Question

数学急！已知sin（a＋派/3）+sina＝-(4根号3)/5，－派/2<a<0，则cosa

Answer 1

sin（a＋派/3）+sina＝-(4√3)/5
∴sinacosπ/3+cosasinπ/3+sina=-4√3/5
∴3/2*sina+√3/2*cosa=-4√3/5
两边同时除以√3
∴√3/2sina+1/2*cosa=-4/5
∴sin(a+π/6)=-4/5
∵-π/2<a<0 ∴-π/3<a+π/6<π/6
∴cos(a+π/6)=√[1-sin²(a+π/6)]=3/5
∴cosa=cos[(a+π/6)-π/6]
=cos(a+π/6)cosπ/6+sin(a+π/6)sinπ/6
=3/5*√3/2-4/5*1/2
=(3√3-4)/10

Answer 2

sin（a＋派/3）+sina＝-(4根号3)/5
1/2*sina+√3/2*cosa+sina＝-(4根号3)/5
3/2*sina+√3/2*cosa＝-(4根号3)/5
√3(√3/2*sina+1/2*cosa)＝-(4根号3)/5
√3/2*sina+1/2*cosa＝-4/5
sin（a＋派/6）=-4/5
∵－派/2<a<0
∴cos（a＋派/6）=3/5
所以cosa=cos[（a＋派/6）-派/6]=cos（a＋派/6）cos派/6+sin（a＋派/6）sin派/6
=√3/2*cos（a＋派/6)+1/2*sin（a＋派/6）
=√3/2*(3/5)+1/2*(-4/5)
=(3√3-4)/10

Answer 3

sinθ+sinφ = 2 sin[(θ+φ)/2] cos[(θ-φ)/2]
You know?

Answer 4

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