Using cofuncrion, transform the original to the following form.

cos(sinθ) > cos(π/2-cosθ)

In the domain of 0 ≤ θ ≤ π/2, cosine function is decreasing. What we need to do is to prove sinθ< π/2-cosθ

Since \sinθ+\cosθ=\sqrt{2}\sin(θ+\dfrac{\pi}{4})\leq \sqrt{2} < \dfrac{\pi }{2}

Therefore,

\sinθ < \dfrac{\pi }{2}-\cos θ

Then

\cos\sinθ >\cos(\dfrac{\pi }{2}-\cos θ )

\cos\sinθ >\sin\cos θ