sin（a+兀／3）+sina=-4根号3／5,-兀／2＜a＜0,则cosa

Question

sin（a+兀／3）+sina=-4根号3／5,-兀／2＜a＜0,则cosa

Answer 1

解析：
已知sin（a+兀／3）+sina=-4根号3／5,那么：
sina*cos(π/3)+cosa*sin(π/3)+sina=-4根号3/5
(3/2)*sina+cosa*(根号3/2)=-4根号3/5
(根号3/2)*sina+cosa*(1/2)=-4/5
sina*cos(π/6)+cosa*sin(π/6)=-4/5
即得：sin(a+π/6)=-4/5
因为-π/2＜a＜0,那么：-π/3＜a+π/6＜π/6
所以由sin(a+π/6)=-4/5<0,可知角a+π/6是第四象限角
那么：cos(a+π/6)>0
所以解得：cos(a+π/6)=根号[1-sin²(a+π/6)]=3/5
那么：cosa
=cos[(a+π/6)-π/6]
=cos(a+π/6)*cos(π/6)+ sin(a+π/6)*sin(π/6)
=(3/5)*(根号3/2) + (-4/5)*(1/2)
=(3根号3 -4)/10