Understanding arcsin inequalities












1














Assuming this is correct:



$arcsin x neq 0$



$sin(0) neq x$



$0neq x$



Following the same logic, why is this incorrect?



$arcsin(x+frac{1}{3}) geq 0$



$sin (0) geq x+frac{1}{3}$



$0 geq x + frac{1}{3}$



$x leq -frac{1}{3}$



$xin(-infty; -frac{1}{3}] $



The correct answer should be:



$xin[-frac{1}{3}; infty)$



Why is this incorrect? Can I use the $ sin(0) $ trick only for when there's $neq$ sign, thus cannot be used in inequalities?



I do have one more question, by the way. Is this correct?



$ arcsin(expression) geq 0 $



$ expressiongeq 0 $



and



$arccos (expression) geq 0 $



$ expressiongeq 0 $



Thanks






trigonometry







share|cite|improve this question





























    1














    Assuming this is correct:



    $arcsin x neq 0$



    $sin(0) neq x$



    $0neq x$



    Following the same logic, why is this incorrect?



    $arcsin(x+frac{1}{3}) geq 0$



    $sin (0) geq x+frac{1}{3}$



    $0 geq x + frac{1}{3}$



    $x leq -frac{1}{3}$



    $xin(-infty; -frac{1}{3}] $



    The correct answer should be:



    $xin[-frac{1}{3}; infty)$



    Why is this incorrect? Can I use the $ sin(0) $ trick only for when there's $neq$ sign, thus cannot be used in inequalities?



    I do have one more question, by the way. Is this correct?



    $ arcsin(expression) geq 0 $



    $ expressiongeq 0 $



    and



    $arccos (expression) geq 0 $



    $ expressiongeq 0 $



    Thanks






    trigonometry







    share|cite|improve this question



























      1












      1








      1







      Assuming this is correct:



      $arcsin x neq 0$



      $sin(0) neq x$



      $0neq x$



      Following the same logic, why is this incorrect?



      $arcsin(x+frac{1}{3}) geq 0$



      $sin (0) geq x+frac{1}{3}$



      $0 geq x + frac{1}{3}$



      $x leq -frac{1}{3}$



      $xin(-infty; -frac{1}{3}] $



      The correct answer should be:



      $xin[-frac{1}{3}; infty)$



      Why is this incorrect? Can I use the $ sin(0) $ trick only for when there's $neq$ sign, thus cannot be used in inequalities?



      I do have one more question, by the way. Is this correct?



      $ arcsin(expression) geq 0 $



      $ expressiongeq 0 $



      and



      $arccos (expression) geq 0 $



      $ expressiongeq 0 $



      Thanks






      trigonometry







      share|cite|improve this question















      Assuming this is correct:



      $arcsin x neq 0$



      $sin(0) neq x$



      $0neq x$



      Following the same logic, why is this incorrect?



      $arcsin(x+frac{1}{3}) geq 0$



      $sin (0) geq x+frac{1}{3}$



      $0 geq x + frac{1}{3}$



      $x leq -frac{1}{3}$



      $xin(-infty; -frac{1}{3}] $



      The correct answer should be:



      $xin[-frac{1}{3}; infty)$



      Why is this incorrect? Can I use the $ sin(0) $ trick only for when there's $neq$ sign, thus cannot be used in inequalities?



      I do have one more question, by the way. Is this correct?



      $ arcsin(expression) geq 0 $



      $ expressiongeq 0 $



      and



      $arccos (expression) geq 0 $



      $ expressiongeq 0 $



      Thanks






      trigonometry




      trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 '18 at 23:02







      weno

















      asked Nov 21 '18 at 22:42









      wenoweno

      1078




      1078






















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          The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).



          Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
          $$
          -frac{1}{3}le xlefrac{2}{3}
          $$
          There's no way the solution set is $[-1/3;infty)$




          For the arccosine you have to take into account that it is decreasing, so $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.






          share|cite|improve this answer





















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            0














            The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).



            Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
            $$
            -frac{1}{3}le xlefrac{2}{3}
            $$
            There's no way the solution set is $[-1/3;infty)$




            For the arccosine you have to take into account that it is decreasing, so $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.






            share|cite|improve this answer


























              0














              The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).



              Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
              $$
              -frac{1}{3}le xlefrac{2}{3}
              $$
              There's no way the solution set is $[-1/3;infty)$




              For the arccosine you have to take into account that it is decreasing, so $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.






              share|cite|improve this answer
























                0












                0








                0






                The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).



                Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
                $$
                -frac{1}{3}le xlefrac{2}{3}
                $$
                There's no way the solution set is $[-1/3;infty)$




                For the arccosine you have to take into account that it is decreasing, so $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.






                share|cite|improve this answer












                The arcsine is an increasing function, so $arcsin agearcsin b$ is equivalent to $age b$ (provided $a$ and $b$ are in the domain of the arcsine).



                Thus your inequality $arcsin(x+1/3)ge0=arcsin 0$ becomes $x+1/3ge0$, that is, $xge-1/3$. Taking into account that $-1le x+1/3le 1$, we get $-4/3le xle 2/3$, so combining the inequalities yields
                $$
                -frac{1}{3}le xlefrac{2}{3}
                $$
                There's no way the solution set is $[-1/3;infty)$




                For the arccosine you have to take into account that it is decreasing, so $arccos agearccos b$ becomes $ale b$. Since $0=arccos(1)$, the inequality $arccos xge 0$ becomes $xle1$. This is anyway obvious, because $arccos xge0$ for every $xin[-1;1]$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 '18 at 23:03









                egregegreg

                179k1485202




                179k1485202






























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