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Let $\{\phi_n\}_{n\in\mathbb{N}}$ be a sequence of real numbers, such that : $\phi_n\in(-\frac{\pi}{2},\frac{\pi}{2})$ for all $n.$

My question is : On what condition on $\phi_n$ we can find (or we can prove unconditionally that there is ) $\alpha>0$ such that : $$\forall n\in\mathbb{N}\;\;\; \cos(\phi_n)\ge\alpha $$

Thanks

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    @JohnBentin, Thanks for your comment, the condition $\phi_n\in(-\pi/2,\pi/2)$ is enough ? – 2017-02-04
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    No, its not enough as for example $\cos(\pi/2-1/n)\to\cos(\pi/2)=0$ from continuity of $\cos$. – 2017-02-04
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    The condition you need is that there exists an $\epsilon>0$ so that $\phi_n<\pi/2-\epsilon$ and $\phi_n>-\pi/2+\epsilon$ for all $n$. You could for example guarantee this by requiring $\phi_n$ converges *in* $(-\pi/2,\pi/2)$. – 2017-02-04
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    @s.harp, thanks for your comment. – 2017-02-04

2 Answers 2

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Hint: It's useful to draw the functions to see intuitively the possible solution.

In the picture below, red line is $g(x)=\alpha$, blue graph is $f(x)=\cos{x}$. Anything in between the intersections of these two functions will satisfy $\cos{x}\geq \alpha$.

It also becomes evident that $1 \geq \alpha$, otherwise there are no sequences satisfying the condition.

enter image description here

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For example, we could take $-\frac12\pi+\epsilon\leqslant\phi_n\leqslant \,\frac12\pi-\epsilon$ for some $\epsilon>0$ and all $n\in \Bbb N$.