Question

高中任意角的三角函数那节的定义法和公式一是什么东西哦、、、

如题、具体题目是下面那个、你跟我说什么是定义法和公式一就行、要是把题也解了就更好了、、

用定义法、公式一求下列角的三个三角函数值
①-17π/3 ②21π/4 ③-23π/6 ④1500°

Answer 1

①-17π/3 =-18π/3+π/3=-6π+π/3
所以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

②21π/4 =16π/4+5π/4=4π+5π/4
sin(21π/4)=sin(4π+5π/4)=sin(5π/4)=sin(π+π/4)=-sinπ/4=-√2/2
cos(21π/4)=cos(5π/4)=cos(π+π/4)=-cosπ/4=-√2/2
tan(21π/4)=tan(5π/4)=tan(π+π/4)=tanπ/4=1

③-23π/6 =-24π/6+π/6=-4π+π/6
sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=1/2
cos(π/6))=√3/2 tan(π/6)=√3/3

④1500°=1440°+60°=360°×4+60°
sin1500°=sin(360°×4+60°)=sin60°（结果同①）