已知cos(π/4-a)=-3/5
所以cos(a-π/4)=-3/5
π/2≤a≤π
所以π/4≤a-π/4≤3π/4
所以sin(a-π/4)=√[1-(-3/5)²]=4/5
所以cosa=cos[(a-π/4)+π/4]
=cos(π/4-a)cos(π/4)-sin(a-π/4)sin(π/4)
=(-3/5)*(√2/2)-(4/5)*(√2/2)
=-√2/10
已知cos(π/4-a)=-3/5
所以cos(a-π/4)=-3/5
π/2≤a≤π
所以π/4≤a-π/4≤3π/4
所以sin(a-π/4)=√[1-(-3/5)²]=4/5
所以cosa=cos[(a-π/4)+π/4]
=cos(π/4-a)cos(π/4)-sin(a-π/4)sin(π/4)
=(-3/5)*(√2/2)-(4/5)*(√2/2)
=-√2/10