Question

Найти значение выражения
Cos(arccos 1/2 + arccos -1/7)
arcctg 1/3 + arctg 2

Answer 1

cos(arccos1/2+arccos(-1/7))=cos(arccos(1/2*√(1-1/48)-1/7√(1-1/4))=
=cos(arccos(1/2*4√3/7-1/7*√3/2))=cos(arccos(4√3/14-√3/14))=
=cos(arccos(3√3/14)=3√3/14
arccosx+arccosy=arccos(x√(1-y²)+y√(1-x²))

arcctg1/3+arctg2=arctg2+arctg3=arctg((2+3)/(1-2*3))+π=arctg(-1)+π=-π/4+π=3π/4
arcctgx=arctg(1/x)
arctgx+arctgy=arctg((x+y)/(1-xy)),xy>1 ,x>0,y>0