Результат вычислений

$sin(a+b)=cosa*sinb+sina*cosb\\ cos(2arccos\frac{\sqrt{2}}{2}) *sin(2arctg2)+sin(2arccos\frac{\sqrt{2}}{2})*cos(2arctg2)\\$
теперь так как
$cos(arccosa)=a\\ 1)cos(2arccos\frac{\sqrt{2}}{2})=2cos^2(arccos\frac{\sqrt{2}}{2})-1=0\\ 2)sin(2arctg2)=2sin(arctg2)cos(arctg2)\\ sina=1- \frac{1}{1+tg^2(2arctg2)}=\frac{4}{\sqrt{5}}\\ cos(arctg2)=\frac{1}{\sqrt{5}}\\ sin(2arctg2)=\frac{4}{5}\\ 3)sin(2arccos\frac{\sqrt{2}}{2})=2sin(arccos\frac{\sqrt{2}}{2})*cos(arccos\frac{\sqrt{2}}{2})=\sqrt{2}*\frac{\sqrt{2}}{2}=1\\ 4)cos(2arctg2)=-0.6\\ 10*(0*\frac{4}{5}+1*-0.6)=-6$