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Trigonometric and Oscillatory Examples in Several Variables: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

The circular curve defeats the equality form of the vector-valued mean value theorem

Statement refuted

Refuted claim: if mN with m1, if a<b are real, and if f:[a,b]Rm is continuous on [a,b] and differentiable on (a,b), then there is some ξ(a,b) such that

f(b)f(a)=(ba)f(ξ).

The claim is false already for m=2. The curve

f:[0,2π]R2,f(t)=(cost,sint),

is continuous on its interval and differentiable in its interior, its endpoint increment is zero, and f(t)2=1 for every t(0,2π). Hence no ξ(0,2π) satisfies the displayed equality.

Facts & Assumptions

Given: The curve f(t)=(cost,sint) on [0,2π].

[L1]

The functions sin and cos are differentiable on R, with (sint)=cost and (cost)=sint; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L2]

Let mN with m1, let AR, and let cA be a limit point of A. A map g:ARm is differentiable at c if and only if every component is differentiable there, and its derivative then has the component derivatives as coordinates (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L3]

Let mN with m1, let A be a subspace of a metric space, and let g:ARm. The map g is continuous at a point of A if and only if every component is continuous there (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

[L4]

For a real-valued function on AR, differentiability at a limit point cA implies continuity there (A function differentiable at c is continuous at c).

[L5]

For every real t, sin2t+cos2t=1 (Parity and the Pythagorean identity for sine and cosine).

[L7]
[L8]

Counterexample

technique · direct
1.1

By [L1] and [L4], both components of f are continuous on [0,2π], so f is continuous there by [L3].

L1L3L4
1.2

Periodicity and the values at zero give f(2π)=f(0)=(1,0), so f(2π)f(0)=(0,0).

L1L6algebra
1.3

Componentwise differentiation gives f(t)=(sint,cost) for every t(0,2π).

L1L2
1.4

Since π>0, the scalar 2π is nonzero.

L8algebra
2.1

For every t(0,2π), f(t)2=sin2t+cos2t=1.

step 1.3L5L7algebra
3.1

If the refuted equality held at some ξ(0,2π), step 1.2 would give (0,0)=2πf(ξ); step 1.4 would then force f(ξ)=(0,0), contradicting step 2.1. Thus no such ξ exists.

step 1.2step 2.1step 1.4algebra

Remarks

The obstruction is geometric: the curve returns to its initial point while its velocity never vanishes. The scalar mean value theorem is not weakened by this example; the failure is the demand that one point encode a vector increment.

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

FALSE: the mean value equality holds for vector-valued maps

Statement

False claim. Let mN with m1, let a<b be real, and let f:[a,b]Rm be continuous on [a,b] and differentiable on (a,b). Then some ξ(a,b) satisfies

f(b)f(a)=(ba)f(ξ).

Facts & Assumptions

Given: The universal equality claim in the Statement.

[L1]

The curve f(t)=(cost,sint) on [0,2π] is continuous on its interval and differentiable in its interior, its endpoint increment is zero, and f(t)2=1 for every t(0,2π); hence no ξ(0,2π) satisfies the claimed equality (The circular curve defeats the equality form of the vector-valued mean value theorem).

[L2]

Let mN with m1, let a<b be real, and let M0 be real. If f:[a,b]Rm is continuous on [a,b], differentiable on (a,b), and f(t)2M throughout (a,b), then f(b)f(a)2M(ba) (The mean value inequality: if f:[a,b]Rm is continuous and differentiable on (a,b) with f2M, then f(b)f(a)2M(ba)).

Refutation

technique · direct
1.1

Fact [L1] supplies an instance with m=2, a=0, and b=2π that satisfies both hypotheses of the false claim but not its conclusion.

L1
2.1

Therefore the universal equality claim is false.

step 1.1
3.1

The failure does not affect the vector-valued mean value inequality: under its derivative-bound hypothesis, the estimate in [L2] remains valid.

L2

Remarks

When m=1, the scalar mean value theorem does give the equality. The circular curve shows that the passage to a vector codomain, not a loss of regularity, is what breaks it.

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sin(xy) and its mixed partial derivatives

Example

For

f:R2R,f(x,y)=sin(xy),

the first partial derivatives and the two mixed partial derivatives exist everywhere and satisfy

xf(x,y)=ycos(xy),yf(x,y)=xcos(xy),

yxf(x,y)=xyf(x,y)=cos(xy)xysin(xy).

These formulas hold without excluding either coordinate axis.

Facts & Assumptions

Given: The function f(x,y)=sin(xy) on R2.

[L1]

The functions sin and cos are differentiable on R, with (sint)=cost and (cost)=sint (The derivatives of sine and cosine are cosine and minus sine).

[L4]

A coordinate partial derivative is the derivative at zero of the corresponding coordinate-line restriction (Directional derivatives and partial derivatives of a map URmRn).

Verification

technique · direct
1.1

Fixing y and differentiating the map xsin(xy) gives xf(x,y)=ycos(xy).

L1L2L4
1.2

Fixing x and differentiating the map ysin(xy) gives yf(x,y)=xcos(xy).

L1L2L4
2.1

Differentiating the formula in step 1.1 with respect to y gives yxf(x,y)=cos(xy)xysin(xy).

step 1.1L1L2L3
2.2

Differentiating the formula in step 1.2 with respect to x gives xyf(x,y)=cos(xy)xysin(xy).

step 1.2L1L2L3
3.1

The formulas in steps 1.1 through 2.2 are defined for every (x,y)R2, including x=0 or y=0, and the two mixed partials agree everywhere.

step 1.1step 1.2step 2.1step 2.2

Remarks

The equality is obtained by direct calculation rather than by invoking Clairaut--Schwarz theorem for continuous second partial derivatives. It is therefore an explicit instance of that theorem, not an application used to determine the common formula.

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xysin(1/(x2+y2)) is differentiable at the origin with unbounded partial derivatives nearby

Example

Define f:R2R by

f(0,0)=0,f(x,y)=xysin ⁣(1x2+y2)when (x,y)(0,0).

Then f is totally differentiable at the origin with Df(0,0)=0. Both partial derivatives exist at every point, but each is unbounded on every neighbourhood of the origin.

Facts & Assumptions

Given: The function f in the Example and s=x2+y2.

[L1]

For every real t, sint1 and cost1 (Parity and the Pythagorean identity for sine and cosine).

[L2]

The functions sin and cos are differentiable on R, with (sint)=cost and (cost)=sint; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L5]

A map is totally differentiable at the origin with derivative zero when f(h)/h20 as h0 through nonzero vectors (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(h2) remainder).

[L6]

The coordinate partial derivatives are the derivatives of the two coordinate-line restrictions (Directional derivatives and partial derivatives of a map URmRn).

[L7]

For (x,y)R2, (x,y)2=x2+y2 (The p-norms xp for rational p1, and x).

[L9]

For every real ε>0 there is a natural n1 with 1/n<ε (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[L10]

Every nonnegative real a has a unique nonnegative square root a1/2, positive when a>0 (Existence and uniqueness of n-th roots: a unique a1/n0 with (a1/n)n=a).

[L11]

Verification

technique · direct
1.1

For all real x,y, 2xyx2+y2=s, since (xy)20.

algebra
1.2

Both coordinate-line restrictions through the origin are identically zero, so xf(0,0)=yf(0,0)=0.

givenL6
1.3

At every point with s>0, the derivative rules give xf=ysin(1/s)2x2ycos(1/s)/s2 and yf=xsin(1/s)2xy2cos(1/s)/s2.

givenL2L3L4algebra
1.4

For k1 put ak=1/4πk. Then ak>0 and ak0: for ε>0, choose N1 with 1/N<4πε2 by [L9]; if kN, then 0<ak2=1/(4πk)<ε2, hence ak<ε.

L9L10L11algebra
2.1

If (x,y)(0,0), then f(x,y)xys/2=(x,y)22/2.

givenstep 1.1L1L7
2.2

Since 1/(2ak2)=2πk, periodicity and the values in [L2] give sin(1/(2ak2))=0 and cos(1/(2ak2))=1; step 1.3 therefore gives xf(ak,ak)=yf(ak,ak)=1/(2ak).

step 1.3L2L8algebra
3.1

Dividing step 2.1 by (x,y)2>0 gives f(x,y)/(x,y)2(x,y)2/20, so f is totally differentiable at the origin with derivative zero.

step 2.1L5algebra
4.1

Given any neighbourhood radius δ>0 and any bound B>0, step 1.4 permits a k with ak<min{δ/2,1/(2B)}; then (ak,ak)2<δ while step 2.2 gives xf(ak,ak)=yf(ak,ak)>B. Thus both partial derivatives are unbounded on every neighbourhood of the origin.

step 1.4step 2.2L7choose

Remarks

Differentiability at one point controls the size of the function's increment there. It does not impose a bound on derivatives at nearby points, and the rapidly oscillating reciprocal phase makes that distinction explicit.

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r2sin(1/r) is differentiable at the origin with a discontinuous gradient

Example

For (x,y)R2 put r=x2+y2 and define

F(0,0)=0,F(x,y)=r2sin(1/r)when r>0.

Then F is totally differentiable at the origin with derivative zero. On the punctured plane,

F(x,y)=(2rsin(1/r)cos(1/r))(xr,yr),

and this gradient is not continuous at the origin.

Facts & Assumptions

Given: The function F in the Example and r=(x,y)2.

[L1]

For (x,y)R2, (x,y)2=x2+y2 (The p-norms xp for rational p1, and x).

[L2]

For every real t, sint1 and sin2t+cos2t=1 (Parity and the Pythagorean identity for sine and cosine).

[L3]

The functions sin and cos are differentiable on R, with (sint)=cost and (cost)=sint; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L6]

A map is totally differentiable at the origin with derivative zero when F(h)/h20 as h0 through nonzero vectors (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(h2) remainder).

[L7]

For a scalar function, the gradient is the vector of its coordinate partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L8]

On (0,), (u1/2)=12u1/2 (Continuity and derivatives of positive-base real powers).

[L9]

If a>0 and qQ, then the real power aq agrees with the rational power; in particular this holds for q=1/2 (The exponential definition of real powers agrees with the existing rational powers).

[L11]

For every real t, sin(t+π)=sint and cos(t+π)=cost (Quarter-turn values and shifts by pi/2 and pi).

[L13]

For every real ε>0 there is a natural n1 with 1/n<ε (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[L14]
[L15]

If a map is totally differentiable at a point, then each partial derivative there equals the total derivative applied to the corresponding standard basis vector (A total derivative computes every directional derivative, and its matrix is the Jacobian).

Verification

technique · direct
1.1

For r>0, F(x,y)/r=rsin(1/r)r; hence this quotient tends to zero as (x,y)(0,0).

givenL1L2algebra
1.2

For k1 put ak=1/(2πk) and bk=1/((2k+1)π). Both are positive. Given ε>0, apply [L13] to 2πε>0 to obtain N1 with 1/N<2πε; then kN gives 0<bk<ak1/(2πN)<ε. Thus ak0 and bk0.

L13L14algebra
1.3

On the punctured plane, applying the derivative of the positive square root to r=(x2+y2)1/2 gives xr=x/r and yr=y/r.

givenL4L5L8L9L10algebra
1.4

For g(r)=r2sin(1/r) on (0,), the derivative rules give g(r)=2rsin(1/r)cos(1/r).

L3L4L5algebra
2.1

Therefore F is totally differentiable at the origin with derivative zero, and F(0,0)=(0,0).

step 1.1L6L7L15
2.2

Since F(x,y)=g(r) for r>0, steps 1.3 and 1.4 and the definition of the gradient give F(x,y)=(2rsin(1/r)cos(1/r))(x/r,y/r).

step 1.3step 1.4L7
3.1

Along the positive x-axis, periodicity gives F(ak,0)=(1,0), while periodicity followed by the shift through π gives F(bk,0)=(1,0).

step 1.2step 2.2L3L11L12algebra
4.1

Both point sequences in step 1.2 approach the origin but their gradient values in step 3.1 are distinct constants, so F has no limit at the origin and is not continuous there.

step 2.1step 1.2step 3.1

Remarks

The factor r2 is strong enough to make F(h)=o(h2) at the origin. Differentiation removes one radial power and exposes the undamped cosine oscillation, which is why the gradient behaves differently from the function.

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The family sin(nx)sin(ny) is uniformly bounded but not equicontinuous

Example

Let K=[0,π]2 with the Euclidean metric. For every natural n1, define

fn:KR,fn(x,y)=sin(nx)sin(ny).

Every fn is continuous, and the family F={fn:n1} is uniformly bounded by 1. It is not equicontinuous at the origin, and hence is not equicontinuous on K.

Facts & Assumptions

Given: The compact square K=[0,π]2 and the family F in the Example.

[L1]

For a nonempty compact metric space K, a family FC(K,R) is equicontinuous at a when every ε>0 admits one δ>0 that works for every fF and every xK; it is uniformly bounded when one M0 bounds f(x) for all f and x (Equicontinuity, pointwise boundedness, and uniform boundedness for families in C(K,R)).

[L2]

For every real t, sin(t)=sint and sint1 (Parity and the Pythagorean identity for sine and cosine).

[L3]

For all reals u,v, sinusinvuv (Sine and cosine are 1-Lipschitz on R).

[L4]
[L5]

For every real ε>0 there is a natural n1 with 1/n<ε (For every ε>0 in a complete ordered field there is a natural n1 with 1/n<ε).

[L6]
[L8]

For (u,v)R2, (u,v)2=u2+v2 (The p-norms xp for rational p1, and x).

Verification

technique · direct
1.1

The set K is a nonempty closed box in R2, so it is compact.

givenL6L7
1.2

For fixed n1 and points (x,y),(u,v)K, the sine bound and Lipschitz estimate give fn(x,y)fn(u,v)nxu+nyv2n(xu,yv)2; hence fn is continuous on K.

givenL2L3L8algebra
1.3

For every n1 and (x,y)K, fn(x,y)1, so F is uniformly bounded.

givenL1L2algebra
1.4

For every n1, fn(0,0)=0, while at pn=(π/(2n),π/(2n))K one has fn(pn)=1.

givenL2L4L6algebra
2.1

Let δ>0. Applying [L5] to δ/π>0 gives some natural n1 with π/n<δ, and then d2(pn,(0,0))=π/(2n)<π/n<δ.

step 1.4L5L6L8choosealgebra
3.1

Taking ε=1/2, steps 1.4 and 2.1 show that every δ>0 admits an fnF and a point pnK within δ of the origin for which fn(pn)fn(0,0)=1>ε. Thus the family is not equicontinuous at the origin.

step 1.1step 1.2step 1.4step 2.1L1

Remarks

Uniform boundedness controls the range of every function in the family. Equicontinuity asks for a common spatial scale, and the oscillation scale 1/n prevents such a scale at the origin.

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The solid generated by rotating y=sinx on [0,π] has volume π2/2

Example

Rotate the region

{(x,y):0xπ, 0ysinx}

about the x-axis. The resulting solid is compact and Jordan measurable, and its volume is

V=π22.

The vanishing endpoint radii are included in the solid and require no separate measurability argument.

Facts & Assumptions

Given: The profile f(x)=sinx on [0,π] and its solid of revolution about the x-axis.

[F1]

If ab and f:[a,b][0,) is continuous, then its solid of revolution about the x-axis is compact and Jordan measurable and has volume πabf(x)2dx (The disc formula for the volume of a solid of revolution).

[L1]

sinπ=0, and sinx>0 for every x with 0<x<π; thus π is the first positive zero of sine, in particular π>0 (Pi is the first positive zero of sine).

[L2]

The functions sin and cos are differentiable on R, with (sinx)=cosx and (cosx)=sinx; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L3]

For a real-valued function on AR, differentiability at a limit point cA implies continuity there (A function differentiable at c is continuous at c).

[L4]

For every real x, sin2x=(1cos2x)/2 (Double-angle and quadratic power-reduction identities).

[L5]

Every continuous real-valued function on a nondegenerate closed interval is Riemann integrable there (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L8]

If a<b, G:[a,b]R is differentiable at every point of [a,b], and G is integrable, then abG=G(b)G(a) (The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a)).

Verification

technique · direct
1.1

Facts [L1], [L2], and [L3] show that f is continuous and nonnegative on [0,π], so [F1] makes the rotated solid compact and Jordan measurable and gives V=π0πsin2xdx.

F1L1L2L3
1.2

Since π>0 by [L1], the interval [0,π] is nondegenerate. The function G(x)=12sin(2x) is differentiable with G(x)=cos(2x), and this derivative is continuous and therefore integrable on [0,π].

L1L2L3L5L9L10
1.3

On the same nondegenerate interval, the constant function 1 is integrable and has integral π.

L1L7algebra
2.1

By power reduction and linearity, followed by the fundamental theorem applied to step 1.2, 0πsin2xdx=120π1dx120πcos(2x)dx=π/212(G(π)G(0))=π/2, because periodicity and [L2] give G(π)=G(0)=0.

step 1.2step 1.3L2L4L6L8L11algebra
3.1

Substituting step 2.1 into the disc formula of step 1.1 gives V=π(π/2)=π2/2.

step 1.1step 2.1algebra

Remarks

The profile radius vanishes at both endpoints, but [F1] permits nonnegative continuous profiles and explicitly includes zero-radius sections. No division by the profile occurs.

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The surface generated by rotating y=sinx on [0,π] has area 2π(2+arsinh1)

Example

Rotate the graph y=sinx, 0xπ, about the x-axis. The resulting surface has area

A=2π(2+arsinh1)=2π(2+log(1+2)).

The profile is positive in the parameter interior and vanishes only at its endpoints, where the generating curve meets the axis of revolution.

Facts & Assumptions

Given: The radius function r(x)=sinx on [0,π] and the surface obtained by rotating its graph about the x-axis.

[F1]

Under the hypotheses of the scalar surface-integral theorem for a surface of revolution, the rotated surface has area 2πabr(s)1+r(s)2ds (The surface of revolution has area 2πabr(s)1+r(s)2ds).

[F2]

Those hypotheses require a<b and r:[a,b][0,) to be C1 on a neighbourhood of [a,b], positive on (a,b), and allowed to vanish only at the endpoints (Scalar surface integrals on a surface of revolution).

[L1]

sinπ=0, and sinx>0 for every x with 0<x<π; thus π is the first positive zero of sine, in particular π>0 (Pi is the first positive zero of sine).

[L2]

The functions sin and cos are differentiable on R, with (sinx)=cosx and (cosx)=sinx; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L3]

For a real-valued function on AR, differentiability at a limit point cA implies continuity there (A function differentiable at c is continuous at c).

[L4]

The function sinh:RR is odd, strictly increasing, and onto; cosh is positive; cosh2vsinh2v=1; and (sinhv)=coshv (Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[L5]

Let IR be order-convex with at least two elements, let f:IR be continuous and injective, and let g:f[I]I be its inverse. If f is differentiable at cI with f(c)0, then g is differentiable at f(c) and g(f(c))=1/f(c) (Derivative of an inverse: if f is continuous and injective on a nondegenerate interval I and differentiable at cI with f(c)0, then the inverse g is differentiable at f(c) with g(f(c))=1/f(c); and if f(c)=0 then g is not differentiable at f(c)).

[L6]

For every real u, arsinhu=log(u+u2+1) (Logarithm formulas for inverse sinh, inverse cosh, and inverse tanh on their natural domains).

[L7]

For every real α, uuα is continuous and differentiable on (0,), with derivative αuα1 (Continuity and derivatives of positive-base real powers).

[L8]

If a>0 and qQ, then the real power aq agrees with the rational power; in particular this holds for q=1/2 (The exponential definition of real powers agrees with the existing rational powers).

[L9]

Every nonnegative real a has a unique nonnegative square root a (Square roots exist: a unique a0 with (a)2=a; the positives are {x2:x0}).

[L12]

Every continuous real-valued function on a nondegenerate closed interval is Riemann integrable there (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L13]

If a<b, G:[a,b]R is differentiable at every point of [a,b], and G is integrable, then abG=G(b)G(a) (The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a)).

[L14]

cos0=1 and cosπ=1 (Quarter-turn values and shifts by pi/2 and pi).

Verification

technique · direct
1.1

Facts [L1], [L2], and [L3] verify the hypotheses in [F2], so [F1] gives A=2π0πsinx1+cos2xdx.

F1F2L1L2L3
1.2

Let uR and v=arsinhu. Facts [L3], [L4], and [L5] give (arsinhu)=1/coshv=1/1+u2, where positivity of coshv and uniqueness in [L9] select the nonnegative square root. Since sinh is an odd bijection, its inverse is odd, so arsinh(u)=arsinhu.

L3L4L5L9algebra
1.3

Since π>0 by [L1], the interval [0,π] is nondegenerate. The function xsinx1+cos2x is continuous on this interval and hence integrable there.

L1L2L3L7L8L9L12algebra
2.1

Define G(x)=12(arsinh(cosx)+cosx1+cos2x). Using step 1.2, the derivative of the positive square root, and the chain and product rules gives G(x)=sinx1+cos2x.

step 1.2L2L7L8L9L10L11algebra
3.1

By the fundamental theorem, step 1.3, and step 2.1, the integral in step 1.1 is G(π)G(0)=2+arsinh1: fact [L14] gives the cosine endpoint values, and the oddness in step 1.2 changes arsinh(1) to arsinh1. Fact [L6] also gives arsinh1=log(1+2).

step 1.2step 1.3step 2.1L6L9L13L14algebra
4.1

Substituting step 3.1 into the area formula of step 1.1 gives A=2π(2+arsinh1)=2π(2+log(1+2)).

step 1.1step 3.1algebra

Remarks

The endpoint zeros satisfy the source theorem's boundary allowance. The integrand stays continuous there, since its square-root factor is at least 1, so no improper-integral convention enters the calculation.

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

FALSE: spherical coordinates are globally injective

Statement

False claim. The spherical-coordinate map

S(r,ϕ,θ)=(rsinϕcosθ, rsinϕsinθ, rcosϕ)

is injective on [0,)×[0,π]×[0,2π].

The angular seam identifies θ=0 with θ=2π. At either polar angle, every azimuth represents the same point, and at radius zero both angles are lost. The Jacobian determinant

detDS(r,ϕ,θ)=r2sinϕ

vanishes on the zero-radius and polar-axis loci.

Facts & Assumptions

Given: The map S:R3R3 displayed in the Statement and its restriction to D=[0,)×[0,π]×[0,2π].

[L1]

The functions sin and cos are differentiable on R, with (sint)=cost and (cost)=sint; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L2]

For every real t, sin2t+cos2t=1 (Parity and the Pythagorean identity for sine and cosine).

[L3]

Sine vanishes exactly at the integer multiples of π, and both sine and cosine have period 2π (The zero sets of sine and cosine and the least positive common period 2 pi).

[L4]

sin(π/2)=1, cos(π/2)=0, sinπ=0, and cosπ=1 (Quarter-turn values and shifts by pi/2 and pi).

[L5]

For a C1 map g:R3R3, its Jacobian matrix consists of its coordinate partial derivatives and its Jacobian determinant is detDg (The Jacobian determinant of a square-dimensional C1 map is the determinant of its Jacobian matrix).

[L6]

Finite componentwise products and composites of C1 Euclidean maps are C1 (Ck Euclidean maps are closed under componentwise algebra and composition).

[L7]
[L8]

For a real-valued function on AR, differentiability at a limit point cA implies continuity there (A function differentiable at c is continuous at c).

Refutation

technique · direct
1.1

The two distinct points (1,π/2,0) and (1,π/2,2π) lie in D, and periodicity gives S(1,π/2,0)=S(1,π/2,2π)=(1,0,0).

givenL1L3L4L7
1.2

At ϕ=0, every θ[0,2π] gives S(r,0,θ)=(0,0,r); at ϕ=π, every such θ gives S(r,π,θ)=(0,0,r); and S(0,ϕ,θ)=(0,0,0) for every pair of angles.

givenL1L3L4algebra
1.3

The coordinate functions of S are finite products and composites of coordinate maps with sine and cosine; their displayed derivatives are continuous, so S is C1.

L1L6L8
2.1

Direct partial differentiation gives the following.

step 1.3L1L5algebra

DS=(sinϕcosθrcosϕcosθrsinϕsinθsinϕsinθrcosϕsinθrsinϕcosθcosϕrsinϕ0)

3.1

Expanding the determinant in step 2.1 and using the Pythagorean identity gives detDS=r2sinϕ.

step 2.1L2algebra
4.1

Step 1.1 already refutes global injectivity. Steps 1.2 and 3.1 show the additional pole and zero-radius identifications and show that the derivative is singular there, since r2sinϕ=0 when r=0, ϕ=0, or ϕ=π.

step 1.1step 1.2step 3.1L3

Remarks

Restricting the radius away from zero, the polar angle away from its endpoints, and the azimuth to a half-open interval removes these particular identifications. The false claim fails because the full closed parameter domain retains every seam and collapsed angular coordinate.

Sources