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verify sin^2(x/2)= tanx-sinx/2tanx
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tan = sin/ cos sin/cos-sin/(2*sin/cos)=sin/cos - sin*cos/(2 sin)=sin/cos - cos/2 =\[ \frac{ \sin }{ \cos } -\frac{ \cos }{ 2 }\]
\[\frac{ 2\sin-\cos^2 }{ 2*\cos }\]
now remove cos from this using trigo formulas
or sin^2x/2=(1-cosx)/2............we know cos2x=1-2sin^2x \[\tan ^2x=\frac{ 1-cosx }{ 1+cosx }\]
1-cosx=(1+cosx)*(tan^2x)
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sin^2(x/2)=(1-cosx)/2=(1+cosx)*tan^2x/2
1+cosx=2cos^2(x/2) (2cos^2(x/2)*tan^2(x/2))/2=sin^2(x/2)
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