China Mathematical Competition

The image of the three intercepting points of $f(x)$ and $y=kx$ is shown in the figure. It is easy to see that the curve and the line are tangent to each other at point $A(\alpha ,-\sin \alpha )$, and $\alpha \in (\pi ,\frac{3\pi }{2})$.

As $f^{\prime }(x)=-\cos x$ for $x\in (\pi ,\frac{3\pi }{2})$, we have $-\cos \alpha =-\frac{\sin \alpha }{\alpha }$, i.e. $\alpha =\tan \alpha$. Then $$\begin{array}{rl}\frac{\cos \alpha }{\sin \alpha +\sin 3\alpha }& =\frac{\cos \alpha }{2\sin 2\alpha \cos \alpha }=\frac{1}{4\sin \alpha \cos \alpha }\\ & =\frac{{\cos }^2\alpha +{\sin }^2\alpha }{4\sin \alpha \cos \alpha }=\frac{1+{\tan }^2\alpha }{4\tan \alpha }\\ & =\frac{1+{\alpha }^2}{4\alpha }.\end{array}$$