Question

For all $\theta$ in $[0,\pi /2]$, show that $\cos \left(\sin \theta \right)>\sin \left(\cos \theta \right)$.

Answer 1

$cos\left(sin\theta \right)7sin\left(cos\theta \right)+\theta ϵ(0,\frac{\pi }{2})$

$cos\left(sin\theta \right)7cos(\frac{\pi }{2}-cos\theta )>0$

$2sin(\frac{\pi }{4}+\frac{1}{2}(sin\theta -cos\theta \left)\right)]=sin[\frac{\pi }{4}-\frac{1}{2}(sin\theta -cos\theta )]>0$

$\to |sin\theta -cos\theta |=\left|\sqrt{2}sin\right(\frac{\pi }{4}\left)\right|\le \sqrt{2}<\frac{\pi }{2}$

$\frac{-\pi }{2}<(sin\theta -cos\theta )<\frac{\pi }{2}$

$\frac{-\pi }{4}<\frac{sin\theta -cos\theta }{2}<\frac{\pi }{4}$

$0<\frac{\pi }{4}+\frac{1}{2}(sin\theta -cos\theta )<\frac{\pi }{4}$

$\therefore sin[\frac{\pi }{4}+\frac{1}{2}(sin\left(\theta \right)-cos\theta \left)\right]>0$

proved $cos\left(sin\theta \right)7sin\left(cos\theta \right)$