Question

$frac{1-8cos^2 frac{ alpha }{2}*sin^2 frac{ alpha }{2} }{cos3 alpha +cos alpha }*(2cos^2 frac{ alpha }{2}-1)$

Answer 1

$frac{1-8Cos^2 frac{x}{2}*Sin^2 frac{x}{2}}{Cos3x+Cosx}*(2Cos^2 frac{x}{2}-1)=$

$= frac{1-2(2Cos frac{x}{2}Sin frac{x}{2})^2}{4Cos^3x-3Cosx+Cosx}*(2( frac{1+Cosx}{2})-1))=$

$= frac{1-2Sin^2x}{2Cosx*(2Cos^2x-1)}*Cosx=$

$= frac{1-2Sin^2x}{2(2Cos^2x-1)}= frac{Cos2x}{2(2* frac{1+Cos2x}{2}-1)}=$

$= frac{Cos2x}{2(1+Cos2x-1)}= frac{Cos2x}{2(Cos2x)}= frac{1}{2}$

Answer 2

$frac{1-2(2sin frac{ alpha }{2}cos frac{ alpha}{2})^2}{2cos frac{3 alpha+ alpha}{2}cos frac{3 alpha- alpha}{2}}*(2cos^2 frac{ alpha}{2}-cos^2 frac{ alpha}{2}-sin^2 frac{ alpha}{2})= \ \ = frac{1-2(sin(2* frac{ alpha}{2}))^2}{2cos2 alpha*cos alpha}*(cos^2 frac{ alpha}{2}-sin^2 frac{ alpha}{2})= \ \ = frac{1-2sin^2 alpha}{2cos2 alpha*cos alpha}*cos(2* frac{ alpha}{2})= \ \ = frac{cos^2 alpha+sin^2 alpha-2sin^2 alpha}{2cos2 alpha}=$
$= frac{cos^2 alpha-sin^2 alpha}{2cos2 alpha}= frac{cos2 alpha}{2cos2 alpha}= frac{1}{2}$