deiteresfp

2021-12-29

How to go from

alexandrebaud43

$2\mathrm{cos}2x+1=2\left(1-2{\mathrm{sin}}^{2}x\right)+1=3-4{\mathrm{sin}}^{2}x=\frac{3\mathrm{sin}x-4{\mathrm{sin}}^{3}x}{\mathrm{sin}x}=\frac{\mathrm{sin}3x}{\mathrm{sin}x}$
Note: $\mathrm{sin}x\ne 0$

Timothy Wolff

Hint for LHS to RHS: Convert $\mathrm{cos}2x$ to one of its three identities, then multiply and divide by $\mathrm{sin}x$
Hint for RHS to LHS: Convert $\mathrm{sin}3x$ to a known identity, then divide by $\mathrm{sin}x$. Once you have that, add and subtract 1 to get a common factor, then you will have a term in ${\mathrm{cos}}^{2}x$ that can be converted into a double angle formula.

karton

We have
$2\mathrm{sin}\alpha \mathrm{cos}\beta =\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)$
so
$\left(2\mathrm{cos}2x+1\right)\mathrm{sin}x=2\mathrm{sin}x\mathrm{cos}2x+\mathrm{sin}x=\mathrm{sin}3x+\mathrm{sin}\left(-x\right)+\mathrm{sin}x=\mathrm{sin}3x$
so assuming that $\mathrm{sin}x\ne 0$
$2\mathrm{cos}2x+1=\frac{\mathrm{sin}3x}{\mathrm{sin}x}$

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