Inequality for $sin(20°)$












2












$begingroup$


Prove that $$frac{1}{3} < sin{20°} < frac{7}{20}$$



Attempt
$$sin60°=3sin20°-4sin^{3}(20°)$$
Taking $sin20°$=x



I got the the equation as
$$8x^3-6x+sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?



Thanks!
Edit-graph of p(x)
enter image description here






trigonometry inequality functional-inequalities







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:03












  • $begingroup$
    @lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:12












  • $begingroup$
    Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:15
















2












$begingroup$


Prove that $$frac{1}{3} < sin{20°} < frac{7}{20}$$



Attempt
$$sin60°=3sin20°-4sin^{3}(20°)$$
Taking $sin20°$=x



I got the the equation as
$$8x^3-6x+sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?



Thanks!
Edit-graph of p(x)
enter image description here






trigonometry inequality functional-inequalities







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:03












  • $begingroup$
    @lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:12












  • $begingroup$
    Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:15














2












2








2


2



$begingroup$


Prove that $$frac{1}{3} < sin{20°} < frac{7}{20}$$



Attempt
$$sin60°=3sin20°-4sin^{3}(20°)$$
Taking $sin20°$=x



I got the the equation as
$$8x^3-6x+sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?



Thanks!
Edit-graph of p(x)
enter image description here






trigonometry inequality functional-inequalities







share|cite|improve this question











$endgroup$




Prove that $$frac{1}{3} < sin{20°} < frac{7}{20}$$



Attempt
$$sin60°=3sin20°-4sin^{3}(20°)$$
Taking $sin20°$=x



I got the the equation as
$$8x^3-6x+sqrt{3} =0$$
But from here I am not able to do anything. Any suggestions?



Thanks!
Edit-graph of p(x)
enter image description here






trigonometry inequality functional-inequalities




trigonometry inequality functional-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 14:19







jayant98

















asked Dec 3 '18 at 13:58









jayant98jayant98

513116




513116








  • 1




    $begingroup$
    Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:03












  • $begingroup$
    @lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:12












  • $begingroup$
    Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:15














  • 1




    $begingroup$
    Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:03












  • $begingroup$
    @lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:12












  • $begingroup$
    Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
    $endgroup$
    – lulu
    Dec 3 '18 at 14:13










  • $begingroup$
    Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
    $endgroup$
    – jayant98
    Dec 3 '18 at 14:15








1




1




$begingroup$
Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
$endgroup$
– lulu
Dec 3 '18 at 14:03






$begingroup$
Hint: if $p(x)$ is your cubic, what is $pleft( frac 13right)$? What is $pleft( frac 7{20}right)$? That, plus a quick sketch of the graph should do it.
$endgroup$
– lulu
Dec 3 '18 at 14:03














$begingroup$
@lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
$endgroup$
– jayant98
Dec 3 '18 at 14:12






$begingroup$
@lulu In that region p(x) is decreasing so p(1/3)>p(x)>p(7/20) and hence $frac{1}{3} <x<frac{7}{20}$
$endgroup$
– jayant98
Dec 3 '18 at 14:12














$begingroup$
Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
$endgroup$
– lulu
Dec 3 '18 at 14:13




$begingroup$
Well, I think this is what you are trying to prove, no? I don't see what argument you are making other than to restate the question.
$endgroup$
– lulu
Dec 3 '18 at 14:13












$begingroup$
Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
$endgroup$
– lulu
Dec 3 '18 at 14:13




$begingroup$
Sketch the graph! What can you say about the roots of $p(x)$? Can you roughly describe their locations?
$endgroup$
– lulu
Dec 3 '18 at 14:13












$begingroup$
Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
$endgroup$
– jayant98
Dec 3 '18 at 14:15




$begingroup$
Yes this is what I was trying to prove. I wanted affirmation from your side only to be only sure about my steps. But, thanks for help.
$endgroup$
– jayant98
Dec 3 '18 at 14:15










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $p(x)$ denote the cubic.



Since $$lim_{xto - infty}p(x)=-infty<0quad quad p(0)>0quad quad p(.5)<0quad quadlim_{xto + infty}p(x)=+infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$



Of course $sin(20^{circ})$ is positive and, since $0<20<30$ we see that $sin(20^{circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$pleft( frac 13 right)>0 quad & quad pleft( frac 7{20}right)<0$$ so the root we care about must be between $frac 13$ and $frac {7}{20}$ and we are done.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Mostly for fun, here's a way to show that



    $$6left(1over3right)-8left(1over3right)^3ltsqrt3lt6left(7over20right)-8left(7over20right)^3$$



    with only a small amount of multi-digit arithmetic:



    $$6left(1over3right)-8left(1over3right)^3=2-{8over27}lt2-{8over28}=2-{2over7}={12over7}$$



    and $12^2=144lt147=3cdot7^2$, which gives the first inequality, while



    $$6left(7over20right)-8left(7over20right)^3={7over10}left(3-left(7over10right)^2 right)={7over10}left(300-49over100 right)gt{7over10}cdot{250over100}={7over4}$$



    and $7^2=49gt48=3cdot4^2$ gives the second inequality.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for another way to attempt such inequalities.
      $endgroup$
      – jayant98
      Dec 3 '18 at 16:43











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let $p(x)$ denote the cubic.



    Since $$lim_{xto - infty}p(x)=-infty<0quad quad p(0)>0quad quad p(.5)<0quad quadlim_{xto + infty}p(x)=+infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$



    Of course $sin(20^{circ})$ is positive and, since $0<20<30$ we see that $sin(20^{circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$pleft( frac 13 right)>0 quad & quad pleft( frac 7{20}right)<0$$ so the root we care about must be between $frac 13$ and $frac {7}{20}$ and we are done.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Let $p(x)$ denote the cubic.



      Since $$lim_{xto - infty}p(x)=-infty<0quad quad p(0)>0quad quad p(.5)<0quad quadlim_{xto + infty}p(x)=+infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$



      Of course $sin(20^{circ})$ is positive and, since $0<20<30$ we see that $sin(20^{circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$pleft( frac 13 right)>0 quad & quad pleft( frac 7{20}right)<0$$ so the root we care about must be between $frac 13$ and $frac {7}{20}$ and we are done.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $p(x)$ denote the cubic.



        Since $$lim_{xto - infty}p(x)=-infty<0quad quad p(0)>0quad quad p(.5)<0quad quadlim_{xto + infty}p(x)=+infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$



        Of course $sin(20^{circ})$ is positive and, since $0<20<30$ we see that $sin(20^{circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$pleft( frac 13 right)>0 quad & quad pleft( frac 7{20}right)<0$$ so the root we care about must be between $frac 13$ and $frac {7}{20}$ and we are done.






        share|cite|improve this answer











        $endgroup$



        Let $p(x)$ denote the cubic.



        Since $$lim_{xto - infty}p(x)=-infty<0quad quad p(0)>0quad quad p(.5)<0quad quadlim_{xto + infty}p(x)=+infty$$ we see that one root is negative, a second lies between $0$ and $.5$ and the third is greater than $.5$



        Of course $sin(20^{circ})$ is positive and, since $0<20<30$ we see that $sin(20^{circ})<.5$ Thus the root we want is the middle one of the three. We then remark that $$pleft( frac 13 right)>0 quad & quad pleft( frac 7{20}right)<0$$ so the root we care about must be between $frac 13$ and $frac {7}{20}$ and we are done.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 14:40

























        answered Dec 3 '18 at 14:35









        lulululu

        39.8k24778




        39.8k24778























            1












            $begingroup$

            Mostly for fun, here's a way to show that



            $$6left(1over3right)-8left(1over3right)^3ltsqrt3lt6left(7over20right)-8left(7over20right)^3$$



            with only a small amount of multi-digit arithmetic:



            $$6left(1over3right)-8left(1over3right)^3=2-{8over27}lt2-{8over28}=2-{2over7}={12over7}$$



            and $12^2=144lt147=3cdot7^2$, which gives the first inequality, while



            $$6left(7over20right)-8left(7over20right)^3={7over10}left(3-left(7over10right)^2 right)={7over10}left(300-49over100 right)gt{7over10}cdot{250over100}={7over4}$$



            and $7^2=49gt48=3cdot4^2$ gives the second inequality.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for another way to attempt such inequalities.
              $endgroup$
              – jayant98
              Dec 3 '18 at 16:43
















            1












            $begingroup$

            Mostly for fun, here's a way to show that



            $$6left(1over3right)-8left(1over3right)^3ltsqrt3lt6left(7over20right)-8left(7over20right)^3$$



            with only a small amount of multi-digit arithmetic:



            $$6left(1over3right)-8left(1over3right)^3=2-{8over27}lt2-{8over28}=2-{2over7}={12over7}$$



            and $12^2=144lt147=3cdot7^2$, which gives the first inequality, while



            $$6left(7over20right)-8left(7over20right)^3={7over10}left(3-left(7over10right)^2 right)={7over10}left(300-49over100 right)gt{7over10}cdot{250over100}={7over4}$$



            and $7^2=49gt48=3cdot4^2$ gives the second inequality.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for another way to attempt such inequalities.
              $endgroup$
              – jayant98
              Dec 3 '18 at 16:43














            1












            1








            1





            $begingroup$

            Mostly for fun, here's a way to show that



            $$6left(1over3right)-8left(1over3right)^3ltsqrt3lt6left(7over20right)-8left(7over20right)^3$$



            with only a small amount of multi-digit arithmetic:



            $$6left(1over3right)-8left(1over3right)^3=2-{8over27}lt2-{8over28}=2-{2over7}={12over7}$$



            and $12^2=144lt147=3cdot7^2$, which gives the first inequality, while



            $$6left(7over20right)-8left(7over20right)^3={7over10}left(3-left(7over10right)^2 right)={7over10}left(300-49over100 right)gt{7over10}cdot{250over100}={7over4}$$



            and $7^2=49gt48=3cdot4^2$ gives the second inequality.






            share|cite|improve this answer









            $endgroup$



            Mostly for fun, here's a way to show that



            $$6left(1over3right)-8left(1over3right)^3ltsqrt3lt6left(7over20right)-8left(7over20right)^3$$



            with only a small amount of multi-digit arithmetic:



            $$6left(1over3right)-8left(1over3right)^3=2-{8over27}lt2-{8over28}=2-{2over7}={12over7}$$



            and $12^2=144lt147=3cdot7^2$, which gives the first inequality, while



            $$6left(7over20right)-8left(7over20right)^3={7over10}left(3-left(7over10right)^2 right)={7over10}left(300-49over100 right)gt{7over10}cdot{250over100}={7over4}$$



            and $7^2=49gt48=3cdot4^2$ gives the second inequality.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 3 '18 at 15:58









            Barry CipraBarry Cipra

            59.4k653125




            59.4k653125












            • $begingroup$
              Thanks for another way to attempt such inequalities.
              $endgroup$
              – jayant98
              Dec 3 '18 at 16:43


















            • $begingroup$
              Thanks for another way to attempt such inequalities.
              $endgroup$
              – jayant98
              Dec 3 '18 at 16:43
















            $begingroup$
            Thanks for another way to attempt such inequalities.
            $endgroup$
            – jayant98
            Dec 3 '18 at 16:43




            $begingroup$
            Thanks for another way to attempt such inequalities.
            $endgroup$
            – jayant98
            Dec 3 '18 at 16:43


















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