Asher Olsen

2022-04-01

Prove that
$16{\mathrm{cos}}^{5}A-20{\mathrm{cos}}^{3}A+5\mathrm{cos}A=\mathrm{cos}5A$

Charlie Haley

Use
$2\mathrm{cos}x={e}^{ix}+{e}^{-ix}$
Then
$32{\mathrm{cos}}^{5}x={e}^{5ix}+5{e}^{3ix}+10{e}^{-ix}+5{e}^{-3ix}+{e}^{-5ix}$
$=2\mathrm{cos}5x+10\mathrm{cos}3x+20\mathrm{cos}x$
Moreover
$8{\mathrm{cos}}^{3}x={e}^{3ix}+3{e}^{ix}+3{e}^{-ix}+{e}^{3ix}=2\mathrm{cos}3x+6\mathrm{cos}x$
Therefore
$16{\mathrm{cos}}^{5}x-20{\mathrm{cos}}^{3}x+5\mathrm{cos}x=\left(\mathrm{cos}5x+5\mathrm{cos}3x+10\mathrm{cos}x\right)-5\left(\mathrm{cos}3x+3\mathrm{cos}x\right)+5\mathrm{cos}x=\mathrm{cos}5x$
This is actually backwards, so here's a different proof:
$\mathrm{cos}5x+i\mathrm{sin}5x={\left(\mathrm{cos}x+i\mathrm{sin}x\right)}^{5}$
$={\mathrm{cos}}^{5}x+5i{\mathrm{cos}}^{4}x\mathrm{sin}x-10{\mathrm{cos}}^{3}x{\mathrm{sin}}^{2}x-10i{\mathrm{cos}}^{2}x{\mathrm{sin}}^{3}x+5\mathrm{cos}x{\mathrm{sin}}^{4}x+i{\mathrm{sin}}^{5}x$
Taking the real part
$\mathrm{cos}5x={\mathrm{cos}}^{5}x-10{\mathrm{cos}}^{3}x\left(1-{\mathrm{cos}}^{2}x\right)+5\mathrm{cos}x{\left(1-{\mathrm{cos}}^{2}x\right)}^{2}$
and you can finish.

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