Question

Вычислите sin t и cos t, если:
а) t= 13П/6
б) t= -8П/3
в) t=23П/6
г) t=-11П/3
д) t=9П/4

Answer 1

1) 13π/6=2π+π/6
sin(2π+π/6)=sinπ/6=1/2; cos(2π+π/6)=cosπ/6=√3/2
2) 8π/3=2π+2π/3
sin(-(2π+2π/3))=-sin(2π+2π/3)=-sin2π/3=-sin(π-π/3)=-sinπ/3=-√3/2;
cos(-(2π+2π/3))=cos(2π+2π/3)=cos2π/3=-cos(π-π/3)=-cosπ/3=-1/2;
3) 23π/6=4π-π/6
sin(4π-π/6)=-sinπ/6=-1/2; cos(4π-π/6)=cosπ/6=√3/2
4) 11π/3=4π-π/3
sin(-(4π-π/3))=-sin(4π-π/3)=sinπ/3=√3/2;
cos(-(4π-π/3))=cos(4π-π/3)=cosπ/3=1/2;
5) 9π/4=2π+π/4
sin(2π+π/4)=sinπ/4=√2/2; cos(2π+π/4)=cosπ46=√2/2

cosα - функция четная
sinα - функция нечетная