TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-02

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$T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$

Statement

For every $n \in N$ and real $θ$ , $T_{n} (\cos θ) = \cos (n θ), U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ) .$ In particular, for $n \geq 1$ and $0 \leq j \leq n$ , $T_{n} (\cos (j π / n)) = (- 1)^{j}$ . The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, The principle of mathematical induction.

Facts & Assumptions

Given: A natural $n$ and a real $θ$ .

Proof

technique · induction

1.1

The two identities follow directly from the initial polynomial values at $n = 0$ and $n = 1$ .

base

1.2

Assume the identities at $n$ and $n - 1$ .

2.1

The recurrences and the addition formulas give the usual second-order recurrences for $\cos ((n + 1) θ)$ and $\sin ((n + 2) θ)$ .

step 1.2algebra

3.1

Hence the identities hold at $n + 1$ , so induction proves them for every natural $n$ ; substituting $θ = j π / n$ gives the stated alternating values.

discharge-induction∎

Depends on

Used by

Dependency tree · two levels

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Sources

NIST Digital Library of Mathematical Functions, Chapter 18 (standard reference, not scraped)

Tn(cos⁡θ)=cos⁡(nθ)Tn​(cosθ)=cos(nθ) and Un(cos⁡θ)sin⁡θ=sin⁡((n+1)θ)Un​(cosθ)sinθ=sin((n+1)θ) for every n∈Nn∈N