题目
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)
提问时间:2020-08-10
答案
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
=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
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