Jameson Lucero

2022-07-04

By chance I found:
$2\mathrm{sin}\left(2m+1\right)x-\mathrm{sin}\left(2m-1\right)x=\mathrm{sin}\left(x\right)\left(1+2\mathrm{cos}\left(2mx\right)+2\sum _{k=1}^{m}\mathrm{cos}\left(2kx\right)\right)$

Immanuel Glenn

Expert

You may write, for any real number $x\notin \mathbb{Z}\pi$
$\begin{array}{rl}\sum _{k=1}^{m}\mathrm{cos}\left(2kx\right)& =\mathrm{\Re }\sum _{k=1}^{m}{e}^{2ikx}\\ & =\mathrm{\Re }\left({e}^{2ix}\frac{{e}^{2imx}-1}{{e}^{2ix}-1}\right)\\ & =\mathrm{\Re }\left({e}^{2ix}\frac{{e}^{imx}\left({e}^{imx}-{e}^{-imx}\right)}{{e}^{ix}\left({e}^{ix}-{e}^{-ix}\right)}\right)\\ & =\mathrm{\Re }\left({e}^{i\left(m+1\right)x}\frac{\mathrm{sin}\left(mx\right)}{\mathrm{sin}\left(x\right)}\right)\\ & =\mathrm{\Re }\left(\left(\mathrm{cos}\left(\left(m+1\right)x\right)+i\mathrm{sin}\left(\left(m+1\right)x\right)\right)\frac{\mathrm{sin}\left(mx\right)}{\mathrm{sin}\left(x\right)}\right)\\ & =\frac{\mathrm{cos}\left(\left(m+1\right)x\right)}{\mathrm{sin}\left(x\right)}\mathrm{sin}\left(mx\right).\end{array}$
Thus
$\begin{array}{rl}& \mathrm{sin}\left(x\right)\left(1+2\mathrm{cos}\left(2mx\right)+2\sum _{k=1}^{m}\mathrm{cos}\left(2kx\right)\right)\\ & =\mathrm{sin}\left(x\right)+2\mathrm{sin}\left(x\right)\mathrm{cos}\left(2mx\right)+2\mathrm{cos}\left(\left(m+1\right)x\right)\mathrm{sin}\left(mx\right)\end{array}$
and, using $2\mathrm{sin}a\mathrm{cos}b=\mathrm{sin}\left(a+b\right)+\mathrm{sin}\left(a-b\right)$, we get
$\begin{array}{rl}& \mathrm{sin}\left(x\right)\left(1+2\mathrm{cos}\left(2mx\right)+2\sum _{k=1}^{m}\mathrm{cos}\left(2kx\right)\right)\\ & =\mathrm{sin}\left(x\right)+\mathrm{sin}\left(\left(2m+1\right)x\right)-\mathrm{sin}\left(\left(2m-1\right)x\right)+\mathrm{sin}\left(\left(2m+1\right)x\right)-\mathrm{sin}\left(x\right)\\ & =2\mathrm{sin}\left(\left(2m+1\right)x\right)-\mathrm{sin}\left(\left(2m-1\right)x\right)\end{array}$
as announced.

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