问题描述: 求下列角的三个三角函数值.(1)-17π/3(2)21π/4(3)-23/6(4)1500° 1个回答 分类:数学 2014-10-01 问题解答: 我来补答 -17π/3=-6π+π/3sin(-17π/3)=sin(-6π+π/3)=sin(π/3)=√3/2cos(-17π/3)=cos(π/3)=1/2tan(-17π/3)=tan(π/3)=√321π/4=6π-3π/4sin(21π/4)=sin(6π-3π/4)=-√2/2cos(21π/4)=cos(6π-3π/4)=-√2/2tan(21π/4)=tan(6π-3π/4)=1-23π/6=-4π+π/6sin(-23π/6)=sin(-4π+π/6)=sin(π/6)=1/2cos(-23π/6)=cos(π/6)=√3/2tab(-23π/6)=tan(π/6)=√3/31500°=360°×4+60°sin1500°=sin(360°×4+60°)=sin60°=√3/2cos1500°=cos60°=1/2tan1500°=tan60°=√3 展开全文阅读