Sin(arctg2+arctg3)=?

Question

Sin(arctg2+arctg3)=?

Answer 1

Sin2x=2tgx/(1+tg²x) 
Тогда получаем: 

sin(2arctg3)=2tg(arctg3)/(1+tg²(arctg3))=2*3/(1+9)=6/10=0,6