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I was studying transformations of finite products of trig functions into sums, and empirically observed that the following curious identity appears to hold for all non-negative integer $m$: $$\prod_{n=0}^m \sin\left(\frac z{2^n}\right)=\frac1{2^m}\sum _{n=0}^{2^m-1} (-1)^{\large t_n} \sin\left(\frac{\pi\,m}2+\frac{2\,n+1}{2^m}\,z\right),\tag{$\diamond$}$$ where $t_n$ is the Thue–Morse sequence.$^{[1]}$$\!^{[2]}$$\!^{[3]}$ How can we prove this identity?

Vladimir Reshetnikov
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    Just as a guess, I suspect there's a relatively straightforward induction proof based on the recursive definition of the Thue-Morse sequence; multiply every term on the right by $\sin(z/2^{m+1})$ and use product-to-sum identities to turn each term into two with the appropriate signs. – Steven Stadnicki May 26 '19 at 02:22
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    BTW, the product on the left is related to the Fourier transform of the Fabius function and its piecewise-polynomial approximations, and the Thue–Morse sequence is closely related to the same function as well. – Vladimir Reshetnikov May 26 '19 at 02:32

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