Question

№6.2 пожалуйста срочно надо

Answer 1

sin (-2π)=sin0=0
sin(-π/2)=-sinπ/2=-1
sin(-3π/2)=-sin3π/2=-(-1)=1
sin(-π)=-sinπ=0
cos (-2π)=cos0=1
cos(-π/2)=cosπ/2=0
cos(-3π/2)=cos3π/2=0
cos(-π)=cosπ=-1
tg(-2π)=tg0=0
tg(-π/2)=не сущ
tg(-3π/2)=tg3π/2=не сущ
tg(-π)=tg0=0