Question

求y=3cosx-根号3sinx 的值域 要过程

Answer 1

y=2√3(cosx*√3/2-sinx*1/2)
=2√3(cosxcosπ/6-sinxsinπ/6)
=2√3cos(x+π/6)
所以值域[-2√3,2√3]

Answer 2

y=3cosx-√3sinx
=√3/2(√3/2cosx-1/2sinx)
=√3/2(sinπ/3cosx-cosπ/3sinx)
=√3/2sin(π/3-x)
因为sin(π/3-x)∈[-1,1]
所以，y∈[-√3/2,√3/2].

Answer 3

令t=根号3sinx ,则y=3(1-t^2/9)-t，t属于[0，根号3]
配方知函数在t属于[0，根号3]是减函数，
所以值域为[2-根号3,3]

Answer 4

y=2√3[(√3/2)cosx-(1/2)sinx]
=2√3cos(π/6+x)
-2√3<=y<=2√3
值域[-2√3 , 2√3]