Find the solution of $\lfloor {\mathrm{sin}}^{-1}(x)\rfloor >\lfloor {\mathrm{cos}}^{-1}(x)\rfloor $

Leonel Contreras
2022-06-25
Answered

Find the solution of $\lfloor {\mathrm{sin}}^{-1}(x)\rfloor >\lfloor {\mathrm{cos}}^{-1}(x)\rfloor $

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pheniankang

Answered 2022-06-26
Author has **22** answers

Noting that $\lfloor \mathrm{arcsin}x\rfloor =-2,-1,0,1$ and that $\lfloor \mathrm{arccos}x\rfloor =0,1,2,3$ we have

$\begin{array}{rl}\lfloor \mathrm{arcsin}x\rfloor >\lfloor \mathrm{arccos}x\rfloor & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\lfloor \mathrm{arcsin}x\rfloor =1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\lfloor \mathrm{arccos}x\rfloor =0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}1\le \mathrm{arcsin}x<2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0\le \mathrm{arccos}x<1\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}1\le x\le 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}1<x\le 1\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{\mathrm{sin}1\le x\le 1}\end{array}$

$\begin{array}{rl}\lfloor \mathrm{arcsin}x\rfloor >\lfloor \mathrm{arccos}x\rfloor & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\lfloor \mathrm{arcsin}x\rfloor =1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\lfloor \mathrm{arccos}x\rfloor =0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}1\le \mathrm{arcsin}x<2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0\le \mathrm{arccos}x<1\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}1\le x\le 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}1<x\le 1\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{\mathrm{sin}1\le x\le 1}\end{array}$

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