Sin^4x-cos^4x если tgx/2=0.5

$tg\frac{x}{2}=0,5\\\\sin^4x-cos^4x=(sin^2x-cos^2x)(sin^2x+cos^2x)=\\\\=sin^2x-cos^2x=(sinx-cosx)(sinx+cosx)=\\\\=\left (\frac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}}- \frac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} \right )\left ( \frac{2tg\frac{x}{2}}{1+tg^2\frac{x}{2}} + \frac{1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} \right )=\\\\=\frac{2tg\frac{x}{2}-1+tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} \cdot \frac{2tg\frac{x}{2}+1-tg^2\frac{x}{2}}{1+tg^2\frac{x}{2}} =$

$=\frac{2\cdot 0,5-1+0,25}{1+0,25} \cdot \frac{2\cdot 0,5+1-0,25}{1+0,25}$ $=0,2\cdot 1,4=0,28$