∵0≤φ≤π/2
∴0≤sinφ≤1,0≤cosφ≤1
且sinφ+cosφ
=√2(√2/2sinφ+√2/2cosφ)
=√2sin(φ+π/4)≤√2
又√2<π/2
∴sinφ+cosφ<π/2
∴0≤sinφ<π/2-cosφ<π/2
∵函数y=cosx在[0,π/2]上为减函数
∴cos(sinφ)>cos[π/2-cosφ]=sin(cosφ)
即cos(sinφ)>sin(cosφ)