Dale Tate

2022-06-26

It's easy to show that
$\mathrm{sin}A-\mathrm{sin}B=2\mathrm{cos}\left(\frac{A+B}{2}\right)\mathrm{sin}\left(\frac{A-B}{2}\right),$
$\mathrm{sin}\left(2k+1\right)x-\mathrm{sin}\left(2k-1\right)x=2\mathrm{cos}2kx\cdot \mathrm{sin}x.$

rioolpijpgp

Expert

The double angle formula gives ${\mathrm{sin}}^{2}\left(kx\right)=\frac{1-\mathrm{cos}\left(2kx\right)}{2}$
So we have
$\sum _{k=1}^{n}{\mathrm{sin}}^{2}\left(kx\right)=\sum _{k=1}^{n}\frac{1-\mathrm{cos}\left(2kx\right)}{2}=\frac{n}{2}-\frac{1}{2}\sum _{k=1}^{n}\mathrm{cos}\left(2kx\right)$
$=\frac{n}{2}-\frac{1}{4\mathrm{sin}x}\sum _{k=1}^{n}2\mathrm{cos}\left(2kx\right)\mathrm{sin}x$
$=\frac{n}{2}-\frac{1}{4\mathrm{sin}x}\sum _{k=1}^{n}\left[\mathrm{sin}\left(2k+1\right)x-\mathrm{sin}\left(2k-1\right)x\right]$
$=\frac{n}{2}-\frac{1}{4\mathrm{sin}x}\left[\mathrm{sin}\left(2n+1\right)x-\mathrm{sin}x\right]$
$=\frac{n}{2}-\frac{1}{4\mathrm{sin}x}\left[2\mathrm{cos}\left(n+1\right)x\mathrm{sin}nx\right]$
$=\frac{n}{2}-\frac{\mathrm{cos}\left(n+1\right)x\mathrm{sin}nx}{2\mathrm{sin}x}$
Provided $\mathrm{sin}x\ne 0$, i.e. x is not an integer multiple of $\pi$. Though in these cases the equality is still true if taken as a limit.

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