Question

please help with verifying :
sin^2x/2-2cosx=cos^2 x/2

Answer 1

\[\frac{ \sin ^{2}x }{ 2-2\cos x }=\frac{ 1 }{ 2 }*\frac{\sin ^{2}x }{1-\cos x }*\frac{1+\cos x }{1+\cos x }\]
\[=\frac{ 1 }{ 2 }*\frac{ \sin ^{2}x \left( 1+\cos x \right) }{ 1-\cos ^{2}x }\]
you can solve and use
\[1-\cos ^{2}x=\sin ^{2}x and 1+\cos x=2\cos ^{2}\frac{ x }{ 2 }\]

Answer 2

thanks for the help!!! I never thought about splitting at the beginning into two fractions :P thanks!

Answer 3

yw