Question

sin ( arctg 2 + arctg 3 )

Answer 1

$sin(a+b)=sina*cosb+sinb*cosa\ sin(arctg2+arctg3)=sin(arctg2)*cos(arctg3)+sin(arctg3)*cos(arctg2)=\ \ arctg3=x\ tgx=3\ frac{sinx}{cosx}=3\ frac{sinx}{sqrt{1-sin^2x}}=3\ 9(1-sin^2x)=sin^2x\ 9=10sin^2x\ sinx=frac{3}{sqrt{10}}\ cosx=frac{1}{sqrt{10}}\ x=arcsin(frac{3}{sqrt{10}})\ x=arccos(frac{1}{sqrt{10}})$
$arctg2=y\ tgy=2\ frac{siny}{cosy}=2\ sin^2y=4(1-sin^2y)\ 5sin^2y=4\ siny=frac{2}{sqrt{5}}\ cosy=frac{1}{sqrt{5}}\ y=arcsinfrac{2}{sqrt{5}}\ y=arccosfrac{1}{sqrt{5}}$
Теперь все подставим с условием что 
cos(arccosa)=a
sin(arcsinb)=b

$frac{2}{sqrt{5}}*frac{1}{sqrt{10}}+frac{3}{sqrt{10}}*frac{1}{sqrt{5}}=frac{sqrt{2}}{2}$