Question

доказать тождество
 1+2cosx+cos2x делить 1+cos2x-2cosx=-ctg^2x/2

Answer 1

$frac{1+2cosx+cos2x}{1+cos2x-2cosx}=frac{1+2cosx+2cos^2x-1}{1+2cos^2x-1-2cosx}=frac{2cosx+2cos^2x}{2cos^2x-2cosx}=frac{1+cosx}{cosx-1}=\\frac{1+cos^2(frac{x}{2})-sin^2(frac{x}{2})}{cos^2(frac{x}{2})-sin^2(frac{x}{2})-1}=frac{sin^2(frac{x}{2})+cos^2(frac{x}{2})+cos^2(frac{x}{2})-sin^2(frac{x}{2})}{cos^2(frac{x}{2})-sin^2(frac{x}{2})-sin^2(frac{x}{2})-cos^(frac{x}{2})}=frac{2cos^2(frac{x}{2})}{-2sin^2(frac{x}{2})}=-ctg^2(frac{x}{2})$