illusiia

2020-11-12

Solve the equation $\frac{\left(2\mathrm{sin}\theta \mathrm{sin}2\theta \right)}{\mathrm{cos}\theta +\mathrm{cos}3\theta }=\mathrm{tan}\left(\theta \right)\mathrm{tan}\left(2\theta \right)$

### Answer & Explanation

tafzijdeq

$\frac{\left(2\mathrm{sin}\theta \mathrm{sin}2\theta \right)}{\mathrm{cos}\theta +\mathrm{cos}3\theta }$
$=\frac{2\mathrm{sin}\left(\theta \right)\mathrm{sin}\left(2\theta \right)}{-2\mathrm{cos}\left(\theta \right)+4{\mathrm{cos}}^{3}\left(\theta \right)}$
$=\frac{2\mathrm{sin}\left(\theta \right)\mathrm{sin}\left(2\theta \right)}{2\mathrm{cos}\left(\theta \right)\left(-1+{\mathrm{cos}}^{2}\left(\theta \right)\right)}$
$=\frac{2\mathrm{sin}\left(\theta \right)\mathrm{sin}\left(2\theta \right)}{\mathrm{cos}\left(\theta \right)\left(2{\mathrm{cos}}^{2}\left(\theta \right)-1\right)}$
Use the following identity: $\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}=\mathrm{tan}\left(x\right)$
$=\frac{\mathrm{sin}\left(2\theta \right)\mathrm{tan}\left(\theta \right)}{\left(2{\mathrm{cos}}^{2}\left(\theta \right)-1\right)}$
$=\frac{\mathrm{sin}\left(2\theta \right)\mathrm{tan}\left(\theta \right)}{\mathrm{cos}\left(2\theta }$
Use the following identity: $\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}=\mathrm{tan}\left(x\right)$
$=\frac{\mathrm{tan}\left(\theta \right)}{tan\left(2\theta }$
Therefore
$\frac{2\mathrm{sin}\theta \mathrm{sin}2\theta }{\mathrm{cos}\theta +\mathrm{cos}3\theta }=\mathrm{tan}\left(\theta \right)\mathrm{tan}\left(2\theta \right)$

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