sin(-21π/4) cos(-20π/3) tan(-19π/6)=

问题描述:

sin(-21π/4) cos(-20π/3) tan(-19π/6)=
sin(-21π/4) cos(-20π/3) tan(-19π/6)
分别=?
要详细过程 解释下~~~~~~~~~~~~~~~~~~~~~~~~
1个回答 分类:数学 2014-10-25

问题解答:

我来补答
sin(-21π/4)
=sin(-6兀+3兀/4)
=sin(3兀/4)
=sin(兀-兀/4)
=sin(兀/4)
=√2/2
cos(-20π/3)
=cos(-6兀-2兀/3)
=cos(-2兀/3)
=cos(2兀/3)
=cos(兀-兀/3)
=-cos(兀/3)
=-1/2
tan(-19兀/6)
=tan(-4兀+5兀/6)
=tan(5兀/6)
=tan(兀-兀/6)
=-tan(兀/6)
=-√3/3
 
 
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