Q:

Interpolation with the roots of orthogonal polynomials & Spectral expansion

I'm a bit confused about the relationships between these two approximation methods mentioned in the title. 

Does this kind of interpolation also belongs to the field of spectral methods? 
Are the Lagrange interpolants we get from using the roots of orthogonal polynomials also orthogonal? 
It's likely to get mixed with these two methods, could someone please clarify their differences?

Edit:
Let me use the Chebyshev polynomials as an example:
(1) Using the Chebyshev polynomials as basis functions, then $f(x)$ is approximated as
\begin{equation}
f(x)\simeq\sum\limits_{n=0}^Na_nT_n (x)
\end{equation}
(2) Interpolating at the $(N+1)$ roots, $x_0, x_1, ..., x_k,...,x_N$, of the Chebyshev polynomial $T_{N+1}(x)$, then the interpolation of $f(x)$ is
\begin{equation}
f(x)\simeq P_N(x)=\sum\limits_{k=0}^N f(x_k)L_k(x)
\end{equation}
where $L_k(x)$ is the interpolant function at $x=x_k$.

A:

I hope I understood the question correctly. They try to compute exactly the same thing, so they really are equivalent. I'll use Chebyshev polynomials because they are easy to analyze.
Given a function $f(x)$ on $[-1,1]$, the spectral interpolant is the truncation of
$$ \begin{aligned}
f(x) &= \sum_{n\geq0} \bar a_n T_n(x), \\
\bar a_n &= \frac{1+[n>0]}{\pi}\int_{-1}^1 f(x)T_n(x)\frac{dx}{\sqrt{1-x^2}} \\&= \frac{1+[n>0]}{\pi} \int_{0}^\pi f(\cos\theta)\cos(n\theta)\,\mathrm{d}\theta.
\end{aligned} $$
The $N$-th degree Lagrange interpolant, using the roots of $T_{N+1}(x)$ is given by 
$$ a_n = \frac{1+[n>0]}{N+1}\sum_{k=0}^N f(x_k) T_n(x_k), \qquad x_k = \cos\frac{\pi (k+\frac12)}{N+1}. $$
This uses the fact that $\sum_{k=0}^N T_m(x_k)T_n(x_k)$ is zero when $m\neq n$, $m,n\leq N$.
The formula for $a_n$ is nothing but the discrete cosine transform (type-II) applied to the function values at $x_k$, due to $T_n(x_k) = \cos \pi n(k+\frac12)/(N+1)$.
These are not, strictly speaking, the same, though.
The formula for $a_n$ is a trapezoidal rule approximation to the Fourier cosine integral in the formula for the exact coefficients $\bar a_n$. The trapezoidal rule is known to be exponentially accurate for smooth periodic functions (Trefethen-Weideman 2014), which $f(\cos\theta)$ is. Since for a spectral interpolant you would still have to evaluate the integral somehow, the Lagrange interpolant with the roots as nodes is just a way of evaluating that integral.
I tried to compute the exact difference between $a_n$ and $\bar a_n$, by expanding the sum for $a_n$ using the full series for $f$, and using the identity 
$$ \sum_{k=0}^{N} \cos(j\theta_k)\cos(n\theta_k) = \frac{N+1}{2}\big( [N+1\setminus j-n](-1)^{(j-n)N/(N+1)} + [N+1\setminus j+n](-1)^{(j+n)N/(N+1)}\big), $$
and for a complex-differentiable function $g(\theta)=f(\cos\theta)$ that is holomorphic in the region of the complex plane $|\Im \theta|<\alpha$, heuristically the error appears to be something on the order
$$ a_n - \bar a_n \sim |\bar a_{N+1-n}| \lesssim e^{-\alpha(N+1-n)}. $$
So for smooth functions this error decays very quickly and becomes negligible, so the Lagrange and the spectral interpolants can be considered identical.
Edit. What is the relationship between $\sum a_nT_n(x)$ and $\sum f(x_k) \ell_k(x)$?
Let $a_n$ be defined as above, let $f_1(x) = \sum_{n=0}^{N} a_n T_n(x)$, and let $f_2(x) = \sum_{k=0}^N f(x_k) \ell_k(x)$, where $\ell_k(x) = \prod_{j\neq k} (x-x_j)/(x_k-x_j)$.
Both $f_1(x)$ and $f_2(x)$ are polynomials in $x$ of degree $N$, by construction.
Using the above definition of $a_n$, together with
$$T_n(x_k) = \cos(n\theta_k), \qquad \theta_k = \pi(k+\tfrac12)/(N+1) $$
we can check that $f_1(x_k) = f(x_k)$, using the identity
$$ \sum_{n=0}^{N} \frac{1+[n>0]}{N+1} \cos(n\theta_j)\cos(n\theta_k) = [j=k]. $$
Therefore $f_1(x)$ and $f_2(x)$ are (non-identically-zero) polynomials of degree $N$ that pass through the same $N+1$ points, and therefore are the same polynomial.
Using the form of the interpolating polynomial in terms of $a_n$ makes the relationship with the Chebyshev series of the function $f(x)$ clearer than the Lagrange interpolation form.

