急已知函数f(x)=sinx+sin(x－π/3),在△ABC中.角A,B,C的对边分

题目

急已知函数f(x)=sinx+sin(x－π/3),在△ABC中.角A,B,C的对边分
已知函数f(x)=sinx+sin(x－π/3),
①求f(x)的单调递增区间
②在△ABC中.角A,B,C的对边分别为abc,已知f(A)=√3/2,a=√3b试判断△ABC的形状

答案

1
f(x)=sinx+sin(x-π/3)
=(3/2)sinx-[(√3)cosx]/2
=(√3)sin(x-π/6)
2kπ-π/2≤x-π/6≤2kπ+π/2
单调递增区间{x|2kπ-(π/3)≤x≤2kπ+2π/3}
2
f(A)=(√3)/2
A=π/3
sinA=(√3)/2
asinB=bsinA
B=π/6
C=90°
△ABC为RT三角形