设β-π/4为A,α+β为B,则α+π/4为B-A
COS(α+π/4)=COS(B-A)=COSBCOSA+SINBSINA
COSA=1/3,SIN²A=8/9.且π/4<A<3/4π,所以SINA=2√2/3
SINB=4/5,COS²B=9/25,且π<B<3π/2,所以COSB=-3/5
原式=-3/5×1/3+4/5×2√2/3
=(8√2-3)/15
设β-π/4为A,α+β为B,则α+π/4为B-A
COS(α+π/4)=COS(B-A)=COSBCOSA+SINBSINA
COSA=1/3,SIN²A=8/9.且π/4<A<3/4π,所以SINA=2√2/3
SINB=4/5,COS²B=9/25,且π<B<3π/2,所以COSB=-3/5
原式=-3/5×1/3+4/5×2√2/3
=(8√2-3)/15