Roots of the Chebyshev polynomials of the second kind It is known that the roots of  Chebyshev polynomials of the second kind, denote it by $U_n(x)$,  are in the interval $(-1,1)$. I have noticed that, by looking at the low values of $n$, the roots of $(1-x)U_n(x)+U_{n-1}(x)$ are in the interval $(-2,2)$. However, I don't have a clear idea how to start proving this, could anyone help me please?
*PS I have asked this question on StackExchange and set a bounty but still have not got any answer for it. see this link 
 A: The key is to write $U_n(x)$ in terms of $x$ explicitly. Let $x = \cos t$, then $\sin t = \sqrt{1-x^2}$, where we fixed a branch of square root so that $\sqrt{-1} = i$; the value of $U_n$ is independent of the choice. Then $e^{it} = x + i \sqrt{1 - x^2}$. For $|x| \ge 1$, we can thus write $e^{it} = x - \sqrt{x^2 - 1}$.
For $|x| \ge 1$, let $y = x - \sqrt{x^2 -1}$, $A = 1-\sqrt{x^2 -1}$ and $B = (1-x)y + 1 $. Observe that $0 < y \le 1$. By analytic continuation, we can write
$$(1-x)U_n(x) + U_{n-1}(x) = \frac{ A y^{-(n+1)} -B y^n }{2 \sqrt{x^2 -1}}.$$
For $x > \sqrt{2}$, $A < 0$ and $B > 0$. So it's pretty clear the function has no zero there.
Similar argument can be made for $x \le -1$.
A: My previous answer shows that if all roots are real, then they must be contained with $[-1,\sqrt{2})$. To show all roots are real, use lemma 6.3.9 from the book Analytic Theory of Polynomials, which states that 
If $P,Q$ are monic polynomials of degree $n,m$ respectively, with $n \ge 3$ and $m < n$, and $R = (z-\alpha)P - \beta Q$, with $\alpha, \beta \in \mathbb{R}$ and $\beta  >0$. Then $P,Q$ have strictly interlacing zeros iff $P,R$ do. (Personally I don't see why $n = 2$ fails.)
We know that consecutive Chebyshev polynomials have strictly interlacing zeros by Proposition 6.3.10 (due to recurrence relation), so the conditions are satisfied. Note the condition on $m < n$ isn't really much more general, since interlacing implies $m = n-1$.
Real stability is a powerful tool. It also appears in the proof of Gurvits' lower bound for permanent and the proof of Kadison-Singer conjecture.
