# Solve the problem. sin2x - sin4x

Solve the problem.
$\mathrm{sin}2x-\mathrm{sin}4x$
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minotaurafe
$\mathrm{sin}2x-\mathrm{sin}4x=\mathrm{sin}2x-2\mathrm{sin}2x.\mathrm{cos}2x=\mathrm{sin}2x\left(1-2\mathrm{cos}2x\right)$
$=\mathrm{sin}2x\left(1-\mathrm{cos}2x+1\right)=2\mathrm{sin}x\mathrm{cos}x\left(1-2\mathrm{cos}2x\right)$
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Ismael Molina
$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha \mathrm{sin}\beta$
$\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta -\mathrm{cos}\alpha \mathrm{sin}\beta -$
$\mathrm{sin}\left(\alpha +\beta \right)-\mathrm{sin}\left(\alpha -\beta \right)=2\mathrm{cos}\alpha \mathrm{sin}\beta$ , assume $\alpha +\beta =p$ and $\alpha -\beta =q$
$\alpha +\beta =p$
$\alpha -\beta =q+2\alpha =p+q=>\alpha =1/2\left(p+q\right)$
$\alpha +\beta =p$
$\alpha -\beta =q-2\beta =p-q=>\beta =1/2\left(p-q\right)$
$\therefore \mathrm{sin}p-\mathrm{sin}q=2\mathrm{cos}1/2\left(p+q\right)\mathrm{sin}1/2\left(p-q\right)$
$\mathrm{sin}2x-\mathrm{sin}4x=2\mathrm{cos}1/2\left(2x+4x\right)\mathrm{sin}1/2\left(2x-4x\right)$
$=2\mathrm{cos}1/2\left(6x\right)\mathrm{sin}1/2\left(-2x\right)$
$=2\mathrm{cos}3x\ast \mathrm{sin}-x=>\mathrm{sin}\left(-x\right)=-\mathrm{sin}x$
$=-2\mathrm{cos}3x\mathrm{sin}x$