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

zevs34 zevs34    1   17.06.2019 23:10    8

Ответы
Kamaria086 Kamaria086  02.10.2020 04:11
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
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