Chebyshev polynomials

Recall $$\cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \alpha.$$ We then have $$\cos((n+1)\theta)=\cos \theta \cos n\theta-\sin \theta \sin n\theta\\ \cos((n-1)\theta)=\cos \theta \cos n\theta+\sin \theta \sin n\theta\\ \cos((n+1)\theta)=2\cos \theta \cos n\theta-\cos ((n-1)\theta)\quad (*)$$ We claim that for each non-negative integer $n$, there exist integers $c_i$ such that $$\cos n\theta=\sum_{i=0}^n c_i\cos^i\theta$$ (can be proved by induction). Namely $$\cos(n\theta)=T_n(\cos \theta).$$ Let $x=\cos \theta$. Then we have Chebyshev polynomials  $$T_n(x)=\cos(n\arccos x)$$  for $x\in [-1,1]$ (since $\arccos x$ is only defined in $[-1,1]$).

The equation $(*)$ also leads to Chebyschev recurrence relation $$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).$$ In fact, each $T_n$ is actually an algebraic polynomial of degree n: $$T_0(x)=\cos(0)=1\\ T_1(x)=\cos(\arccos x)=x\\ T_2(x)=\cos (2\arccos x)=\cos^2 y-\sin^2 y=\cos^2 y-(1-\cos^2 y)=2x^2-1,\quad \cos y=x\\ T_3(x)=\cos (3\arccos x)=4x^3-3x$$ They can also be derived using the recurrence relation.

Application
Errors in Polynomial Interpolation

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