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Let $n \in \mathbb{Z}$ be an integer such that $| \cos n | < \frac{1}{4}$. Prove that $| \cos (n+1) | > \frac{1}{4}$.

I tried to use Lagrange for the difference $\cos(n+1)-\cos n$, hoping that I can prove that $|\cos (n+1) -\cos n| > \frac{1}{2}$.

We can write $\cos (n+1)- \cos n = - \sin \xi_n$ for some $\xi _n \in (n, n+1)$. But, nothing.

If we make a graph it's easy to see why the conclusion is true, but I don't know how to write it in a "formal language".

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  • $\begingroup$ well, you can just write how it is for the min an max possible case obeying |cos n|<1/4 $\endgroup$ – Fabrice NEYRET Nov 4 '15 at 19:57
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By the cosine addition trig identity:

$$\cos(n+1)=\cos n \cos 1-\sin n \sin 1 \tag{1}$$

where

$$|\cos n \cos 1|=\cos1\cdot|\cos n|<(\cos1)\cdot\tfrac{1}{4}<0.1351$$

and since $$\cos^2n+\sin^2n=1 \implies |\sin n|>\sqrt{1-\tfrac{1}{16}}=\frac{\sqrt{15}}{4}$$ we also have $$|\sin n\sin 1|=\sin1\cdot|\sin n|>(\sin1)\cdot\frac{\sqrt{15}}{4}>0.8147$$

So the first term in equation (1) has magnitude less than 0.1351, while the second term has magnitude greater than 0.8147, hence:

$$|\cos(n+1)|>0.8147-0.1351=0.6796$$

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