tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=

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tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin

tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=
tan(5π+a)=-2且cosa>0则sin(a-π)=
sin(-π/3)+2sin4/3π+3sin2/3π=

tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=
sin(a-π)= （根号5）分之2
sin(-π/3)+2sin4/3π+3sin2/3π= 自己算吧,太难写了,应该不难