# Show that $\inf\limits_{|x|\geq 1} 1-\frac{\sin(x)}{x}=1-\sin(1)\geq \frac{1}{7}$ [closed]

I am interested in showing that \begin{eqnarray} \inf\limits_{|x|\geq 1} \left(1-\frac{\sin(x)}{x}\right)=1-\sin(1)\geq \frac{1}{7}. \end{eqnarray}

## closed as off-topic by Carl Mummert, projectilemotion, Juniven, Leucippus, NamasteMar 3 '17 at 1:26

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Hint. We have that for $|x|\geq 2$, $$1-\frac{\sin(x)}{x}\geq 1-\frac{|\sin(x)|}{|x|}\geq 1-\frac{|\sin(x)|}{2}\geq 1-\frac{1}{2}=\frac{1}{2}$$ Now show that the function $f(x)=1-\frac{\sin(x)}{x}$ (which is even) is increasing in $[1,2]$.

• Another (actually just slightly weaker) way could be showing that it has no extremal points in the interior, and then confronting the values at the boundary gives the claim – b00n heT Nov 12 '16 at 19:04

Proof of $1-\sin(1)\geq \frac{1}{7}$:

Using Taylor's expansion we get,

$$1-\sin(1) = \frac{1}{3!} - \frac{1}{5!} + \frac{1}{7!} - \frac{1}{9!} +-\cdots \geq \frac{1}{3!} - \frac{1}{5!} = \frac{19}{120} > \frac{1}{7}.$$

Proof of first equality:

Will prove the claim mentioned by Mr. Robert Z, that is $f(x)=1-\frac{\sin(x)}{x}$ is increasing in $[1,2]$.

$f'(x)>0 \iff \tan(x) > x, x \in [1,2],$ which is true by taylor series expansion for $\tan(x)$.