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=

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