Greater of the angles $alpha=2tan^{-1}(2sqrt{2}-1)$, $beta=3sin^{-1}frac{1}{3}+sin^{-1}frac{3}{5}$












2












$begingroup$



Find the greater of the two angles $alpha=2tan^{-1}(2sqrt{2}-1)$ and $beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}$




My Attempt
$$
alpha=2tan^{-1}(2sqrt{2}-1)=2tan^{-1}(1.82)>2tan^{-1}sqrt{3}=2.frac{pi}{3}\
implies boxed{alpha>frac{2pi}{3}=0.67pi}\
beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}=3sin^{-1}(0.33)+sin^{-1}(0.6)\
=sin^{-1}(0.8)+sin^{-1}(0.6)<sin^{-1}(1)+sin^{-1}(0.7=1/sqrt{2})\
boxed{beta<frac{pi}{2}+frac{pi}{4}=frac{3pi}{4}=0.75pi}
$$



I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to solve this ?






trigonometry maxima-minima inverse-function







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    2












    $begingroup$



    Find the greater of the two angles $alpha=2tan^{-1}(2sqrt{2}-1)$ and $beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}$




    My Attempt
    $$
    alpha=2tan^{-1}(2sqrt{2}-1)=2tan^{-1}(1.82)>2tan^{-1}sqrt{3}=2.frac{pi}{3}\
    implies boxed{alpha>frac{2pi}{3}=0.67pi}\
    beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}=3sin^{-1}(0.33)+sin^{-1}(0.6)\
    =sin^{-1}(0.8)+sin^{-1}(0.6)<sin^{-1}(1)+sin^{-1}(0.7=1/sqrt{2})\
    boxed{beta<frac{pi}{2}+frac{pi}{4}=frac{3pi}{4}=0.75pi}
    $$



    I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to solve this ?






    trigonometry maxima-minima inverse-function







    share|cite|improve this question









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      2












      2








      2


      1



      $begingroup$



      Find the greater of the two angles $alpha=2tan^{-1}(2sqrt{2}-1)$ and $beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}$




      My Attempt
      $$
      alpha=2tan^{-1}(2sqrt{2}-1)=2tan^{-1}(1.82)>2tan^{-1}sqrt{3}=2.frac{pi}{3}\
      implies boxed{alpha>frac{2pi}{3}=0.67pi}\
      beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}=3sin^{-1}(0.33)+sin^{-1}(0.6)\
      =sin^{-1}(0.8)+sin^{-1}(0.6)<sin^{-1}(1)+sin^{-1}(0.7=1/sqrt{2})\
      boxed{beta<frac{pi}{2}+frac{pi}{4}=frac{3pi}{4}=0.75pi}
      $$



      I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to solve this ?






      trigonometry maxima-minima inverse-function







      share|cite|improve this question









      $endgroup$





      Find the greater of the two angles $alpha=2tan^{-1}(2sqrt{2}-1)$ and $beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}$




      My Attempt
      $$
      alpha=2tan^{-1}(2sqrt{2}-1)=2tan^{-1}(1.82)>2tan^{-1}sqrt{3}=2.frac{pi}{3}\
      implies boxed{alpha>frac{2pi}{3}=0.67pi}\
      beta=3sin^{-1}dfrac{1}{3}+sin^{-1}dfrac{3}{5}=3sin^{-1}(0.33)+sin^{-1}(0.6)\
      =sin^{-1}(0.8)+sin^{-1}(0.6)<sin^{-1}(1)+sin^{-1}(0.7=1/sqrt{2})\
      boxed{beta<frac{pi}{2}+frac{pi}{4}=frac{3pi}{4}=0.75pi}
      $$



      I am stuck with the above result which does not seem to give enough information to decide which one is greater, wht's the easiest way to solve this ?






      trigonometry maxima-minima inverse-function




      trigonometry maxima-minima inverse-function






      share|cite|improve this question













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      asked Dec 16 '18 at 20:45









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          $begingroup$

          Note that $sqrt{2}approx1.414$.



          $(2sqrt{2}-1)^2=8-4sqrt{2}+1=9-4sqrt{2}>3Leftrightarrow6>4sqrt{2}Leftrightarrow $ after raising to the second power $Leftrightarrow 36>32$ which is true and $2sqrt{2}-1>sqrt{3}$



          We know that $tandfrac{pi}{3}=sqrt{3}$, we get that $(2sqrt{2}-1)>tandfrac{pi}{3}$, then $2tan^{-1}(2sqrt{2}-1)>dfrac{2pi}{3}$, because the inverse of the tangent is an increasing function.



          So, $alpha>dfrac{2pi}{3}$



          Note that $sin3theta=3sintheta-4sin^2theta$, then $3theta=sin^{-1}[3sintheta-4sin^3theta]$.



          We have that $theta=sin^{-1}dfrac13$ and $sintheta=dfrac13$



          So, $$beta=3sin^{-1}dfrac13+sin^{-1}dfrac35$$
          $$=sin^{-1}left[3left(dfrac13right)-4left(dfrac13right)^3right]+sin^{-1}dfrac35$$
          $$=sin^{-1}dfrac{23}{27}+sin^{-1}dfrac{3}{5}$$



          Note that $dfrac{23}{27}<dfrac{sqrt{3}}{2}$ and $dfrac35<dfrac{sqrt{3}}{2}$, because $(23cdot2)^2<(27sqrt{3})^2$ and $(3cdot2)^2<(5sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.



          So, $sin^{-1}dfrac{23}{27}+sin^{-1}dfrac35<sin^{-1}dfrac{sqrt{3}}{2}+sin^{-1}dfrac{sqrt{3}}{2}=dfrac{pi}{3}+dfrac{pi}{3}=dfrac{2pi}{3}$, because the inverse of the sine function is strictly increasing.



          Therefore, $$alpha>dfrac{2pi}{3}, beta<dfrac{2pi}{3}$$
          $$mbox{So, }alpha>beta$$






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            1 Answer
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            1 Answer
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            active

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            2












            $begingroup$

            Note that $sqrt{2}approx1.414$.



            $(2sqrt{2}-1)^2=8-4sqrt{2}+1=9-4sqrt{2}>3Leftrightarrow6>4sqrt{2}Leftrightarrow $ after raising to the second power $Leftrightarrow 36>32$ which is true and $2sqrt{2}-1>sqrt{3}$



            We know that $tandfrac{pi}{3}=sqrt{3}$, we get that $(2sqrt{2}-1)>tandfrac{pi}{3}$, then $2tan^{-1}(2sqrt{2}-1)>dfrac{2pi}{3}$, because the inverse of the tangent is an increasing function.



            So, $alpha>dfrac{2pi}{3}$



            Note that $sin3theta=3sintheta-4sin^2theta$, then $3theta=sin^{-1}[3sintheta-4sin^3theta]$.



            We have that $theta=sin^{-1}dfrac13$ and $sintheta=dfrac13$



            So, $$beta=3sin^{-1}dfrac13+sin^{-1}dfrac35$$
            $$=sin^{-1}left[3left(dfrac13right)-4left(dfrac13right)^3right]+sin^{-1}dfrac35$$
            $$=sin^{-1}dfrac{23}{27}+sin^{-1}dfrac{3}{5}$$



            Note that $dfrac{23}{27}<dfrac{sqrt{3}}{2}$ and $dfrac35<dfrac{sqrt{3}}{2}$, because $(23cdot2)^2<(27sqrt{3})^2$ and $(3cdot2)^2<(5sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.



            So, $sin^{-1}dfrac{23}{27}+sin^{-1}dfrac35<sin^{-1}dfrac{sqrt{3}}{2}+sin^{-1}dfrac{sqrt{3}}{2}=dfrac{pi}{3}+dfrac{pi}{3}=dfrac{2pi}{3}$, because the inverse of the sine function is strictly increasing.



            Therefore, $$alpha>dfrac{2pi}{3}, beta<dfrac{2pi}{3}$$
            $$mbox{So, }alpha>beta$$






            share|cite|improve this answer









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              2












              $begingroup$

              Note that $sqrt{2}approx1.414$.



              $(2sqrt{2}-1)^2=8-4sqrt{2}+1=9-4sqrt{2}>3Leftrightarrow6>4sqrt{2}Leftrightarrow $ after raising to the second power $Leftrightarrow 36>32$ which is true and $2sqrt{2}-1>sqrt{3}$



              We know that $tandfrac{pi}{3}=sqrt{3}$, we get that $(2sqrt{2}-1)>tandfrac{pi}{3}$, then $2tan^{-1}(2sqrt{2}-1)>dfrac{2pi}{3}$, because the inverse of the tangent is an increasing function.



              So, $alpha>dfrac{2pi}{3}$



              Note that $sin3theta=3sintheta-4sin^2theta$, then $3theta=sin^{-1}[3sintheta-4sin^3theta]$.



              We have that $theta=sin^{-1}dfrac13$ and $sintheta=dfrac13$



              So, $$beta=3sin^{-1}dfrac13+sin^{-1}dfrac35$$
              $$=sin^{-1}left[3left(dfrac13right)-4left(dfrac13right)^3right]+sin^{-1}dfrac35$$
              $$=sin^{-1}dfrac{23}{27}+sin^{-1}dfrac{3}{5}$$



              Note that $dfrac{23}{27}<dfrac{sqrt{3}}{2}$ and $dfrac35<dfrac{sqrt{3}}{2}$, because $(23cdot2)^2<(27sqrt{3})^2$ and $(3cdot2)^2<(5sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.



              So, $sin^{-1}dfrac{23}{27}+sin^{-1}dfrac35<sin^{-1}dfrac{sqrt{3}}{2}+sin^{-1}dfrac{sqrt{3}}{2}=dfrac{pi}{3}+dfrac{pi}{3}=dfrac{2pi}{3}$, because the inverse of the sine function is strictly increasing.



              Therefore, $$alpha>dfrac{2pi}{3}, beta<dfrac{2pi}{3}$$
              $$mbox{So, }alpha>beta$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Note that $sqrt{2}approx1.414$.



                $(2sqrt{2}-1)^2=8-4sqrt{2}+1=9-4sqrt{2}>3Leftrightarrow6>4sqrt{2}Leftrightarrow $ after raising to the second power $Leftrightarrow 36>32$ which is true and $2sqrt{2}-1>sqrt{3}$



                We know that $tandfrac{pi}{3}=sqrt{3}$, we get that $(2sqrt{2}-1)>tandfrac{pi}{3}$, then $2tan^{-1}(2sqrt{2}-1)>dfrac{2pi}{3}$, because the inverse of the tangent is an increasing function.



                So, $alpha>dfrac{2pi}{3}$



                Note that $sin3theta=3sintheta-4sin^2theta$, then $3theta=sin^{-1}[3sintheta-4sin^3theta]$.



                We have that $theta=sin^{-1}dfrac13$ and $sintheta=dfrac13$



                So, $$beta=3sin^{-1}dfrac13+sin^{-1}dfrac35$$
                $$=sin^{-1}left[3left(dfrac13right)-4left(dfrac13right)^3right]+sin^{-1}dfrac35$$
                $$=sin^{-1}dfrac{23}{27}+sin^{-1}dfrac{3}{5}$$



                Note that $dfrac{23}{27}<dfrac{sqrt{3}}{2}$ and $dfrac35<dfrac{sqrt{3}}{2}$, because $(23cdot2)^2<(27sqrt{3})^2$ and $(3cdot2)^2<(5sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.



                So, $sin^{-1}dfrac{23}{27}+sin^{-1}dfrac35<sin^{-1}dfrac{sqrt{3}}{2}+sin^{-1}dfrac{sqrt{3}}{2}=dfrac{pi}{3}+dfrac{pi}{3}=dfrac{2pi}{3}$, because the inverse of the sine function is strictly increasing.



                Therefore, $$alpha>dfrac{2pi}{3}, beta<dfrac{2pi}{3}$$
                $$mbox{So, }alpha>beta$$






                share|cite|improve this answer









                $endgroup$



                Note that $sqrt{2}approx1.414$.



                $(2sqrt{2}-1)^2=8-4sqrt{2}+1=9-4sqrt{2}>3Leftrightarrow6>4sqrt{2}Leftrightarrow $ after raising to the second power $Leftrightarrow 36>32$ which is true and $2sqrt{2}-1>sqrt{3}$



                We know that $tandfrac{pi}{3}=sqrt{3}$, we get that $(2sqrt{2}-1)>tandfrac{pi}{3}$, then $2tan^{-1}(2sqrt{2}-1)>dfrac{2pi}{3}$, because the inverse of the tangent is an increasing function.



                So, $alpha>dfrac{2pi}{3}$



                Note that $sin3theta=3sintheta-4sin^2theta$, then $3theta=sin^{-1}[3sintheta-4sin^3theta]$.



                We have that $theta=sin^{-1}dfrac13$ and $sintheta=dfrac13$



                So, $$beta=3sin^{-1}dfrac13+sin^{-1}dfrac35$$
                $$=sin^{-1}left[3left(dfrac13right)-4left(dfrac13right)^3right]+sin^{-1}dfrac35$$
                $$=sin^{-1}dfrac{23}{27}+sin^{-1}dfrac{3}{5}$$



                Note that $dfrac{23}{27}<dfrac{sqrt{3}}{2}$ and $dfrac35<dfrac{sqrt{3}}{2}$, because $(23cdot2)^2<(27sqrt{3})^2$ and $(3cdot2)^2<(5sqrt{3})^2$, which is equivalent to $2116<2187$ and $36<75$ which is true.



                So, $sin^{-1}dfrac{23}{27}+sin^{-1}dfrac35<sin^{-1}dfrac{sqrt{3}}{2}+sin^{-1}dfrac{sqrt{3}}{2}=dfrac{pi}{3}+dfrac{pi}{3}=dfrac{2pi}{3}$, because the inverse of the sine function is strictly increasing.



                Therefore, $$alpha>dfrac{2pi}{3}, beta<dfrac{2pi}{3}$$
                $$mbox{So, }alpha>beta$$







                share|cite|improve this answer












                share|cite|improve this answer



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                answered Dec 16 '18 at 21:57









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