TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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$T_n(\cos\theta)=\cos(n\theta)$ and $U_n(\cos\theta)\sin\theta=\sin((n+1)\theta)$ for every $n\in\mathbb N$

Statement

For every $n\in\mathbb N$ and real $\theta$ , $T_n(\cos\theta)=\cos(n\theta),\qquad U_n(\cos\theta)\sin\theta=\sin((n+1)\theta).$ In particular, for $n\ge1$ and $0\le j\le n$ , $T_n(\cos(j\pi/n))=(-1)^j$ . 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 $\theta$ .

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 $\cos((n+1)\theta)$ and $\sin((n+2)\theta)$ .

step 1.2algebra

3.1

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

discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 46 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Tn(cos⁡θ)=cos⁡(nθ)T_n(\cos\theta)=\cos(n\theta)Tn​(cosθ)=cos(nθ) and Un(cos⁡θ)sin⁡θ=sin⁡((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta)Un​(cosθ)sinθ=sin((n+1)θ) for every n∈Nn\in\mathbb Nn∈N