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prove that : there exists a deleted neighborhood of $x=0$ such that : $$ x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5 $$

MyTry:

let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$ And :

$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$

Now what ?

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  • $\begingroup$ The inequality is flipped for $x<0.$ $\endgroup$ – zhw. Jun 9 '17 at 16:40
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We'll prove that your inequality is true for all $0<x<\frac{\pi}{2}$.

Indeed, $$f'(x)=\cos{x}-1+\frac{x^2}{2};$$ $$f''(x)=x-\sin{x}$$ and $$f'''(x)=1-\cos{x}>0.$$ Thus, $$f''(x)>f''(0)=0,$$ which gives $$f'(x)>f'(0)=0$$ and $$f(x)>f(0)=0.$$ Now, $$g'(x)=\cos{x}-1+\frac{x^2}{2}-\frac{x^4}{24};$$ $$g''(x)=-\sin{x}+x-\frac{x^3}{3}=-f(x)<0.$$ Thus, $$g'(x)<g'(0)=0,$$ which gives $$g(x)<g(0)=0$$ and we are done!

By the way, for $x<0$ your inequality is wrong.

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You have\begin{align*}\sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-\frac{x^9}{9!}+\frac{x^{11}}{11!}-\cdots\\&=x-\frac{x^3}6+\frac{x^5}{5!}\left(1-\frac{x^2}{6\times7}\right)+\frac{x^9}{9!}\left(1-\frac{x^2}{10\times11}\right)+\cdots\\&>x-\frac{x^3}6\end{align*}if $0<x<\sqrt{42}$. A similar argument proves that $\displaystyle\sin x<x-\frac{x^3}6+\frac{x^5}{120}$ in some interval $(0,\varepsilon)$.

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  • $\begingroup$ In fact, $\sin x<x-x^3/6+x^5/120$ for all positive $x$. $\endgroup$ – Lord Shark the Unknown Jun 9 '17 at 18:44
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From $$3^k\,\sin\frac{x}{3^k}-3^{k-1}\,\sin\frac{x}{3^{k-1}}=4\cdot3^{k-1}\,\sin^3\frac{x}{3^k},$$ by summing the telescoping series, $$x-\sin x=4\,\sum^\infty_{k=1}3^{k-1}\,\sin^3\frac{x}{3^k}.$$ Since $\sin x<x$ for $x>0$, the RHS is $\le x^3/6$.
BTW, the inequality for $\sin x$ can be shown in a similar way, because $$\frac{\sin x}{x}=\prod^\infty_{k=1}\cos\frac{x}{2^k}<1$$ for $x\neq0$.

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Well, $f(0)=0$ and $$f'(x)=\cos x-1+\frac{x^2}2.$$ If we could show that $f'(x)>0$ when $x>0$, then it would follow that $f'(x)>0$ when $x>0$.

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    $\begingroup$ There might be a small typo. $\endgroup$ – Zain Patel Jun 9 '17 at 18:42
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Taking my answer in Is this really equal to sin x? one more step:

If you start with $\sin'(x) = \cos(x), \cos'(x) = -\sin(x), \sin(0) = 0, \cos(0) = 1, \sin^2(x)+\cos^2(x) = 1$, you can proceed like this (not original with me):

$$\sin(x) =\int_0^x \cos(t)dt \le\int_0^x dt =x $$ $$\cos(x)-\cos(0) =\int_0^x -\sin(t) dt =-\int_0^x \sin(t) dt \ge-\int_0^x t dt =-\frac{x^2}{2}\\ \text{ so } \cos(x) \ge 1-\frac{x^2}{2} $$ $$\sin(x) =\int_0^x \cos(t)dt \ge\int_0^x (1-\frac{t^2}{2})dt =x-\frac{x^3}{6} $$ $$\cos(x)-\cos(0) =\int_0^x -\sin(t) dt =-\int_0^x \sin(t) dt \ge-\int_0^x (t-\frac{t^3}{6}) dt =-\frac{x^2}{2}+\frac{x^4}{24}\\ \text{ so } \cos(x) \le 1-\frac{x^2}{2}+\frac{x^4}{24} $$ $$\sin(x) =\int_0^x \cos(t)dt \le\int_0^x (1-\frac{x^2}{2}+\frac{x^4}{24})dt =x-\frac{x^3}{6}+\frac{x^5}{120} $$

By induction you can derive the power series for sin and cos and show that they are enveloping (the sum is between any two consecutive sums).

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