求下列角的三个三角函数值.（1）-17π／3（2）21π／4（3）-23／6（4）1500°

网友回答

-17π/3=-6π+π/3
sin(-17π/3)=sin(-6π+π/3)=sin(π/3)=√3/2
cos(-17π/3)=cos(π/3)=1/2
tan(-17π/3)=tan(π/3)=√3
21π/4=6π-3π/4
sin(21π/4)=sin(6π-3π/4)=-√2/2
cos(21π/4)=cos(6π-3π/4)=-√2/2
tan(21π/4)=tan(6π-3π/4)=1
-23π/6=-4π+π/6
sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=1/2
cos(-23π/6)=cos(π/6)=√3/2
tab(-23π/6)=tan(π/6)=√3/3
1500°=360°×4+60°
sin1500°=sin(360°×4+60°)=sin60°=√3/2
cos1500°=cos60°=1/2
tan1500°=tan60°=√3
======以下答案可供参考======
供参考答案1:
函数在哪