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ylucho
please help with verifying : sin^2x/2-2cosx=cos^2 x/2
\[\frac{ \sin ^{2}x }{ 2-2\cos x }=\frac{ 1 }{ 2 }*\frac{\sin ^{2}x }{1-\cos x }*\frac{1+\cos x }{1+\cos x }\] \[=\frac{ 1 }{ 2 }*\frac{ \sin ^{2}x \left( 1+\cos x \right) }{ 1-\cos ^{2}x }\] you can solve and use \[1-\cos ^{2}x=\sin ^{2}x and 1+\cos x=2\cos ^{2}\frac{ x }{ 2 }\]
thanks for the help!!! I never thought about splitting at the beginning into two fractions :P thanks!