Question Number 85442 by Power last updated on 22/Mar/20
Commented by Power last updated on 22/Mar/20
![[x]−integer part=trunc(x) {x}−fractional part=frac(x)](https://www.tinkutara.com/question/Q85449.png)
$$\left[\boldsymbol{\mathrm{x}}\right]−\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{part}}=\boldsymbol{\mathrm{trunc}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\left\{\boldsymbol{\mathrm{x}}\right\}−\boldsymbol{\mathrm{fractional}}\:\boldsymbol{\mathrm{part}}=\boldsymbol{\mathrm{frac}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$
Answered by mr W last updated on 22/Mar/20
![since 0^0 is not defined, sin x≠0 ⇒x≠kπ ⇒cos x≠±1 cos x≠0 ⇒x≠(k+(1/2))π ⇒sin x≠±1 since sin x≠±1 and cos x≠±1, ⇒[sin x]=0, [cos x]=0 ⇒{sin x}=sin x, {cos x}=cos x {sin x}^([sin x]) =(sin x)^0 =1 {cos x}^([cos x]) =(cos x)^0 =1 i.e. {sin x}^([sin x]) ={cos x}^([cos x]) is true. ⇒solution is x∈R ∧ x≠(k+(1/2))π ∧ x≠kπ](https://www.tinkutara.com/question/Q85510.png)
$${since}\:\mathrm{0}^{\mathrm{0}} \:{is}\:{not}\:{defined}, \\ $$$$\mathrm{sin}\:{x}\neq\mathrm{0}\:\Rightarrow{x}\neq{k}\pi\:\Rightarrow\mathrm{cos}\:{x}\neq\pm\mathrm{1} \\ $$$$\mathrm{cos}\:{x}\neq\mathrm{0}\:\Rightarrow{x}\neq\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi\:\Rightarrow\mathrm{sin}\:{x}\neq\pm\mathrm{1} \\ $$$${since}\:\mathrm{sin}\:{x}\neq\pm\mathrm{1}\:{and}\:\mathrm{cos}\:{x}\neq\pm\mathrm{1}, \\ $$$$\Rightarrow\left[\mathrm{sin}\:{x}\right]=\mathrm{0},\:\left[\mathrm{cos}\:{x}\right]=\mathrm{0} \\ $$$$\Rightarrow\left\{\mathrm{sin}\:{x}\right\}=\mathrm{sin}\:{x},\:\left\{\mathrm{cos}\:{x}\right\}=\mathrm{cos}\:{x} \\ $$$$\left\{\mathrm{sin}\:{x}\right\}^{\left[\mathrm{sin}\:{x}\right]} =\left(\mathrm{sin}\:{x}\right)^{\mathrm{0}} =\mathrm{1} \\ $$$$\left\{\mathrm{cos}\:{x}\right\}^{\left[\mathrm{cos}\:\:{x}\right]} =\left(\mathrm{cos}\:{x}\right)^{\mathrm{0}} =\mathrm{1} \\ $$$${i}.{e}.\:\left\{\mathrm{sin}\:{x}\right\}^{\left[\mathrm{sin}\:{x}\right]} =\left\{\mathrm{cos}\:{x}\right\}^{\left[\mathrm{cos}\:\:{x}\right]} \:{is}\:{true}. \\ $$$$ \\ $$$$\Rightarrow{solution}\:{is} \\ $$$${x}\in{R}\:\wedge\:{x}\neq\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)\pi\:\wedge\:{x}\neq{k}\pi \\ $$
Commented by Power last updated on 22/Mar/20
Commented by Power last updated on 22/Mar/20
$$\mathrm{sir}\:\mathrm{mr}\:\mathrm{W}\:\:? \\ $$
Commented by Power last updated on 22/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 22/Mar/20
$${what}'{s}\:{your}\:{problem}? \\ $$
Commented by Power last updated on 22/Mar/20
![why [sinx]=0 [cosx]=0 ? please explain that](https://www.tinkutara.com/question/Q85517.png)
$$\mathrm{why}\:\left[\mathrm{sinx}\right]=\mathrm{0}\:\:\:\:\left[\mathrm{cosx}\right]=\mathrm{0}\:\:?\:\:\mathrm{please}\:\mathrm{explain}\:\mathrm{that} \\ $$
Commented by Power last updated on 22/Mar/20
$$\mathrm{sinx}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Power last updated on 22/Mar/20
$$\mathrm{cosx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 22/Mar/20
![−1< sin x <1 ⇒[sin x]=ipart(sin x)=trunc(sin x)=0 −1< cos x <1 ⇒[cos x]=ipart(cos x)=trunc(cos x)=0](https://www.tinkutara.com/question/Q85521.png)
$$−\mathrm{1}<\:\mathrm{sin}\:{x}\:<\mathrm{1}\:\Rightarrow\left[\mathrm{sin}\:{x}\right]={ipart}\left(\mathrm{sin}\:{x}\right)={trunc}\left(\mathrm{sin}\:{x}\right)=\mathrm{0} \\ $$$$−\mathrm{1}<\:\mathrm{cos}\:{x}\:<\mathrm{1}\:\Rightarrow\left[\mathrm{cos}\:{x}\right]={ipart}\left(\mathrm{cos}\:{x}\right)={trunc}\left(\mathrm{cos}\:{x}\right)=\mathrm{0} \\ $$
Commented by Power last updated on 22/Mar/20
$$\mathrm{sir}\:\:\mathrm{ipart}\:\:? \\ $$
Commented by mr W last updated on 22/Mar/20
$${ipart}\left({x}\right)={integer}\:{part}\:{of}\:{x} \\ $$
Commented by Power last updated on 22/Mar/20
Commented by mr W last updated on 22/Mar/20
![[−(1/2)]=ipart(−(1/2))=0 {−(1/2)}=fpart(−(1/2))=−(1/2) this is what i applied, see below. you may accept it or refuse it. i won′t discuss with you, because there are different definitions for [−(1/2)]. if you take an other definition than i, you should present your own solution.](https://www.tinkutara.com/question/Q85529.png)
$$\left[−\frac{\mathrm{1}}{\mathrm{2}}\right]={ipart}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\left\{−\frac{\mathrm{1}}{\mathrm{2}}\right\}={fpart}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${this}\:{is}\:{what}\:{i}\:{applied},\:{see}\:{below}. \\ $$$${you}\:{may}\:{accept}\:{it}\:{or}\:{refuse}\:{it}. \\ $$$${i}\:{won}'{t}\:{discuss}\:{with} \\ $$$${you},\:{because}\:{there}\:{are}\:{different} \\ $$$${definitions}\:{for}\:\left[−\frac{\mathrm{1}}{\mathrm{2}}\right].\:{if}\:{you}\:{take}\:{an} \\ $$$${other}\:{definition}\:{than}\:{i},\:{you}\:{should} \\ $$$${present}\:{your}\:{own}\:{solution}. \\ $$
Commented by mr W last updated on 22/Mar/20
Commented by Power last updated on 22/Mar/20
$$\mathrm{okey}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$