解:
由于:
sinxcosx
=sinxcosx+(1-1)
=sinxcosx+[sin^2(x)+cos^2(x)]-1
={2sinxcosx+[sin^2(x)+cos^2(x)]-1}/2
=[(sinx+cosx)^2-1]/2
设T=sinx+cosx
则:
y
=(sinx+1)(cosx+1)
=sinxcosx+1+(sinx+cosx)
=[(sinx+cosx)^2-1]/2 +1+(sinx+cosx)
=[T^2-1]/2+1+T
=(1/2)T^2+T+(1/2)
=(1/2)(T+1)^2
由于:
T
=sinx+cosx
=√2sin(x+兀/4)
又X属于[-兀/6,兀/2]
则:(X+兀/4)属于[兀/12,3兀/4]
则:T属于[(√3+1)/2,√2]
则:
当T=(√3+1)/2时,Y取最小值=(6+3√3)/4
当T=√2时,Y取最大值=(3+2√2)/2