# Existence of Chebyshev Polynomials of the Second Kind

## Theorem

There exists a Chebyshev polynomial of the second kind for all natural numbers $n$.

## Proof

For $n = 0$:

 $\ds \map \sin { \paren {0+1} \theta}$ $=$ $\ds \map \sin \theta$ $\ds$ $=$ $\ds 1 \cdot \sin \theta$
$\map {U_0} x = 1$, $U_0 \in \Bbb P$

For $n = 1$:

 $\ds \map \sin { \paren {1+1} \theta}$ $=$ $\ds \map \sin {2 \theta}$ $\ds$ $=$ $\ds 2 \sin \theta \cos \theta$ Double Angle Formula for Sine $\ds$ $=$ $\ds 2 \cos \theta \cdot \sin \theta$
$\map {U_1} x = 2x$, $U_1 \in \Bbb P$

Assume $\map {U_n} {\cos \theta}\sin \theta = \map \sin { \paren {n+1} \theta }$.

 $\ds \map {U_{n+1} } {\cos \theta} \sin \theta$ $=$ $\ds \map \sin { \paren {n+1+1} \theta }$ $\ds$ $=$ $\ds \map \sin { \paren {n+1} \theta + \theta }$ $\ds$ $=$ $\ds \map \sin { \paren {n+1} \theta} \map \cos \theta + \map \cos { \paren {n+1} \theta } \map \sin \theta$ Sine of Sum $\ds$ $=$ $\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta + \map \cos { \paren {n+1} \theta } \map \sin \theta - \map \sin { \paren {n+1} \theta} \map \cos \theta$ $\ds$ $=$ $\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta - \paren { \map \sin { \paren {n+1} \theta} \map \cos \theta - \map \cos { \paren {n+1} \theta } \map \sin \theta }$ $\ds$ $=$ $\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta - \map \sin { \paren {n+1} \theta - \theta }$ Sine of Difference $\ds$ $=$ $\ds 2 \map \cos \theta \map \sin { \paren {n+1} \theta} - \map \sin { \paren {n-1+1} \theta }$ $\ds$ $=$ $\ds 2 \map \cos \theta \map {U_n} {\cos \theta} \map \sin \theta - \map {U_{n-1} } {\cos \theta} \sin \theta$ by hypothesis $\ds$ $=$ $\ds \paren { 2 \map \cos \theta \map {U_n} {\cos \theta} - \map {U_{n-1} } {\cos \theta} } \cdot \map \sin \theta$
$\map {U_{n+1} } x = 2x \map {U_n} x - \map {U_{n - 1} } x$, $U_{n+1} \in \Bbb P$

By the Second Principle of Mathematical Induction, $U_n \in \Bbb P$ for all natural numbers $n$.

$\blacksquare$