Question

急！！！用定义法，公式一以及计算器等求下列角的三个三角函数值。

（1）- 17π/ 3 （2）21π /4 （3）- 23π /6 （4）1500° 【请帮我解决这几道题，最好把公式也写一下，我就是不会用公式！】

Answer 1

sin(2kπ+α)=sinα，cos(2kπ+α)=cosπ，tan(2kπ+α)=tanα，k∈Z
sin(π+α)=-snα，cos(π+α)=-cosα，tan(π+α)=tanα.

(1)sin(-17π/3)=sin(-6π+π/3)=sinπ/3=√3/2，
cos(-17π/3)=cosπ/3=1/2，
tan(-17π/3)=tanπ/3=√3；
(2)sin21π/4=sin[4π+(π++π/4)]=sin(π+π/4)=-sinπ/4=-√2/2，
cos21π/4=cos(π+π/4)=-cosπ/4=-√2/2，
tan21π/4=tan(π+π/4)=tanπ/4=1；
(3)sin(-23π/4)=sin(-6π+π/4)=sinπ/4=√2/2，
cos(-23π/4)=cosπ/4=√2/2，
tan(-23π/4)=tanπ/4=1；
(4)sin1500° =sin(4×360° +60° )=sin60° =√3/2,
cos1500° =cos60° =1/2,
tan1500° =tan60° =√3。

Answer 2

sin(2kπ+α)=sinα，cos(2kπ+α)=cosπ，tan(2kπ+α)=tanα，k∈Z
sin(π+α)=-snα，cos(π+α)=-cosα，tan(π+α)=tanα.

(1)sin(-17π/3)=sin(-6π+π/3)=sinπ/3=√3/2，
cos(-17π/3)=cosπ/3=1/2，
tan(-17π/3)=tanπ/3=√3；
(2)sin21π/4=sin[4π+(π++π/4)]=sin(π+π/4)=-sinπ/4=-√2/2，
cos21π/4=cos(π+π/4)=-cosπ/4=-√2/2，
tan21π/4=tan(π+π/4)=tanπ/4=1；
(3)sin(-23π/4)=sin(-6π+π/4)=sinπ/4=√2/2，
cos(-23π/4)=cosπ/4=√2/2，
tan(-23π/4)=tanπ/4=1；
(4)sin1500° =sin(4×360° +60° )=sin60° =√3/2,
cos1500° =cos60° =1/2,
tan1500° =tan60° =√3。

Answer 3

1. sin-17π/3=sin(-6π+π/3)=sinπ/3=√3/2
同理cos-17π/3=cosπ/3=1/2 tan- 17π/ 3 =sin(-17π/3)/(cos-17π/3)=√3
2. sin21π /4=sin(5π+π/4)=sin(π+π/4)=-sinπ/4=-√2/2 同理cos21π /4=-cosπ/4=-√2/2
tan21π /4=1
3.sin- 23π /6=sin(-4π+π/6)=sinπ/6=1/2 同理cos- 23π /6=cosπ/6=√3/2 tan- 23π /6= √3/3
4. sin1500°=sin(4π +60°)=sin60°= √3/2 同理cos1500°=cos60°=1/2
tan1500°=sin1500°/cos1500°=√3