Question

Результат вычислений $10*sin(2arccos frac{ sqrt{2}}{2} +2arctg2)$

Answer 1

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