sin【a+(π/3)】+sina=(-4√3)/5,-π/2<a<0,cos【a+(2π/3)】=
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sin(a+π/3) + sina
=sinacosπ/3+cosasinπ/3+sina
=3/2sina+√3/2cosa
=√3(1/2sina+1/2cosa)
=√3(sinacosπ/6+cosasinπ/6)
= √3sin(a+π/6)
由sin(a+π/3) + sina= -4√3/5,得sin(a+π/6) = -4/5
cos(a+2π/3) = cos(a+π/6+π/2) = -sin(a+π/6)
cos(a+2π/3)= 4/5
=sinacosπ/3+cosasinπ/3+sina
=3/2sina+√3/2cosa
=√3(1/2sina+1/2cosa)
=√3(sinacosπ/6+cosasinπ/6)
= √3sin(a+π/6)
由sin(a+π/3) + sina= -4√3/5,得sin(a+π/6) = -4/5
cos(a+2π/3) = cos(a+π/6+π/2) = -sin(a+π/6)
cos(a+2π/3)= 4/5
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