PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Polar-coordinate forms of two Cartesian expressions

Statement

Let $r > 0$ and $θ \in R$ , and put $x = r \cos θ$ and $y = r \sin θ$ . Then

$\frac{x y}{x^{2} + y^{2}} = \frac{1}{2} \sin (2 θ) .$

If also $\cos θ \neq 0$ , equivalently $x \neq 0$ , then

$\frac{y (x^{2} + y^{2})}{x} = r^{2} \tan θ .$

Thus the Cartesian functions $x y / (x^{2} + y^{2})$ and $y (x^{2} + y^{2}) / x$ become, on their stated domains, a purely angular function and a quadratic radial factor times an angular function, respectively.

Facts & Assumptions

Given: A real $r > 0$ , a real angle $θ$ , and $x = r \cos θ$ , $y = r \sin θ$ .

[L1]

For every real $t$ , $\sin^{2} t + \cos^{2} t = 1$ (Parity and the Pythagorean identity for sine and cosine).

[L2]

For every real $t$ , $\sin (2 t) = 2 \sin t \cos t$ (Double-angle and quadratic power-reduction identities).

[L3]

Where $\cos t \neq 0$ , $\tan t = \sin t / \cos t$ (Tangent, cotangent, secant, and cosecant on their exact natural domains).

Proof

technique · direct

1.1

The Pythagorean identity gives $x^{2} + y^{2} = r^{2} (\cos^{2} θ + \sin^{2} θ) = r^{2}$ , which is nonzero because $r > 0$ .

givenL1algebra

2.1

Therefore $x y / (x^{2} + y^{2}) = r^{2} \sin θ \cos θ / r^{2} = \sin θ \cos θ = \frac{1}{2} \sin (2 θ)$ .

step 1.1L2algebra

2.2

If $\cos θ \neq 0$ , then $x = r \cos θ \neq 0$ and $y (x^{2} + y^{2}) / x = (r \sin θ) r^{2} / (r \cos θ) = r^{2} \tan θ$ .

step 1.1L3algebra

3.1

Steps 2.1 and 2.2 prove both identities on exactly the domains stated.

step 2.1step 2.2∎

Remarks

The first identity has no radial dependence, whereas the second carries the factor $r^{2}$ . This difference explains why setting either expression equal to zero at the origin produces markedly different behaviour along rays.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, sections 8.3 and 11.4 (standard reference, not scraped)
University of Toronto MAT237, section 2.1 Differentiation of real-valued functions (standard reference, not scraped)