Trigonometric definition of Chebyshev polynomials

Remarks

Proof idea: the cases $n = 0$ and $n = 1$ are clear. The sum formula for the cosine gives $\cos((n + 1)\theta) + \cos((n - 1)\theta) = 2 \cos \theta \cos(n \theta)$. This is the recursion of the $T_n$ evaluated at $x = \cos \theta$, so induction on $n$ gives the identity. Uniqueness holds because two polynomials that agree on the infinite set $[-1, 1]$ are equal.

Example: $\cos(3\theta) = 4 \cos^3 \theta - 3 \cos \theta$, which matches $T_3 = 4x^3 - 3x$.

Many texts take the identity as the definition of $T_n$ and derive the recursion from it. Outside $[-1, 1]$ the same polynomial satisfies $T_n(\cosh t) = \cosh(n t)$ for real $t$, by the same induction.

Going deeper

For a student

The identity explains most properties of $T_n$ at once. It gives $\abs{T_n} \le 1$ on $[-1, 1]$, the zeros $\cos \frac{(2k - 1)\pi}{2n}$ and the extreme points $\cos \frac{k\pi}{n}$. It also turns expansions in $T_n$ on $[-1, 1]$ into cosine series in $\theta$, so Fourier methods apply to Chebyshev series. The composition rule $T_m(T_n(x)) = T_{mn}(x)$ follows from $\cos(m \cdot n\theta) = \cos(mn\theta)$.