How to prove that at least one of the absolute values of sin(n) , sin(n+1),sin(n+2) is greater than 0.5. I tried to split it to cases but i cant handle sin(1),...sin(29) maybe i should work with radians. In the first section I proved that the series sin(n)/sqrt(n) converges in condition.
Can anyone help how to prove it?
Thank you in advance.
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First, let's try to find for what values of $x$ do we have $|\sin x|>0.5$. We know that the values for which $|\sin x|=0.5$ are $x=\pi/6+k\pi$ and $x=5\pi/6+k\pi$, with $k \in \mathbb{Z}$. Therefore, each $x$ in an interval of the form $[\pi/6+k\pi,5\pi/6+k\pi]$ will satisfy the condition $|\sin x|>0.5$.
The "gap" between two consecutive intervals like these is: $$\pi/6+(k+1)\pi-(5\pi/6+k\pi)=2\pi/6$$ Since $\pi<4$, the "gap" between two of these intervals is lower than $2$. Therefore, $3$ consecutive intergers cannot be in the same gap.
Moreover, $\pi>3$ so the length of an interval is greater than $2$. Which means if we take $3$ consecutive integers, they can't be in two different gaps.
The conclusion follows: between $n$, $n+1$ and $n+2$, one of them is in an interval of the form $[\pi/6+k\pi,5\pi/6+k\pi]$. So one of the numbers $|\sin (n)|$, $|\sin (n+1)|$ and $|\sin (n+2)|$ is greater than $0.5$.
If we look at the $\sin x$ graph and draw two horizontal lines $y=\pm 0.5$, we can see only when $x\in [n\pi - \frac{1}{6}\pi, n\pi +\frac{1}{6}\pi ], |\sin x| \leq 0.5$, otherwise, $|\sin x |>0.5$. In other words,these are shorter sections of length $\frac{\pi}{3} (|\sin x| \leq 0.5$ )separated by longer sections of length $\frac{2\pi}{3}(|\sin x| >0.5)$. It follows that if n is in a shorter section, n+1 will be in the following longer section and |$\sin (n+2)|$>0.5; If n is in a longer section, $|\sin n|>0.5$. Therefore, at least one of $\sin n$ or $\sin (n+2)$ has its absolute value larger than 0.5.