SOLUTION: please verify {{{(Sin^2x)/(2-2Cos(x)) = Cos^2}}}{{{(x/2)}}}

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Question 240826: please verify %28Sin%5E2x%29%2F%282-2Cos%28x%29%29+=+Cos%5E2%28x%2F2%29
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
%28Sin%5E2x%29%2F%282-2Cos%28x%29%29+=+Cos%5E2%28x%2F2%29

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We will work only with the left sides.
Using the identity Sin%5E2phi%2BCos%5E2phi=1, 
we have Sin%5E2phi=1-Cos%5E2phi and so we
replace Sin%5E2x by 1-Cos%5E2x
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%281-Cos%5E2x%29%2F%282-2Cos%28x%29%29+=+Cos%5E2%28x%2F2%29
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Factor the numerator as the difference perfect 
squares, and factor 2 out of the denominator:
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%28+%281-Cos%28x%29%29%281%2BCos%28x%29%29%29%2F%282%281-Cos%28x%29%29%29+=+Cos%5E2%28x%2F2%29
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Cancel the %281-Cos%28x%29%29 factors:
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%28x%2F2%29

1%2F2%281%2BCos%28x%29%29+=+Cos%5E2%28x%2F2%29
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Now we use the identity Cos%282phi%29=2Cos%5E2phi-1
by letting phi=x%2F2.  This gives Cos%282%28x%2F2%29%29=2Cos%5E2x%2F2-1
or Cos%28x%29=2Cos%5E2%28x%2F2%29-1. So we substitute that
for Cos%28x%29
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1%2F2%22%28%221%2B2Cos%5E2%28x%2F2%29-1+%22%29%22+%22%22+=+Cos%5E2%28x%2F2%29

1%2F2%22%28%222Cos%5E2%28x%2F2%29%22%29%22+%22%22+=+Cos%5E2%28x%2F2%29

1%2Fcross%282%29%22%28%22cross%282%29Cos%5E2%28x%2F2%29%22%29%22+%22%22+=+Cos%5E2%28x%2F2%29

Cos%5E2%28x%2F2%29%22=%22Cos%5E2%28x%2F2%29

Edwin