三角函数求值,谢谢给好评
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三角函数求值,谢谢给好评
2.(1) cos210°=cos(180°+30°=-cos30°=-(√3/2);
(3)cos(-π/6)=cos(π/6)=√3/2;
(5)cos(-11π/9)=cos(11π/9)=cos(π+2π/9)=-cos(2π/9);
(7) tan632°24'=tan632.4° =tan(2*360°--87.6°)=-tan87.6°=-23.86.
2.(1)cos(-17π/4)=cos(17π/4)=cos[4π+π/4)=cosπ/4=√2/2;
(2)sin(-2160°52'=-sin(6*360°+0.87°)=-sin(0.87) =-0.015 .
(3)cos(-π/6)=cos(π/6)=√3/2;
(5)cos(-11π/9)=cos(11π/9)=cos(π+2π/9)=-cos(2π/9);
(7) tan632°24'=tan632.4° =tan(2*360°--87.6°)=-tan87.6°=-23.86.
2.(1)cos(-17π/4)=cos(17π/4)=cos[4π+π/4)=cosπ/4=√2/2;
(2)sin(-2160°52'=-sin(6*360°+0.87°)=-sin(0.87) =-0.015 .