The Problems of Ukrainian Authors

Let an angle $\phi \in (0,\frac{\pi }{2})$ such that $\sin \phi =\frac{\sqrt{5}}{3}$. Then $\cos \phi =\frac{2}{3}$ and from the recurrent relation we have: $u_{n+1}=2u_n\cos \phi -u_{n-1}$. Moreover, $u_1=\sin \phi$, $u_2=\frac{2}{\sqrt{5}}\sin \phi \cos \phi =\frac{1}{\sqrt{5}}\sin 2\phi$.

Suppose that for every $n\in \mathbb{N}$ it holds: $u_n=\frac{1}{\sqrt{5}}\sin n\phi$. Then by mathematical induction we can get:

$u_{n+1}=\frac{2}{\sqrt{5}}\sin n\phi \cos \phi -\frac{1}{\sqrt{5}}\sin (n-1)\phi =\frac{1}{\sqrt{5}}\sin (n+1)\phi$, which is required.

The proof follows from the properties of sine function.