cos3/2x*sin3/2x-cos1/2x*sin1/2x
=1/2*sin3x-1/2*sinx
=1/2*(sin3x-sinx)
=1/2*[sin(2x+x)-sin(2x-x)]
=1/2*(sin2xcosx+cos2xsinx-sin2xcosx+cos2xsinx)
=sinxcos2x那求單調(diào)增區(qū)間和最值呢f(x)=sinxcos2x=sinx(1-2sin²x)=-2sin³x+sinx那么f'(x)=-6sin²xcosx+cosx=cosx(1-6sin²x)=cosx(6cos²x-5)令f'(x)≥0，那么cosx(6cos²x-5)≥0cosx(cosx+√30/6)(cosx-√30/6)≥0所以-√30/6≤cosx≤0，或√30/6≤cosx≤1那么2kπ+π/2≤x≤2kπ+arccos(-√30/6)(k∈Z)或2kπ+2π-arccos(-√30/6)≤x≤2kπ+3π/2(k∈Z)或2kπ-arccos(√30/6)≤x≤2kπ+arccos(√30/6)(k∈Z)即單調(diào)遞增區(qū)間為上述的三個(gè)區(qū)間f(x)max=f(2kπ+3π/2)=1，f(x)min=f(2kπ+π/2)=-1