Garrett Black

2022-06-29

Prove $\mathrm{sin}2x+\mathrm{sin}4x+\mathrm{sin}6x=4\mathrm{cos}x\mathrm{cos}2x\mathrm{sin}3x$

Dustin Durham

Expert

To solve this problem, we can use the identities:
$\mathrm{sin}A+\mathrm{sin}B=2\mathrm{sin}\frac{A+B}{2}\mathrm{cos}\frac{A-B}{2},$
$\mathrm{cos}A+\mathrm{cos}B=2\mathrm{cos}\frac{A+B}{2}\mathrm{cos}\frac{A-B}{2},$
and
$\mathrm{sin}2\varphi =2\mathrm{sin}\varphi \mathrm{cos}\varphi .$
Going back to the question,
$\text{LHS}=\mathrm{sin}2x+\mathrm{sin}4x+\mathrm{sin}6x\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}3x\mathrm{cos}x+\mathrm{sin}6x\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}3x\mathrm{cos}x+2\mathrm{sin}3x\mathrm{cos}3x\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}3x\left(\mathrm{cos}x+\mathrm{cos}3x\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{sin}3x×2\mathrm{cos}2x\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=4\mathrm{cos}x\mathrm{cos}2x\mathrm{sin}3x\phantom{\rule{0ex}{0ex}}=\text{RHS}.$
Hence, proved.

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