求下列三角函数值: ①-17π/3 ②21π/4 ③-23π/6 ④1500°

tity819 1年前 已收到1个回答 举报

beermusic 幼苗

共回答了20个问题采纳率:95% 举报

-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
tan(-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

1年前

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