6arctg√3 -4arcsin (-1/√2) 2arctg1+3arcsin(-1/2) - вопрос №63412213125332867001 от ShiQualence 14.01.2021 11:10

Question

Алгебра: 6arctg√3 -4arcsin (-1/√2) 2arctg1+3arcsin(-1/2) 5arctg(-√3)-3arccos(-√2/2)

ShiQualence · Answer

1) = 6*  pi/3  - 4*(- pi/4) = 2 pi + pi = 3pi; 2) = 2* pi/4 + 3*(-[i/6) = pi/2 - pi/2 = 0;3)  = 5*(-pi/3) - 3*3pi/4 = - 5pi/3 - 9 pi/4 = = - 20pi/12 - 27 pi/12= - 47 pi/12...