# Prove Prove sin5A−sin A+sin2A=sin2A(2cos3A+1)

Prove

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Obiajulu

Notice that if you distribute the sin2A on the right side of $\mathrm{sin}5A-\mathrm{sin}A+\mathrm{sin}2A=\mathrm{sin}2A\left(2\mathrm{cos}3A+1\right)$ you get $\mathrm{sin}5A-\mathrm{sin}A+\mathrm{sin}2A=2\mathrm{sin}2A\mathrm{cos}3A+\mathrm{sin}2A$. Since both sides have $\mathrm{sin}2A$, this means we need to rewrite $\mathrm{sin}5A-\mathrm{sin}A=2\mathrm{sin}2A\mathrm{cos}3A.$
The Sum-To-Product Formula for a difference of sines is $\mathrm{sin}u-\mathrm{sin}v=2\mathrm{cos}\left(\frac{u+v}{2}\right)\mathrm{sin}\left(\left(\frac{u-v}{2}\right).$
Therefore, if . Using the Sum-to-Product Formula then gives:
$\mathrm{sin}5A-\mathrm{sin}A+\mathrm{sin}2A$
$=2\mathrm{cos}\frac{5A+A}{2}\mathrm{sin}\frac{5A-A}{2}+\mathrm{sin}2A$  Use the Sum-to-Product formula
$=2\mathrm{cos}\frac{6A}{2}\mathrm{sin}\frac{4A}{2}+\mathrm{sin}2A$  Simplify.
$=2\mathrm{cos}3A\mathrm{sin}2A+\mathrm{sin}2A$  Simplify.
$=\mathrm{sin}2A\left(2\mathrm{cos}3A+1\right)$  Factor out $\mathrm{sin}2A$.