求函數(shù)y=2cosxsinx(x+π/3)-根號(hào)3sin^2x+sinxcosx的周期,最值和單調(diào)區(qū)間

數(shù)學(xué)人氣：540 ℃時(shí)間：2020-04-11 03:52:23

優(yōu)質(zhì)解答

解
原式=2cosx(sinx*1/2+cosx√3/2)-√3sin²x+sinxcosx
=cosxsinx+√3cos²x-√3sin²x+sinxcosx
=2sinxcosx+√3(cosx²-sin²x)
=sin2x+√3cos2x
=2sin(2x+π/3)
周期為2π/2=π
當(dāng)sin(2x+π/3)=1時(shí),有最大值2
當(dāng)sin(2x+π/3)=-1時(shí),有最小值-2
sinx在x∈[2kπ-π/2,2kπ+π/2]上單增
所以sin(2x+π/3)在x∈[kπ-π/12,kπ+π/12]上單調(diào)遞增
sinx在x∈[2kπ+π/2,2kπ+3π/2]上單減
所以sin(2x+π/3)在x∈[kπ+π/12,kπ+7π/12]上單調(diào)遞減