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

Question

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}

Can someone please help me with this?

Answer 1

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 $[1,2]$.

Answer 2

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 $\tan(x)$.