If I write the Chebyshev polynomial of the first kind like this:
$T_n(x)=\cos(n\cos^{-1}x)$ for $x$ in $[-1,1]$.
It is clear that if $x=1$ then:
$T_n(1)=\cos(n\cos^{-1}1)=\cos(0)=1$ for all $n$, which is true.
But if I write it in a different way like this:
$T_n(\cos\theta)=\cos(n\theta)$ for $\theta$ in $[0,\pi]$. Then for $\theta=0$ I have: $$T_n(\cos0)=T_n(1)=\cos(n1)=\cos(n)$$ Which is not what I should have.
What am I missing?
Thank you in advance.
Why $T_n(1)=\cos(n1)$? Note that $T_n(\cos\theta)=\cos(n\theta)$, instead it seems you applied something like $T_n(\theta)=\cos(n\theta)$. It should be $$T_n(\cos(0))=T_n(1)=\cos(n\cos^{-1}(1))=\cos(n0)=1.$$