Trigonometric inequalities with substitution











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Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$



I tried with substitution $tan frac{x}{2} = t$



$sin x = frac{2t}{t^2+1}$



$t>frac{x}{2}$



$2t>x$



$t^2+1>1+frac{x^2}{4}$



After this I am unable to do it
Please help me
Thank you










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  • NB both the original and derived identities are only valid for $x > 0$.
    – Travis
    Nov 13 at 9:55















up vote
0
down vote

favorite












Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$



I tried with substitution $tan frac{x}{2} = t$



$sin x = frac{2t}{t^2+1}$



$t>frac{x}{2}$



$2t>x$



$t^2+1>1+frac{x^2}{4}$



After this I am unable to do it
Please help me
Thank you










share|cite|improve this question
























  • NB both the original and derived identities are only valid for $x > 0$.
    – Travis
    Nov 13 at 9:55













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$



I tried with substitution $tan frac{x}{2} = t$



$sin x = frac{2t}{t^2+1}$



$t>frac{x}{2}$



$2t>x$



$t^2+1>1+frac{x^2}{4}$



After this I am unable to do it
Please help me
Thank you










share|cite|improve this question















Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$



I tried with substitution $tan frac{x}{2} = t$



$sin x = frac{2t}{t^2+1}$



$t>frac{x}{2}$



$2t>x$



$t^2+1>1+frac{x^2}{4}$



After this I am unable to do it
Please help me
Thank you







trigonometry






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edited Nov 13 at 9:39









MathFun123

459216




459216










asked Nov 13 at 9:34









Ashwini

11




11












  • NB both the original and derived identities are only valid for $x > 0$.
    – Travis
    Nov 13 at 9:55


















  • NB both the original and derived identities are only valid for $x > 0$.
    – Travis
    Nov 13 at 9:55
















NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55




NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55










1 Answer
1






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We know that, $$1-t^4<1$$



$Longrightarrow frac{1}{1+t^2}>1-t^2$



$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$



$Longrightarrow sin(x)>x-frac{x^3}{4}$



Hence proved.



Hope it helps:)






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    up vote
    0
    down vote













    We know that, $$1-t^4<1$$



    $Longrightarrow frac{1}{1+t^2}>1-t^2$



    $Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$



    $Longrightarrow sin(x)>x-frac{x^3}{4}$



    Hence proved.



    Hope it helps:)






    share|cite|improve this answer

























      up vote
      0
      down vote













      We know that, $$1-t^4<1$$



      $Longrightarrow frac{1}{1+t^2}>1-t^2$



      $Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$



      $Longrightarrow sin(x)>x-frac{x^3}{4}$



      Hence proved.



      Hope it helps:)






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        We know that, $$1-t^4<1$$



        $Longrightarrow frac{1}{1+t^2}>1-t^2$



        $Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$



        $Longrightarrow sin(x)>x-frac{x^3}{4}$



        Hence proved.



        Hope it helps:)






        share|cite|improve this answer












        We know that, $$1-t^4<1$$



        $Longrightarrow frac{1}{1+t^2}>1-t^2$



        $Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$



        $Longrightarrow sin(x)>x-frac{x^3}{4}$



        Hence proved.



        Hope it helps:)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 13 at 9:45









        Crazy for maths

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