Prove that $\vert\sin(x)\vert + \vert\sin(x-1)\vert \ge \sin(1)$

While looking into the convergence of the series $\sum_{n=1}^{\infty}\frac{\sin(n)}{n}$ I stumbled into the inequality $\vert\sin(n)\vert + \vert\sin(n-1)\vert \ge \sin(1)$ for all $n\in\mathbb{R}$. (In specific it was used at the bottom of here to prove conditional convergence). The inequality was given without proof so I assume it should be somewhat trivial, however I cannot seem to get my head around it. Can anyone provide a proof of this?

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$1 = x - (x-1)$, so \begin{align*} \vert \sin 1\vert &= \vert\sin\left(x - (x-1)\right)\vert\\ &= \vert\sin x \cos(x-1) - \cos x \sin(x-1)\vert\\ &\le \vert\sin x \cos(x-1)\vert + \vert\cos x\sin(x-1)\vert\\ &\le \vert\sin x \vert + \vert\sin(x-1)\vert, \end{align*} since $\vert \cos(x-1)\vert$ and $\vert \cos x\vert$ are at most $1$.
$\sin(1) = \sin(x + (1-x)) = \sin(x)\cos(1-x)+\sin(1-x)\cos(x) \le \vert\sin(x)\vert+\vert\sin(x-1)\vert$