}\sin(2^{m+1}x)=P\cdot \sin x \Rightarrow P=\frac{\sin(2^{m+1}x)}{2^{m+1}\sin x}$$