Proof that $\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$ for $x\ge 0$

Question

Is there any proof of the identity $$\sin x=4^{-x/\pi}\sum_{n=0}^\infty (-1)^n\binom{4x/\pi}{2n+1}$$ where $x\ge 0$?

If we replace $4x/\pi$ by $x$ and write the binomial in terms of the falling factorial $(x)_n$, we get $$\sum_{n=0}^\infty (-1)^n \frac{(x)_{2n+1}}{(2n+1)!},$$ which is remarkably similar to the power series for $\sin x$ centered at $0$: $$\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}.$$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\left.\sin\pars{x} \vphantom{A^{A}}\,\right\vert_{\ x\ \geq\ 0}\,\,\, =\,\,\, 4^{-x/\pi}\sum_{n = 0}^{\infty}\pars{-1}^{n} {4x/\pi \choose 2n + 1}}:\ {\Large ?}}$.

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\begin{align} &\bbox[5px,#ffd]{4^{-x/\pi} \sum_{n = 0}^{\infty}\pars{-1}^{n}{4x/\pi \choose 2n + 1}} = -\,4^{-x/\pi}\,\,\,\ic\sum_{n = 0}^{\infty}\ic^{2n + 1} {4x/\pi \choose 2n + 1} \\[5mm] = &\ -\,4^{-x/\pi}\,\,\,\ic\sum_{n = 1}^{\infty}\ic^{n} {4x/\pi \choose n}{1 - \pars{-1}^{n} \over 2} = 4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n} {4x/\pi \choose n} \\[5mm] = &\ 4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n} \,{\Gamma\pars{4x/\pi + 1} \over \Gamma\pars{n + 1}\Gamma\pars{4x/\pi - n + 1}} \\[5mm] = &\ 4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}\ic^{n}\ {\Gamma\pars{4x/\pi + 1} \over \Gamma\pars{n + 1}} \ \times \\[1mm] &\ \phantom{4^{-x/\pi}\,\,\,\Im\sum_{n = 1}^{\infty}} {\Gamma\pars{n - 4x/\pi}\sin\pars{\pi\bracks{n - 4x/\pi}} \over \pi} \\[5mm] = &\ -\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im\sum_{n = 1}^{\infty}\pars{-\ic}^{n}\ {\Gamma\pars{n - 4x/\pi}\Gamma\pars{4x/\pi + 1} \over \Gamma\pars{n + 1}} \\[5mm] = &\ -\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im \sum_{n = 1}^{\infty}\pars{-\ic}^{n} \int_{0}^{1}t^{n - 4x/\pi - 1}\,\,\pars{1 - t}^{4x/\pi} \,\,\,\dd t \\[5mm] = &\ -\,{4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\,\Im \int_{0}^{1}t^{-4x/\pi - 1}\,\,\pars{1 - t}^{4x/\pi} \sum_{n = 1}^{\infty} \pars{-\ic t}^{n}\,\,\,\dd t \\[5mm] = &\ {4^{-x/\pi}\,\,\sin\pars{4x} \over \pi}\ \underbrace{\int_{0}^{1} {t^{-4x/\pi}\,\,\pars{1 - t}^{4x/\pi} \over 1 + t^{2}}\dd t} _{\ds{4^{x/\pi}\,\pi\,\csc\pars{4x}\color{red}{\sin\pars{x}}}} \\[5mm] = &\ \bbx{\sin\pars{x}} \\ & \end{align}

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The last integral is straightforward evaluated with the change $\ds{{1 - t \over t} \mapsto t}$.

Answer 2

Start with: $$ (1+z)^A = \sum_{n=0}^{\infty}\binom{A}{n}z^n $$Split this into even and odd pieces: $$ (1+z)^A = \sum_{n=0}^{\infty}\binom{A}{2n+1}z^{2n+1}+ \sum_{n=0}^{\infty}\binom{A}{2n}z^{2n} $$Playing around with the even-only piece (exercise!), we have $$ \sum_{n=0}^{\infty}\binom{A}{2n}z^{2n}=\frac{1}{2} \left((1-z)^A+(1+z)^A\right) $$So, $$ \sum_{n=0}^{\infty}\binom{A}{2n+1}z^{2n} =\frac{(1+z)^A-(1-z)^A}{2 z}; $$put $z=i$, $A=4x/\pi$ and try and simplify it using Euler's Formula.