= Solution
{c}
Use the <Chebyshev polynomial> identities $T_0=1$, $T_1=x$, and $T_{j+1}=2xT_j-T_{j-1}$. Induction shows that $T_j$ has degree $j$ with leading coefficient $2^{j-1}$ for $j\geq1$. The trigonometric definition also gives
$$
T_n'(\cos\theta)=n\frac{\sin(n\theta)}{\sin\theta}.
$$
Therefore
$$
f_n(x)=\frac{2^{1-n}}n(x^2-1)T_n'(x).
$$
The leading coefficient is $2^{1-n}n^{-1}\cdot n2^{n-1}=1$, and the degree is $n+1$. The endpoints are roots through $x^2-1$; the other $n-1$ points $\cos(s\pi/n)$ are roots through $T_n'$. They are distinct, so this monic polynomial is exactly $\pi_n$.
Differentiating the product $\pi_n(x)=(x-x_s)\pi_{n,s}(x)$ at $x_s$ gives $\pi_n'(x_s)=\pi_{n,s}(x_s)=w_s^{-1}$. For an interior node, differentiate the trigonometric form using $d/dx=-(\sin\theta)^{-1}d/d\theta$:
$$
\pi_n'(\cos\theta)=2^{1-n}
\left[\cot\theta\sin(n\theta)+n\cos(n\theta)\right].
$$
At $\theta_s=s\pi/n$, this is $2^{1-n}n(-1)^s$. At either endpoint, taking the limit makes the two terms equal, doubling the value: $\pi_n'(x_s)=2^{2-n}n(-1)^s$ for $s=0,n$.
After removing the common factor $2^{n-1}/n$, the <barycentric interpolation formula> therefore uses
$$
\boxed{w_s=(-1)^s\delta_s,\qquad
\delta_s=1\ (1\leq s<n),\quad \delta_0=\delta_n=\tfrac12.}
$$
These <Chebyshev-Lobatto node> weights differ in magnitude by only a factor of two.
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