Question

求下列各三角函数值: (1).sin(-11π/6) (2).cos(-17π/3) （3).tan(-17π)/6) (4).tan(-31π/4) 详解

Answer 1

解：(1).sin(-11π/6)=sin(π/6-2π)=sin(π/6)=1/2
(2).cos(-17π/3)=cos(-5π/3) =cos(π/3)=1/2
（3).tan(-17π)/6)=tan(π/6-3π)=tan(π/6)=(√3)/3
(4).tan(-31π/4) =tan(π/4-8π)=tan(π/4)=1

Answer 2

(1).sin(-11π/6)=1/2, (2).cos(-17π/3) =1/2
（3).tan(-17π)/6)=√3/3 , (4).tan(-31π/4)=1

Answer 3

Sin(-11/6π)
=sin(-π-5/6π)
=-sin(π+5/6π)
=-sinπcos(5/6π)-cosπsin(5/6π)
=1/2
（2）Cos(-17/3π)
=cos(17/3π)
=cos(4π+5/3π)
=cos(5/3π)
=cos(π+2/3π)
=-cos(2/3π)
=1/2
（3）Tan(-17/6π)
=-tan(2π+5/6π)
=-tan(5/6π)
=tan(π/6)
=√3/3
（4）Tan(-31/4π)
=-tan(8π-π/4)
=tan(π/4)
=1