# I've been trying to figure out why this equation is satisfied: 1 &#x2212;

I've been trying to figure out why this equation is satisfied:
$\frac{1-\left(\mathrm{cos}\left(x\right){\right)}^{3}}{{x}^{2}}=\frac{2\cdot \left(\mathrm{sin}\left(\frac{x}{2}\right){\right)}^{2}}{{x}^{2}}\cdot \left(1+\mathrm{cos}\left(x\right)+\left(\mathrm{cos}\left(x\right){\right)}^{2}\right)$
but I can't find the proper trigonometric changes in order to change from one to another. I know that the sine comes from the double angle formula, but I obtain slightly different results in other parts and it's never the same as the formula above.
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floygdarvn
One may observe that, in general,
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
giving
$1-\left(\mathrm{cos}x{\right)}^{3}=\left(1-\mathrm{cos}x\right)\left(1+\mathrm{cos}x+\left(\mathrm{cos}x{\right)}^{2}\right)$
then, by using $2{\mathrm{sin}}^{2}\left(x/2\right)=1-\mathrm{cos}x$ as noticed by the "double-angle" formula $2{\mathrm{sin}}^{2}\left(x/2\right)=1-\mathrm{cos}x$