用公式一求下列角的三个三角函数值：21π/4，-23π/6 ，要有过程
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sin(21π/4)=sin(4π+5π/4)=sin(π+π/4)= - sin(π/4)= - √2/2
sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=0.5
cos(21π/4)=cos(4π+5π/4)=cos(π+π/4)= - cos(π/4)=- √2/2
cos(-23π/6)=cos(-4π+π/6)=cos(π/6)=√3/2
tan(21π/4)=tan(4π+5π/4)=tan(π+π/4)= tan(π/4)=1
tan(-23π/6)=tan(-4π+π/6)=tan(π/6)=√3/3
sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=0.5
cos(21π/4)=cos(4π+5π/4)=cos(π+π/4)= - cos(π/4)=- √2/2
cos(-23π/6)=cos(-4π+π/6)=cos(π/6)=√3/2
tan(21π/4)=tan(4π+5π/4)=tan(π+π/4)= tan(π/4)=1
tan(-23π/6)=tan(-4π+π/6)=tan(π/6)=√3/3
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