Question 370078
Let {{{S=sin(x)}}},{{{C=cos(x)}}} to reduce some of the clutter,
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{{{S^2+C^2=1}}}
{{{S^2C+C^3=C}}}
{{{S^2C=C-C^3}}}
{{{SC=(C-C^3)/S}}}
Now just go backwards.
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{{{S^2+C^2=1}}}
{{{1+C^2/S^2=1/S^2}}}
{{{2(1+C^2/S^2)=2/S^2}}}
{{{2(1+C^2/S^2)=2(1/S^2)(C/C)}}}
{{{2(1+C^2/S^2)=2(C/S)(1/C)(1/S)}}}
{{{2+2(C^2/S^2)=2(C/S)(1/C)(1/S)}}}
Again work backwards.