Deriving a General Expression for $\sin A\times\sin 2A\times\sin 4A\times\sin 8A\dots\times\sin 2^nA$

Question

Hey so I'm currently preparing for an entrance examination, and have found this particular formula of great help while solving questions on cosines of angles in a GP. $$\cos A\times\cos 2A\times\cos 4A\times\cos 8A\dots\cos (2^nA)=\frac{\sin(2^{n+1}A)}{2^n\sin A}$$ I attempted to derive a similar formula for sines of angles in a GP. $$\sin A\times\sin 2A\times\sin 4A\times\sin 8A\dots\sin (2^nA)=(?)$$ I started by multiplying this expression by $\frac{2\cos A}{2\cos A}$ to obtain the following by using the Double-Angle Identity: $$\frac{2\cos A\times\sin A\times\sin 2A\times\sin 4A\dots\sin (2^nA)}{2\cos A}=\frac{\sin^2 2A\times\sin 4A\dots\sin (2^nA)}{2\cos A}$$ I continued this method, attempting to eliminate almost all sines, after which I could use the previously derived expression for cosines.

I obtained this expression in the end, but I'm unsure about how to proceed further. $$\frac{\sin^{n+1}(2^nx)}{2^{\frac{n(n+1)}{2}}\prod_{k=1}^{n}\cos^k(2^{k-1}x)^n}$$

Answer 1

The question is about the expression

$$ B_n(A) := \sin A\cdot\sin 2A\cdot\sin 4A\cdot\sin 8A\cdots\sin (2^nA). \tag1$$

The key is to use the standard identities

$$ \sin A = \frac1{2i}(x - x^{-1}),\;\; \cos A = \frac1{2}(x + x^{-1}),\;\; x:=e^{iA}. \tag2$$

Use this in equation $(1)$ to get

$$ B_n(A) = (2i)^{-n}\prod_{k=0}^n\left(x^{2^k} - x^{-2^k}\right). \tag3 $$

Expand this as a Laurent polynomial in $x$ to get (when $\,n\,$ is even)

$$ B_n(A) = (2i)^{-n-1}(-1)^{n-1}\!\sum_{j=1}^{2^n-1} s_j\left(x^j-x^{-j}\right) \tag4 $$

where $\,s_j\,$ is zero if $j$ is even else

$$ s_{2k+1} = a_k := (-1)^{\text{wt}(k)} \tag5 $$

where $\,a_k\,$ is OEIS sequence A106400 and $\,\text{wt}(k)\,$ is the Hamming weight of the bits of integer $k$.

For example,

$$ B_1(A) = 2^{-1}(\cos A - \cos 3A ) \tag7 $$

$$ B_2(A) =-2^{-2}(\sin A - \sin 3A - \sin 5A + \sin 7A). \tag8 $$

Thus the result for $\,n\,$ even is

$$ B_n(A) = (2i)^{-n}\!\sum_{j=1}^{2^{n+1}-1} s_j \sin(jA). \tag9 $$

The slightly different result for $\,n\,$ odd is

$$ B_n(A) = i(2i)^{-n}\!\sum_{j=1}^{2^{n+1}-1} s_j \cos(jA). \tag{10} $$

Whether this can be accepted as a "general expression" or "closed form" is a matter of opinion.

The explanation of why the result differs depending on $\,n\,$ being even or odd is that each factor of the product is an odd function and thus the product $\,B_n(A)\,$ is an odd or even function depends on the parity of $\,n\,$ itself.

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However, the question is motivated by the example of

$$ C_n(A) := \cos A\cdot\cos 2A\cdot\cos 4A\cdot\cos 8A\cdots\cos (2^nA) \tag{11}$$

using the standard identity

$$ \text{sinc}(2x)/\text{sinc}(x) = \cos(x) $$

to get the general result

$$ C_n(A) = \text{sinc}(2^{n+1}A)/\text{sinc}(A). \tag{12} $$

Thus, consider the function

$$ f(x) \!=\! 1 \!-\! \frac{x^2}{18} \!+\! \frac{19x^4}{16200} \!-\! \frac{583x^6}{42865200} \!+\! \frac{132809x^8}{1311675120000} \!+\! \cdots \tag{13} $$

which satisfies

$$ f(2x)/f(x) = \text{sinc}(x) := \sin(x)/x. \tag{14} $$

This implies

$$ B_n(A) = x^{n+1} 2^{2^n-1}f(2^{n+1}A)/f(A). \tag{15} $$

I don't think that the function $\,f(x)\,$ has a closed form expression although

$$ f(x) = \prod_{k=1}^\infty \text{sinc}(x/2^k). \tag{16} $$

Answer 2

Let $A_n(x)$ be the cosine odd multiple angle sum \begin{align} A_n(x)=\sum_{k=1}^{n}{\cos((2k-1)x)} \end{align} Then \begin{align} B_n(x)= 2^n \sin^{n+1}(x) \prod_{i=1}^n{A_{2^{i-1}}(x)} = \sin(x)\sin(2x)\sin(4x)\sin(8x)...\sin(2^n x) \end{align} is a general expression for your product