Mara Cook

2022-06-26

For all $\theta$ in $\left[0,\pi /2\right]$ I need to show that $\mathrm{cos}\left(\mathrm{sin}\theta \right)>\mathrm{sin}\left(\mathrm{cos}\theta \right)$

pheniankang

Expert

Over $I=\left[0,\frac{\pi }{2}\right]$ we have:
$\mathrm{sin}\left(\theta \right)+\mathrm{cos}\left(\theta \right)=\sqrt{2}\mathrm{sin}\left(\theta +\frac{\pi }{4}\right)\le \sqrt{2}<\frac{\pi }{2}$
hence:
$\begin{array}{}\text{(1)}& \mathrm{\forall }\theta \in I,\phantom{\rule{2em}{0ex}}\mathrm{cos}\theta <\frac{\pi }{2}-\mathrm{sin}\theta .\end{array}$
The LHS of (1) belongs to [0,1], the RHS belongs to $\left[\frac{\pi }{2}-1,\frac{\pi }{2}\right]$; the sine function is increasing over $\left[0,\frac{\pi }{2}\right]$, hence:
$\begin{array}{}\text{(2)}& \mathrm{\forall }\theta \in I,\phantom{\rule{2em}{0ex}}\mathrm{sin}\left(\mathrm{cos}\theta \right)<\mathrm{cos}\left(\mathrm{sin}\theta \right).\end{array}$

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