(According to this website:http://fac-staff.seattleu.edu/difranco/web/Math_371_W11/Files/Chebyshevnodes.pdf)
Between [-1,1], the Chebyshev Nodes are given as:
$x_k = \cos\Big((2k-1)\pi/2n)\Big), k=1,......,n$
and over [a,b] it is given as:
$x_k= 0.5(a+b) +0.5(b-a)\cos\Big((2k-1)(3.14159)/2n)\Big)$
What is the logic behind this transformation?
Similarly, the maximum error over [-1,1] is given as : $1/2^{n-1}$
Over [a,b], why is the error: $(b-a)^{n+1}/2^{2n+1}$?
This is the simple linear transformation $T: [-1,1] \rightarrow [a,b],\quad T(x)=\alpha x + \beta\,$ and $T(-1)=a,\; T(+1)=b.$ Now use $$a=T(-1)=-\alpha + \beta,\quad b=T(+1)=+\alpha + \beta$$ to get the values $$\alpha = \frac{1}{2}(b-a),\quad \beta = \frac{1}{2}(b+a)$$
It is not quite clear what $n$ is in your example (you have non standard indices $1..n,$ but the error scales with $\alpha^{n+1}$ for a polynomial with degree $n+1$ and you get $$\mathrm{err} = \frac{1}{2^n} \alpha^{n+1}=\frac{1}{2^n} \left(\frac{b-a}{2}\right)^{n+1}=\frac{1}{2^{n+1}} \left(b-a\right)^{n+1}$$