f（x）=sin（x+5π/4)+cos（x+3π/4） ①求f（x）的单调递增区间 ②若f（x）=（2√2）/3 si

f(x)=sin(x+5π/4)+cos(x+3π/4),①求f（x）的单调递增区间； ②若f(x)=(2√2)/3,
求 sinx/(1+tanx)的值.
f(x)=sin[π+(x+π/4)]+cos[π+(x-π/4)]=-sin(x+π/4)-cos(x-π/4)=-(√2/2)[(sinx+cosx)+(cosx+sinx)]
=-(√2)(sinx+cosx)=-2sin(x+π/4)
由π/2+2kπ≦x+π/4≦3π/2+2kπ,得单增区间为π/4+2kπ≦x≦5π/4+2kπ,k∈Z.
若f(x)=-2sin(x+π/4)=-(√2)(sinx+cosx)=(2√2)/3,则sinx+cosx=-2/3
故此时sinx/(1+tanx)=sinxcosx/(sinx+cosx)=2sinxcosx/[2(sinx+cosx)]
=[(sinx+cosx)²-1]/[2(sinx+cosx)]=[(-2/3)²-1]/[2(-2/3)]=(5/9)/(4/3)=5/12.