Cameron Russell

2022-01-28

Prove that $\mathrm{sin}\left(x\right)\mathrm{sin}\left(\frac{\pi }{3}+x\right)\mathrm{sin}\left(\frac{\pi }{3}-x\right)=\frac{1}{4}\mathrm{sin}3x$

Troy Sutton

Expert

$\mathrm{sin}3t=3\mathrm{sin}t-4{\mathrm{sin}}^{3}t$
If $\mathrm{sin}3t=\mathrm{sin}3x,3t={180}^{\circ }n+{\left(-1\right)}^{n}3x$ where n is any integer
$t={60}^{\circ }n+{\left(-1\right)}^{n}x$ where $n=-1,0,1$
So, the roots of
$4{\mathrm{sin}}^{3}t-3\mathrm{sin}t+\mathrm{sin}3x=0$
are $\mathrm{sin}t$ where $t={60}^{\circ }n+{\left(-1\right)}^{n}x$ where $n=-1,0,1$
$⇒\mathrm{sin}\left(-{60}^{\circ }-x\right)\mathrm{sin}x\mathrm{sin}\left({60}^{\circ }-x\right)={\left(-1\right)}^{3}\frac{\mathrm{sin}3x}{4}$
$⇔4\mathrm{sin}\left({60}^{\circ }+x\right)\mathrm{sin}x\mathrm{sin}\left({60}^{\circ }-x\right)=\mathrm{sin}3x$

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