已知sin(a+pie⼀3)+sina=-4根号3⼀5,-pie⼀2<a<0,求cosa的值

2025-04-03 22:28:05
推荐回答(1个)
回答1:

∵sin(a+π/3)+sina=-4根号3/5
∴sina/2+(cosa根号3)/2+sina=(-4根号3)/5
∴((sina*根号3)/2+cosa/2)*根号3=(-4根号3)/5
∴sin(a+π/6)=-4/5
又∵-π/2∴-π/3∴cos(a+π/6)=3/5
∴cosa=cos((a+π/6)-π/6)
=(-4+3*根号3)/10