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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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  • $\begingroup$ NB both the original and derived identities are only valid for $x > 0$. $\endgroup$ – Travis Nov 13 '18 at 9:55

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