Solving \frac 58 \cot 36^{\circ}=\cos^3 x without substituting the trig

Solving ${\frac{58}{\mathrm{cot}36}}^{\circ }={\mathrm{cos}}^{3}x$ without substituting the trig values for ${36}^{\circ }$
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otoplilp1
Since
${\mathrm{cos}}^{3}x=\frac{3\mathrm{cos}x+\mathrm{cos}3x}{4}$
we need to prove that:
$\frac{5{\mathrm{cos}36}^{\circ }}{8{\mathrm{sin}36}^{\circ }}=\frac{3{\mathrm{cos}18}^{\circ }+{\mathrm{cos}54}^{\circ }}{4}$
or
$5{\mathrm{cos}36}^{\circ }=3{\mathrm{sin}54}^{\circ }+3{\mathrm{sin}18}^{\circ }+{\mathrm{sin}90}^{\circ }-{\mathrm{sin}18}^{\circ }$
or
$2{\mathrm{cos}36}^{\circ }+2{\mathrm{cos}108}^{\circ }=1$
which is true because
$2{\mathrm{cos}36}^{\circ }+2{\mathrm{cos}108}^{\circ }=\frac{2{\mathrm{sin}36}^{\circ }{\mathrm{cos}36}^{\circ }+2{\mathrm{sin}36}^{\circ }{\mathrm{cos}108}^{\circ }}{{\mathrm{sin}36}^{\circ }}=$
$=\frac{{\mathrm{sin}72}^{\circ }+{\mathrm{sin}144}^{\circ }-{\mathrm{sin}72}^{\circ }}{{\mathrm{sin}36}^{\circ }}=1$
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Jordan Mitchell
$\frac{5}{8}\cdot \mathrm{cot}{36}^{\circ }=\frac{5\mathrm{cos}{36}^{\circ }}{8\mathrm{sin}{36}^{\circ }}=\frac{5{\mathrm{cos}}^{2}{36}^{\circ }}{4\mathrm{cos}{18}^{\circ }}$ Using Proving trigonometric equation $\mathrm{cos}\left({36}^{\circ }\right)-\mathrm{cos}\left({72}^{\circ }\right)=\frac{1}{2}$ $\mathrm{cos}{36}^{\circ }-\left(2{\mathrm{cos}}^{2}{36}^{\circ }-1\right)=\frac{1}{2}⇔5{\mathrm{cos}}^{2}{36}^{\circ }=\left(1+\mathrm{cos}{36}^{\circ }{\right)}^{2}=\left(2{\mathrm{cos}}^{2}{18}^{\circ }{\right)}^{2}$