Mara Cook

Answered

2022-06-26

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

Answer & Explanation

pheniankang

Expert

2022-06-27Added 22 answers

Over $I=[0,\frac{\pi}{2}]$ we have:

$\mathrm{sin}(\theta )+\mathrm{cos}(\theta )=\sqrt{2}\mathrm{sin}(\theta +\frac{\pi}{4})\le \sqrt{2}<\frac{\pi}{2}$

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 $[\frac{\pi}{2}-1,\frac{\pi}{2}]$; the sine function is increasing over $[0,\frac{\pi}{2}]$, hence:

$\begin{array}{}\text{(2)}& \mathrm{\forall}\theta \in I,\phantom{\rule{2em}{0ex}}\mathrm{sin}(\mathrm{cos}\theta )<\mathrm{cos}(\mathrm{sin}\theta ).\end{array}$

$\mathrm{sin}(\theta )+\mathrm{cos}(\theta )=\sqrt{2}\mathrm{sin}(\theta +\frac{\pi}{4})\le \sqrt{2}<\frac{\pi}{2}$

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 $[\frac{\pi}{2}-1,\frac{\pi}{2}]$; the sine function is increasing over $[0,\frac{\pi}{2}]$, hence:

$\begin{array}{}\text{(2)}& \mathrm{\forall}\theta \in I,\phantom{\rule{2em}{0ex}}\mathrm{sin}(\mathrm{cos}\theta )<\mathrm{cos}(\mathrm{sin}\theta ).\end{array}$

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