Question 240826
{{{(Sin^2x)/(2-2Cos(x)) = Cos^2}}}{{{(x/2)}}}
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We will work only with the left sides.
Using the identity {{{Sin^2phi+Cos^2phi=1}}}, 
we have {{{Sin^2phi=1-Cos^2phi}}} and so we
replace {{{Sin^2x}}} by {{{1-Cos^2x}}}
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{{{(1-Cos^2x)/(2-2Cos(x)) = Cos^2}}}{{{(x/2)}}}
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Factor the numerator as the difference perfect 
squares, and factor 2 out of the denominator:
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{{{( (1-Cos(x))(1+Cos(x)))/(2(1-Cos(x))) = Cos^2}}}{{{(x/2)}}}
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Cancel the {{{(1-Cos(x))}}} factors:
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{{{( cross((1-Cos(x)))(1+Cos(x)))/(2cross((1-Cos(x)))) = Cos^2}}}{{{(x/2)}}}

{{{1/2}}}{{{(1+Cos(x)) = Cos^2}}}{{{(x/2)}}}
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Now we use the identity {{{Cos(2phi)=2Cos^2phi-1}}}
by letting {{{phi=x/2}}}.  This gives {{{Cos(2(x/2))=2Cos^2}}}{{{x/2-1}}}
or {{{Cos(x)=2Cos^2}}}{{{(x/2)-1}}}. So we substitute that
for {{{Cos(x)}}}
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{{{1/2}}}{{{"("}}}{{{1+2Cos^2}}}{{{(x/2)-1 }}}{{{")"}}}{{{ "" = Cos^2}}}{{{(x/2)}}}

{{{1/2}}}{{{"("}}}{{{2Cos^2}}}{{{(x/2)}}}{{{")"}}}{{{ "" = Cos^2}}}{{{(x/2)}}}

{{{1/cross(2)}}}{{{"("}}}{{{cross(2)Cos^2}}}{{{(x/2)}}}{{{")"}}}{{{ "" = Cos^2}}}{{{(x/2)}}}

{{{Cos^2}}}{{{(x/2)}}}{{{"="}}}{{{Cos^2}}}{{{(x/2)}}}

Edwin</pre>