Question

函数f(x)=sinx-cos(x+pai/6)的值域为什么是[-根号3，根号3]

Answer 1

原题：函数f(x)=sinx-cos(x+π/6)的值域为什么是[-√3，√3]？

f(x)=sinx-cos(x+π/6)=sinx-sin(x+2π/3)
=sin[(x+π/3)-π/3]-sin[(x+π/3)+π/3]
=-2cos(x+π/3)sin(π/3)=-√3cos(x+π/3)

其值域是[-√3，√3].

希望对你有点帮助！

Answer 2

f(x)=sinx-cos(x+π/6)
＝sinx-根号3/2cosx+1/2sinx
=3/2sinx-根号3/2cosx
=根号3sin(x-π/6)
-1<=sin(x-π/6)<=1
值域是[-根号3，根号3]

Answer 3

解：∵f(x)=sinx-cos(x+π/6)
=sinx-cosx*cos(π/6)+sinx*sin(π/6)
=(3/2)*sinx-(√3/2)*cosx
=√3[(√3/2)*sinx-(1/2)*cox]
=√3sin(x-π/6)
∴f(x)的值域为[-√3，√3]

Answer 4

cos(x+pai/6)=根号3/2*cos(x)-1/2*sin(x)
f(x)=3/2*sin(x)-根号3/2*cos(x)=根号3*tan(x+常数）
所以是-根号3，根号3

Answer 5