若θ∈[0,(π/2)],试比较cos(sinθ)与sin(cosθ)的大小.

[cos(sinθ)-sin(cosθ)]'=-sin(sinθ)cosθ-cos(cosθ)sinθ＜0,所以cos(sinθ)-sin(cosθ)单调递减,所以cos(sinθ)-sin(cosθ)的最小值在θ=π/2的时候取到
当θ=π/2的时候,cos(sinθ)-sin(cosθ)=cos1-sin0=cos1＞0
所以cos(sinθ)＞sin(cosθ),θ∈[0,π/2]