Question

已知函数f(x)=2cos(x-2π/3)+2cosx,x∈[π/2,π]

已知函数f(x)=2cos(x-2π/3)+2cosx,x∈[π/2,π]1.若sinx=4/5，求函数f(x)的值2.求函数f(x)的值域

Answer 1

（1）f(x)=2cos(x-2/3π)+2cosx
=2(cosxcos2/3π+sinxsin2/3π)+2cosx
=√3sinx-cosx+2cosx
=√3sinx+cosx ∵sinx=4/5，∴cosx=±3/5 ∴f(x)=(4√3+3)/5或(4√3-3)/5
(2)f(x)=2(sinxcosπ/3+cosxsinπ/3)
=2sin(x+π/3) -1≤sin(x+π/3)
≤1
∴值域为[-2,2]

Answer 2

f(x)=(sinx+cosx)^2+2(cosx)^2-2=(sinx)^2+(cosx)^2+2sinxcosx+2(cosx)^2-1-1=1+sin2x+cosx2-1=sin2x+cos2x=2^0.5sin(2x+π/4)。故函数周期T=π。