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For $\omega_j\in \left(\frac{(j-1)\pi}{1-R},\frac{j\pi}{1-R}\right)\;(j\geq 1)$, where $R\in (0,1)$, prove or disprove that $$ \mbox{sgn}\left(-\cos(\omega_j)-\omega_j\sin(\omega_j)+q(R-1)\cos[\omega_j(1+R)]\right)=\mbox{sgn}(q)\;, $$ where $q\in\mathbb{R}$.

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I'll write "$x$" for "$q$", for reasons that will become clear.

Since $j$, $\omega_j$, and $R$ are independent of $x$, you're really just comparing $\mathrm{sgn}(x)$ to $\mathrm{sgn}(mx+b)$ for some $m$ and $b$. Effectively, you're asking whether all points $(x,y)$ on the line $y=mx+b$ can be such that $\mathrm{sgn}(y) = \mathrm{sgn}(x)$.

Clearly(?), this situation occurs for, and only for, (strictly) positively-sloped (and non-vertical) lines through the origin. That is, when, and only when, $m>0$ and $b=0$.

Here, $b=-\cos(\omega_j)-\omega_j\sin(\omega_j)$, which may be zero for some isolated $\omega_j$s, but is certainly not zero throughout the full range of $\omega_j$. The value $m = (R-1)\cos(\omega_j(1+R))$ may or may not be positive throughout the range, but this particular point is moot.

So, unless there are additional restrictions or connections going on here, your equation cannot hold for all $q$.

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