f(x)=sin(5π/4-x)-cos(π/4+x)
=sin(π+π/4-x)-cos(π/4+x)
=-sin(π/4-x)-cos(π/4-x)
=-√2(√2/2sin(π/4-x)+√2/2cos(π/4-x))
=-√2sin(π/4-x+π/4)
=-√2sin(π/2-x)
=-√2cosx
∵f(x)遞減
∴cosx遞增
∴2kπ-π<=x<=2kπ
0＜α＜b＜π/2
-π/2<-b<0
-π/2cos(a-b)=3/5
sin(a-b)=-√(1-cos^2(a-b))=-4/5
0cos(a+b)=-3/5
sin(a+b)=√(1-cos^2(a+b))=4/5
cos2b=cos(a+b-(a-b))
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=-3/5*3/5 +4/5*(-4/5)
=-9/25-16/25
=-1
=2cos^2b-1
cosb=0
b=π/2(題意沒說b<=π/2啊,奇了怪了)
f(b)=-√2cosb=0抄錯題了，確實有β<=π/2，大哥第二問是多少啊我答的情況就是β=π/2時的情況謝了。