2022-04-06

How to prove $\mathrm{sin}3\theta =3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta$

Raina Blackburn

We will use the following trigonometric formulas:
$\mathrm{sin}\left(x+y\right)=\mathrm{sin}x\mathrm{cos}y+\mathrm{sin}y\mathrm{cos}x$
$\mathrm{sin}\left(2x\right)=2\mathrm{sin}x\mathrm{cos}x$
$\mathrm{cos}\left(2x\right)=1-2{\mathrm{sin}}^{2}x$
${\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$
So
$\mathrm{sin}\left(2\theta +\theta \right)=\mathrm{sin}\left(2\theta \right)\mathrm{cos}\theta +\mathrm{sin}\theta \mathrm{cos}\left(2\theta \right)$
$=2\mathrm{sin}\theta {\mathrm{cos}}^{2}\theta +\mathrm{sin}\theta \left(1-2{\mathrm{sin}}^{2}\theta \right)$
$=2\mathrm{sin}\theta \left(1-{\mathrm{sin}}^{2}\theta \right)+\mathrm{sin}\theta -2{\mathrm{sin}}^{3}\theta$
$=3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta$

lildeutsch11xq2j

$\mathrm{sin}3\theta =\mathrm{sin}\left(\theta +2\theta \right)$
$=\mathrm{sin}\theta \mathrm{cos}2\theta +\mathrm{cos}\theta \mathrm{sin}2\theta$
$=\mathrm{sin}\theta \left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)+2{\mathrm{cos}}^{2}\theta \mathrm{sin}\theta$
$=3{\mathrm{cos}}^{2}\theta \mathrm{sin}\theta -{\mathrm{sin}}^{3}\theta$
$3\left(1-{\mathrm{sin}}^{2}\theta \right)\mathrm{sin}\theta -{\mathrm{sin}}^{3}\theta$
$=3\mathrm{sin}\theta -4{\mathrm{sin}}^{3}\theta$

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