We have to prove
cos(sinθ)−sin(cosθ)>0
or cos(sinθ)−cos(π2−cosθ)>0
or 2sin(π4+sinθ−cosθ2)
sin(π4+sinθ−cosθ2)>0....(1)
We now show that both the factors on the left hand side of (1) are positive. since,
|sinθ−cosθ|=∣∣∣√2sin(θ−π4)∣∣∣≤√2≤π2
We have −π2<sinθ−cosθ<π2
−π4<sinθ−cosθ2<π4
Add π4 so that 0<π4<sinθ−cosθ2<π2
and therefore sin(π4+sinθ−cosθ2)>0
Similarly we can prove sin(π4−sinθ+cosθ2)>0
Hence (1) holds which is what we wanted to prove.