Question

已知函数f(x)=2cosxsin(x+Ω/3)-根号3sinxsinx+sinxcosx，求f(x)的单调减区间

Answer 1

∵f(x)=sin2x+√3(cosx)^2-√3(sinx)^2=sin2x+√3cos2x=2sin(2x+π/3)
2kπ-π/2<=2x+π/3<=2kπ+π/2==>kπ-5π/12<=x<=kπ+π/12
单调增
2kπ+π/2<=2x+π/3<=2kπ+3π/2==>kπ+π/12<=x<=kπ+7π/12
单调减

Answer 2

原式=cosx(sinx+√3cosx)-√3sin*sin+sinxcosx
=2sinxcosx+√3(cosx*cosx-sinxsinx)
=sin2x+√3cos2x
=2sin(2x+π/3)

Answer 3

解:y=f(x)=(0.5+cosa)sin2x+(0.5cosa+sina+0.5√3)cos2x+0.5cosa+sina-0.5√3
a=Ω/3
1如果Ω=π
则有y'=√5+2√3cos(2x+ψ) <0 tanψ=√3-0.25
2x+ψ∈[2kπ,2kπ+π]
x∈[kπ-0.5ψ,kπ+0.5π-0.5ψ]
2否则同理