Prove that \(\displaystyle{\left({{\sin}^{{2}}{\frac{{{x}}}{{{2}}}}}\right)}\cdot{\frac{{{\sqrt[{{3}}]{{{4}}}}}}{{{3}}}}\cdot{\frac{{{1}}}{{{\sin{{x}}}+{1}-{x}}}}^{{{\frac{{{2}}}{{{3}}}}}}\rbrace{ < }{1}\)

michiiiiiakqm

michiiiiiakqm

Answered question

2022-03-31

Prove that
(sin2x2)4331sinx+1x23}<1

Answer & Explanation

armejantm925

armejantm925

Beginner2022-04-01Added 20 answers

Note that for x[π2,π], then 12sin2x21
Now consider the function f(x)=sinx+1x
For x=π2 then f(x)>0 and for x=π then f(x)<0 so, since f is continuous, there is a zero of f in (π2,π). Let a be that zero.
Now let g(x)=1f(x)23 Note that g(x)>0 since g(x)=(f(x)13)2. If x is in a neighbourhood of a then g(x) has no upper bound. Therefore:
(sin2x2)4331sinx+1x23}12433g(x)
But g(x) has no upper bound so in a small enough neighbourhood of
(sin2x2)4331sinx+1x23}>1

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