Simplified form of $cos^{-1}big[frac{3}{5}cdotcos x+frac{4}{5}cdotsin xbig]$, where...












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Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







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  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34
















1












$begingroup$



Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







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$endgroup$








  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34














1












1








1


0



$begingroup$



Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







share|cite|improve this question











$endgroup$





Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function




trigonometry inverse-function






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edited Dec 10 '18 at 12:54







ss1729

















asked Dec 10 '18 at 11:21









ss1729ss1729

2,04411124




2,04411124








  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34














  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34








2




2




$begingroup$
At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
$endgroup$
– Barry Cipra
Dec 10 '18 at 12:34




$begingroup$
At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
$endgroup$
– Barry Cipra
Dec 10 '18 at 12:34










2 Answers
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Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






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    $begingroup$

    $$-dfrac{3pi}4le xledfrac{3pi}4$$



    $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



    Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



    So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



    Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



    For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



    For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      1












      $begingroup$

      Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



      So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






      share|cite|improve this answer











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        1












        $begingroup$

        Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



        So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



          So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






          share|cite|improve this answer











          $endgroup$



          Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



          So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 12:00

























          answered Dec 10 '18 at 11:32









          Shubham JohriShubham Johri

          5,382818




          5,382818























              0












              $begingroup$

              $$-dfrac{3pi}4le xledfrac{3pi}4$$



              $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



              Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



              So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



              Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



              For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



              For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$-dfrac{3pi}4le xledfrac{3pi}4$$



                $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$-dfrac{3pi}4le xledfrac{3pi}4$$



                  $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                  Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                  So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                  Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                  For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                  For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






                  share|cite|improve this answer









                  $endgroup$



                  $$-dfrac{3pi}4le xledfrac{3pi}4$$



                  $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                  Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                  So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                  Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                  For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                  For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 11:51









                  lab bhattacharjeelab bhattacharjee

                  227k15158277




                  227k15158277






























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