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Trigonometric and Oscillatory Examples in Several Variables

1 · Prerequisites

2 · Summary

The Pythagorean and double-angle identities of Parity and the Pythagorean identity for sine and cosine and Double-angle and quadratic power-reduction identities convert Cartesian expressions into radial and angular factors, while the derivative laws for sine and cosine support explicit partial-derivative calculations. The componentwise rule in The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral supplies the curve analysis, and Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C (K, R)$ fixes the relevant uniformity test. The disc-volume theorem The disc formula for the volume of a solid of revolution and surface-area formula The surface of revolution has area $2 π \int_{a}^{b} r (s) \sqrt{1 + r^{'} (s)^{2}} d s$ turn a sine profile into geometric integrals.

Polar-coordinate forms of two Cartesian expressions identifies the angular and radial structure of two Cartesian expressions. The circular parametrization refutes a vector-valued mean value equality, while $\sin (x y)$ exhibits agreeing mixed partials by calculation. Reciprocal radial phases separate differentiability from bounded or continuous derivatives, and the family $\sin (n x) \sin (n y)$ separates uniform boundedness from equicontinuity. The sine profile yields exact volume and surface area, while the seam, poles, and zero radius of spherical coordinates expose their failure of global injectivity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

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.

5 · Examples, counterexamples and false statements

None yet.