$f(x)=\dfrac{1}{4^{n}}\left( \left( \begin{array}{c} 2n+1 \\ n \end{array} \right)\sin{x}-\left( \begin{array}{c} 2n+1 \\ n-1 \end{array} \right)\sin{3x}+ \cdots +(-1)^n \left( \begin{array}{c} 2n+1 \\ 0 \end{array} \right)\sin{((2n+1)x)}\right)$ diyelim. Öncelikle şu görülüyor ki $f$ fonksiyonu tek fonksiyondur, dolayısıyla da $f(x)=-f(-x)$'dir. $$-f(-x)=-\dfrac{1}{4^{n}}\left( \left( \begin{array}{c} 2n+1 \\ n \end{array} \right)\sin{(-x)}-\left( \begin{array}{c} 2n+1 \\ n-1 \end{array} \right)\sin{(-3x)}+ \cdots +(-1)^n \left( \begin{array}{c} 2n+1 \\ 0 \end{array} \right)\sin{(-(2n+1)x)}\right)$$ $$f(x)=\dfrac{1}{4^{n}}\left( \left( \begin{array}{c} 2n+1 \\ n+1 \end{array} \right)\sin{x}-\left( \begin{array}{c} 2n+1 \\ n+2 \end{array} \right)\sin{3x}+ \cdots +(-1)^n \left( \begin{array}{c} 2n+1 \\ 2n+1 \end{array} \right)\sin{((2n+1)x)}\right)$$ $f(x)-f(-x)=2f(x)$ olduğundan $$2f(x)=\dfrac{(-1)^{n+1}}{4^{n}}\left( \left( \begin{array}{c} 2n+1 \\ 0 \end{array} \right)\sin{(-(2n+1)x)}-\left( \begin{array}{c} 2n+1 \\ 1 \end{array} \right)\sin{(-(2n-1)x)}+ \cdots - \left( \begin{array}{c} 2n+1 \\ 2n+1 \end{array} \right)\sin{((2n+1)x)}\right)$$ $\mathfrak{I}(e^{ix})=\sin{x}$ olduğundan (sanal kısmı), $$f(x)=\mathfrak{I} \left (\dfrac{(-1)^{n+1}}{2^{2n+1}}\left( \left( \begin{array}{c} 2n+1 \\ 0 \end{array} \right)e^{-(2n+1)ix}-\left( \begin{array}{c} 2n+1 \\ 1 \end{array} \right)e^{-(2n-1)ix}+ \cdots - \left( \begin{array}{c} 2n+1 \\ 2n+1 \end{array} \right)e^{(2n+1)ix}\right) \right )$$ olur. Bunu da düzenlersek, $$f(x)=\mathfrak{I} \left (\dfrac{(-1)^{n+1}e^{-(2n+1)ix}}{2^{2n+1}}\left( \left( \begin{array}{c} 2n+1 \\ 0 \end{array} \right)(e^{2ix})^0-\left( \begin{array}{c} 2n+1 \\ 1 \end{array} \right)(e^{2ix})^1+ \cdots - \left( \begin{array}{c} 2n+1 \\ 2n+1 \end{array} \right)(e^{2ix})^{2n+1}\right) \right )$$ $$\Rightarrow f(x)=\mathfrak{I} \left (\dfrac{(-1)^{n+1}e^{-(2n+1)ix}}{2^{2n+1}}\left (- \left(e^{2ix}-1 \right )^{2n+1} \right ) \right )=\mathfrak{I} \left (\dfrac{(-1)^{n}}{2^{2n+1}} \left( e^{ix}-e^{-ix} \right )^{2n+1} \right )$$ $\sin{x}=\dfrac{e^{ix}-e^{-ix}}{2i}$ olduğundan $e^{ix}-e^{-ix}=2i\sin{x}$ olur. Dolayısıyla $$f(x)=\mathfrak{I} \left (\dfrac{(-1)^n}{2^{2n+1}} \left ( 2i\sin{x} \right )^{2n+1} \right )=\mathfrak{I} \left (\dfrac{(-1)^n}{2^{2n+1}} 2^{2n+1}i^{2n+1} \left (\sin{x} \right )^{2n+1} \right )$$ $i^{2n+1}=(i^{2})^n\cdot i=i\cdot (-1)^n$ olduğundan $$f(x)=\mathfrak{I} \left ( i(\sin{x})^{2n+1}\right )=(\sin{x})^{2n+1}$$ bulunur.
Not: Bu eşitliği bir integral sorusunu çözmeye çalışırken fark etmiştim fakat ispatlayamamıştım. Artık ispatı da olduğuna göre sorularda kullanılabilir.
