Question

sin4π/3cos5π/6tan(－4π/3)怎么做,求详细过程

sin4π/3乘以cos5π/6乘以tan(－4π/3)=？

Answer 1

sin cos的周期为2π tan周期为π
sin4π=sin(4π-4π)=sin0=0
3cos5π=3cos(5π-4π)=3cosπ=3*(-1)=-3

6tan(－4π/3)=6tan(－4π/3+π)=6tan(-π/3)=-6tan(π/3)=-6√3

追问

sin4π/3乘以cos5π/6乘以tan(－4π/3)=？

追答

sin（π/2＋α）＝cosα
cos（π/2＋α）＝－sinα

sin4π/3=sin(4π/3-2π)=sin(-2π/3)=-sin(2π/3)=-cos(2π/3-π/2)=-cos(π/6)=-√3/2
cos5π/6=-cos(5π/6-π)=-cos(-π/6)=-cosπ/6=-√3/2

tan(－4π/3)=tan(-4π/3+π)=tan(-π/3)=-tan(π/3)=-√3

sin4π/3乘以cos5π/6乘以tan(－4π/3)=-√3/2*(-√3/2)*(-√3)=-3√3/4

tanα ·cotα＝1
sinα ·cscα＝1
cosα ·secα＝1 sinα/cosα＝tanα＝secα/cscα
cosα/sinα＝cotα＝cscα/secα sin2α＋cos2α＝1
1＋tan2α＝sec2α
1＋cot2α＝csc2α
诱导公式
sin（－α）＝－sinα
cos（－α）＝cosα tan（－α）＝－tanα
cot（－α）＝－cotα
sin（π/2－α）＝cosα
cos（π/2－α）＝sinα
tan（π/2－α）＝cotα
cot（π/2－α）＝tanα
sin（π/2＋α）＝cosα
cos（π/2＋α）＝－sinα
tan（π/2＋α）＝－cotα
cot（π/2＋α）＝－tanα
sin（π－α）＝sinα
cos（π－α）＝－cosα
tan（π－α）＝－tanα
cot（π－α）＝－cotα
sin（π＋α）＝－sinα
cos（π＋α）＝－cosα
tan（π＋α）＝tanα
cot（π＋α）＝cotα

背公式,多做题,

不懂请追问,有帮助请采纳,谢谢.

Answer 2

sin(4π/3)*cos(5π/6)*tan(-4π/3)
=-sin(π/3)*(-cos(π/6))*tan(-π/3)
=(√3/2)*(√3/2)*(-√3)
=-3√3/4