Roots of the Chebyshev polynomials of the second kind. It is known that the roots of the Chebyshev polynomials of the second kind, denote it by $U_n(x)$, are in the interval $(-1,1)$ and they are simple (of multiplicity one). I have noticed that the roots of $U_n{(x)}+U_{n-1}(x)$ (by looking at the law ranks of $n$) also lies in $(-1,1)$, I also noticed that  for $(1-x)U_n{(x)}+U_{n-1}(x)$ the roots lie in  $(-2,2)$. But I don't have any idea how to prove that in general, I wonder, first, if these claims are true? and how can I start proving them?
 A: For your first case:
Since
$U_n(x)
=\frac{\sin((n+1)t)}{\sin(t)}
$
where
$x = \cos(t)
$,
$\begin{array}\\
U_n(x)+U_{n-1}(x)
&=\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\
&=\frac{\sin((n+1)t)+\sin(nt)}{\sin(t)}\\
&=\frac{2\sin((n+1/2)t)\cos(t/2)}{\sin(t)}\\
\end{array}
$
and this is zero when
$t(n+1/2)
=k\pi
$
for some integer $k$,
or
$t
=\frac{k\pi}{n+1/2}
$
for
$1 \le k \le n$.
This gives $n$ real roots,
and that is all
since
$U_n(x)+U_{n-1}(x)$
is of degree $n$.
A: See my answer here by writing chebyshev polynomial explicitly in terms of $x = \cos t$. 
Update: I also show there that the roots are all real.
A: For your second case:
Since
$U_n(x)
=\frac{\sin((n+1)t)}{\sin(t)}
$
where
$x = \cos(t)
$,
$\begin{array}\\
(x-1)U_n(x)+U_{n-1}(x)
&=(\cos(t)-1)\frac{\sin((n+1)t)}{\sin(t)}+\frac{\sin(nt)}{\sin(t)}\\
&=\frac{(\cos(t)-1)\sin((n+1)t)+\sin(nt)}{\sin(t)}\\
\end{array}
$
Putting this into Wolfy,
this has this many roots
from $0$ to $\pi$
for these values of $n$:
$2: 1;
3: 2; 
4: 3;
5:4
 $
This seems to show that
this has $n-1$ real roots.
Since this is a
polynomial of degree $n+1$,
there should be
one complex pair of roots.
This is confirmed by Wolfy
directly by entering
"$(x-1)ChebyshevU[n, x]+ChebyshevU[n-1, x]=0
$"
for various values of $n$.
In all the cases I have tries,
all $n-1$ real roots
are between
$-1$ and $1$.
Other than this,
I haven't been able
to make any real progress
proving the result.
