$If \alpha =\frac{\pi }{19} then \frac{\sin 3\alpha -\sin 23\alpha }{\sin 4\alpha +\sin 16\alpha } =$

$\begin{array}{rcl}\mathrm{\alpha }& =& \frac{\mathrm{\pi }}{19} \Rightarrow 19\mathrm{\alpha }=\mathrm{\pi }\\ & & \frac{\sin 3\mathrm{\alpha }-\sin 23\mathrm{\alpha }}{\sin 4\mathrm{\alpha }+\sin 16\mathrm{\alpha }}\\ & =& \frac{\sin 3\mathrm{\alpha }-\sin (19\mathrm{\alpha }+4\mathrm{\alpha })}{\sin 4\mathrm{\alpha }+\sin (19\mathrm{\alpha }-3\mathrm{\alpha })}\\ & =& \frac{\sin 3\mathrm{\alpha }-\sin (\mathrm{\pi }+4\mathrm{\alpha })}{\sin 4\mathrm{\alpha }+\sin (\mathrm{\pi }+3\mathrm{\alpha })}\\ & =& \frac{\sin 3\mathrm{\alpha }+\sin 4\mathrm{\alpha }}{\sin 4\mathrm{\alpha }+\sin 3\mathrm{\alpha }}\\ & =& 1\end{array}$