avewright/proofwiki-math huggingface dataset:
Theorem: Sine of Integer Multiple of Argument/Formulation 8

:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.

Proof:
From Sine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \map \sin {n \theta}
      | r = \paren {2 \cos \theta } \map \sin {\paren {n - 1 } \theta}  - \map \sin {\paren {n - 2 } \theta}
      | c = 
}}
{{eqn | r = \map \sin {\paren {n - 1 } \theta} \paren {\paren {2 \cos \theta }  - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} } }
      | c = factoring out $\map \sin {\paren {n - 1 } \theta}$
}}
{{end-eqn}}
Therefore $a_0 = 2 \cos \theta$
Once again, from Sine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {n \theta} } {\map \sin {\paren {n - 1 } \theta} }
      | r = 2 \cos \theta  - \dfrac {\map \sin {\paren {n - 2 } \theta} } {\map \sin {\paren {n - 1 } \theta} }
      | c = dividing both sides by $\map \sin {\paren {n - 1 } \theta}$
}}
{{eqn | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - 1 } \theta} } {\map \sin {\paren {n - 2 } \theta} } }
      | c = Move the numerator to the denominator
}}
{{end-eqn}}
In the equations above, let $n = n - k$:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - k - 1 } \theta} }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - k - 1 } \theta} } {\map \sin {\paren {n - k - 2 } \theta} } }
      | c = 
}}
{{eqn | l = \dfrac {\map \sin {\paren {n - k } \theta} } {\map \sin {\paren {n - \paren {k + 1} } \theta} }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {\paren {n - \paren {k + 1 } } \theta} } {\map \sin {\paren {n - \paren {k + 2 } } \theta} } }
      | c = 
}}
{{end-eqn}}
Therefore $a_1 = a_2 = \cdots = a_{n-3} = 2 \cos \theta$
Finally, let $k = n - 3$, then:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \sin {3 \theta} } {\map \sin {2 \theta } }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \sin {2 \theta} } {\map \sin {\theta} } }
      | c = 
}}
{{eqn | r = 2 \cos \theta - \cfrac 1 {\cfrac {2 \map \sin {\theta} \map \cos {\theta} } {\map \sin {\theta} } }
      | c = Double Angle Formula for Sine
}}
{{eqn | r = 2 \cos \theta - \dfrac 1 {2 \map \cos \theta }
      | c = 
}}
{{end-eqn}}
Therefore $a_{n - 2} = 2 \cos \theta$
Therefore:
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
{{qed}}
Theorem: Curl of Vector Field is Solenoidal

Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.
Let $\mathbf V$ be a vector field on $\R^3$:
Then the curl of $\mathbf V$ is a solenoidal vector field.

Proof:
By definition, a solenoidal vector field is one whose divergence is zero.
The result follows from Divergence of Curl is Zero.
{{qed}}
Theorem: Laplacian on Scalar Field is Divergence of Gradient

Let $\R^n$ denote the real Cartesian space of $n$ dimensions.
Let $U$ be a scalar field over $\R^n$.
Let $\nabla^2 U$ denote the laplacian on $U$.
Then:
:$\nabla^2 U = \operatorname {div} \grad U$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator.

Proof:
From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:
{{begin-eqn}}
{{eqn | l = \operatorname {div} \mathbf V
      | r = \nabla \cdot \mathbf V
}}
{{eqn | l = \grad \mathbf U
      | r = \nabla U
}}
{{end-eqn}}
where $\nabla$ denotes the del operator.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis on $\R^n$.
Hence:
{{begin-eqn}}
{{eqn | l = \nabla^2 U
      | o = :=
      | r = \nabla \cdot \paren {\nabla U}
}}
{{eqn | r = \paren {\sum_{k \mathop = 1}^n \mathbf e_k \dfrac \partial {\partial x_k} } \cdot \paren {\sum_{k \mathop = 1}^n \dfrac {\partial U} {\partial x_k} \mathbf e_k}
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \dfrac {\partial^2 U} {\partial {x_k}^2}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Condition for Vector Field to satisfy Laplace's Equation

Let $\mathbf V$ be a vector field over a region of space $R$.
Then:
:$\mathbf V$ is both solenoidal and conservative
{{iff}}:
:$\mathbf V$ is the gradient of a scalar field $F$ over $R$ which satisfies Laplace's equation:
::$\nabla^2 F \equiv 0$

Proof:
=== Sufficient Condition ===
Let $\mathbf V$ be both solenoidal and conservative.
Then from Vector Field is Expressible as Gradient of Scalar Field iff Conservative:
:$\mathbf V = \grad F$
for some scalar field $F$ over $R$.
Because $\mathbf V$ is solenoidal, we have:
:$\operatorname {div} \mathbf V = 0$
that is:
:$\operatorname {div} \grad F = 0$
Hence by Laplacian on Scalar Field is Divergence of Gradient:
:$\nabla^2 F = 0$
where $\nabla^2$ is the Laplacian on $F$.
This holds throughout $R$, and so the equality is an equivalence:
:$\nabla^2 F \equiv 0$
{{qed|lemma}}
Theorem: Condition for Vector Field to satisfy Poisson's Equation

Let $\mathbf V$ be a vector field over a region of space $R$.
Then:
:$\mathbf V$ is conservative but not solenoidal
{{iff}}:
:$\mathbf V$ is the gradient of a scalar field $F$ over $R$ which satisfies Poisson's equation over $R$:
::$\nabla^2 F = \phi$
:where $\phi$ is a function which is not identically zero.

Proof:
=== Sufficient Condition ===
Let $\mathbf V$ be conservative but not solenoidal.
From Vector Field is Expressible as Gradient of Scalar Field iff Conservative:
:$\mathbf V = \grad F$
for some scalar field $F$ over $R$.
Because $\mathbf V$ is not solenoidal, we have:
:$\exists \mathbf v \in R: \operatorname {div} \mathbf v \ne 0$
that is:
:$\operatorname {div} \grad F \ne 0$
for at least some $\mathbf v \in R$.
Hence by Laplacian on Scalar Field is Divergence of Gradient:
:$\nabla^2 F = \phi$
where:
:$\phi$ is not identically zero
:$\nabla^2$ is the Laplacian on $F$.
Hence $F$ satisfies Poisson's equation.
{{qed|lemma}}
Theorem: Cosine of Integer Multiple of Argument/Formulation 8

:$\cos n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-2} - \cfrac 1 {a_{n-1}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$ and
::A terminal $a_{n-1} = \cos \theta$ term.

Proof:
From Cosine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \map \cos {n \theta}
      | r = \paren {2 \cos \theta } \map \cos {\paren {n - 1 } \theta}  - \map \cos {\paren {n - 2 } \theta}
      | c = 
}}
{{eqn | r = \map \cos {\paren {n - 1 } \theta} \paren {\paren {2 \cos \theta }  - \dfrac {\map \cos {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} } }
      | c = factoring out $\map \cos {\paren {n - 1 } \theta}$
}}
{{end-eqn}}
Therefore $a_0 = 2 \cos \theta$
Once again, from Cosine of Integer Multiple of Argument Formulation 4 we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \cos {n \theta} } {\map \cos {\paren {n - 1 } \theta} }
      | r = 2 \cos \theta  - \dfrac {\map \cos {\paren {n - 2 } \theta} } {\map \cos {\paren {n - 1 } \theta} }
      | c = dividing both sides by $\map \cos {\paren {n - 1 } \theta}$
}}
{{eqn | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - 1 } \theta} } {\map \cos {\paren {n - 2 } \theta} } }
      | c = Move the numerator to the denominator
}}
{{end-eqn}}
In the equations above, let $n = n - k$:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \cos {\paren {n - k } \theta} } {\map \cos {\paren {n - k - 1 } \theta} }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - k - 1 } \theta} } {\map \cos {\paren {n - k - 2 } \theta} } }
      | c = 
}}
{{eqn | l = \dfrac {\map \cos {\paren {n - k } \theta} } {\map \cos {\paren {n - \paren {k + 1} } \theta} }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\paren {n - \paren {k + 1 } } \theta} } {\map \cos {\paren {n - \paren {k + 2 } } \theta} } }
      | c = 
}}
{{end-eqn}}
Therefore $a_1 = a_2 = \cdots = a_{n-2} = 2 \cos \theta$
Finally, let $k = n - 2$, then:
{{begin-eqn}}
{{eqn | l = \dfrac {\map \cos {2 \theta} } {\map \cos \theta }
      | r = 2 \cos \theta - \cfrac 1 {\cfrac {\map \cos {\theta} } {\map \cos {0} } }
      | c = 
}}
{{eqn | r = 2 \cos \theta - \dfrac 1 {\map \cos \theta }
      | c = 
}}
{{end-eqn}}
Therefore $a_{n - 1} = \cos \theta$
Therefore:
For $n \in \Z_{>1}$:
:$\cos n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n - 2} - \cfrac 1 {a_{n - 1} } } } } } }$
where $a_0 = a_1 = a_2 = \ldots = a_{n - 2} = 2 \cos \theta$ and
::A terminal $a_{n - 1} = \cos \theta$ term.  
{{qed}}
Theorem: Poisson's Differential Equation for Rotational and Solenoidal Field

Let $R$ be a region of ordinary space.
Let $\mathbf V$ be a vector field over $R$.
Let $\mathbf V$ be both rotational and solenoidal.
Let $\mathbf A$ be a vector field such that $\mathbf V = \curl \mathbf A$.
Then $\mathbf V$ satisfies this version of Poisson's differential equation:
:$\curl \mathbf V = -\nabla^2 \mathbf A \ne \bszero$

Proof:
As $\mathbf V$ is rotational it is not conservative.
Hence from Vector Field is Expressible as Gradient of Scalar Field iff Conservative $\mathbf V$ cannot be the gradient of some scalar field.
However, by definition of rotational vector field:
:$\curl \mathbf V \ne \bszero$
As $\mathbf V$ is solenoidal:
:$\operatorname {div} \mathbf V = 0$
Hence from Divergence of Curl is Zero, for some vector field $\mathbf A$ over $R$:
:$\operatorname {div} \mathbf V = \operatorname {div} \curl \mathbf A = 0$
and so:
:$\mathbf V = \curl \mathbf A$
Then we have:
{{begin-eqn}}
{{eqn | l = \curl \mathbf V
      | r = \curl \curl \mathbf A
      | c = 
}}
{{eqn | r = \grad \operatorname {div} \mathbf A - \nabla^2 \mathbf A
      | c = Curl of Curl is Gradient of Divergence minus Laplacian
}}
{{eqn | r = -\nabla^2 \mathbf A
      | c = {{Defof|Solenoidal Vector Field}}: setting $\operatorname {div} \mathbf A = 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Helmholtz's Theorem

Let $R$ be a region of ordinary space.
Let $\mathbf V$ be a vector field over $R$.
Let $\mathbf V$ be both non-conservative and non-solenoidal.
Then $\mathbf V$ can be decomposed into the sum of $2$ vector fields:
:one being conservative, with scalar potential $S$, but not solenoidal
:one being solenoidal, with vector potential $\mathbf A$, but not conservative.
Thus $\mathbf V$ satisfies the partial differential equations:
{{begin-eqn}}
{{eqn | n = 1
      | l = \operatorname {div} \mathbf V = \operatorname {div} \grad S
      | r = \nabla^2 S \ne 0
      | c = 
}}
{{eqn | n = 2
      | l = \curl \mathbf V = \curl \curl \mathbf A
      | r = \nabla^2 \mathbf A \ne \bszero
      | c = 
}}
{{end-eqn}}
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\curl$ denotes the curl operator
:$\nabla^2$ denotes the Laplacian.

Proof:
Let us write:
:$\mathbf V = \grad S + \curl \mathbf A$
where:
:$S$ is a scalar field
:$\mathbf A$ is a vector field chosen to be solenoidal
{{explain|It is not clear why this is always possible}}
Then:
{{begin-eqn}}
{{eqn | l = \operatorname {div} \mathbf V
      | r = \operatorname {div} \grad S + \operatorname {div} \curl \mathbf A
      | c = 
}}
{{eqn | r = \operatorname {div} \grad S
      | c = Divergence of Curl is Zero
}}
{{eqn | r = \nabla^2 S
      | c = Laplacian on Scalar Field is Divergence of Gradient
}}
{{eqn | o = \ne
      | r = 0
      | c = as $\mathbf V$ is not solenoidal
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \curl \mathbf V
      | r = \curl \grad S + \curl \curl \mathbf A
      | c = 
}}
{{eqn | r = \curl \curl \mathbf A
      | c = Curl of Gradient is Zero
}}
{{eqn | r = \grad \operatorname {div} \mathbf A - \nabla^2 \mathbf A
      | c = Curl of Curl is Gradient of Divergence minus Laplacian
}}
{{eqn | r = -\nabla^2 \mathbf A
      | c = as $\mathbf A$ is solenoidal: $\operatorname {div} \mathbf A = 0$
}}
{{eqn | o = \ne
      | r = \bszero
      | c = as $\mathbf A$ is not conservative
}}
{{end-eqn}}
{{explain|It is not clear how it is certain that $\mathbf A$ is not conservative}}
{{qed}}
{{Namedfor|Hermann Ludwig Ferdinand von Helmholtz|cat = Helmholtz}}
Theorem: Divergence of Product of Scalar Field with Gradient of Scalar Field

Let $R$ be a region of space.
Let $U$ and $W$ be scalar fields over $R$.
Then:
:$\map {\operatorname {div} } {U \grad W} = U \nabla^2 W + \paren {\grad U} \cdot \paren {\grad W}$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\nabla^2$ denotes the Laplacian.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {div} } {U \mathbf A}
      | r = \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U
      | c = Product Rule for Divergence
}}
{{eqn | ll= \leadsto
      | l = \map {\operatorname {div} } {U \grad W}
      | r = \map U {\operatorname {div} \grad W} + \paren {\grad W} \cdot \paren {\grad U}
      | c = substituting $\grad W$ for $\mathbf A$
}}
{{eqn | r = U \nabla^2 W + \paren {\grad U} \cdot \paren {\grad W}
      | c = Laplacian on Scalar Field is Divergence of Gradient
}}
{{end-eqn}}
{{qed}}
Theorem: Divergence of Product of Scalar Field with Curl of Vector Field

Let $R$ be a region of space.
Let $U$ be a scalar field over $R$.
Let $\mathbf A = \curl \mathbf B$ be a vector field over $R$ whose vector potential is $\mathbf B$.
Then:
:$\map {\operatorname {div} } {U \curl \mathbf B} = \paren {\curl \mathbf B} \cdot \paren {\grad U}$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\curl$ denotes the curl operator.

Proof:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {div} } {U \mathbf A}
      | r = \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U
      | c = Product Rule for Divergence
}}
{{eqn | ll= \leadsto
      | l = \map {\operatorname {div} } {U \curl \mathbf B}
      | r = \map U {\operatorname {div} \curl \mathbf B} + \paren {\curl \mathbf B} \cdot \paren {\grad U}
      | c = substituting $\curl \mathbf B$ for $\mathbf A$
}}
{{eqn | r = \paren {\curl \mathbf B} \cdot \paren {\grad U}
      | c = Divergence of Curl is Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Gradient of Newtonian Potential

Let $R$ be a region of space.
Let $S$ be a Newtonian potential over $R$ defined as:
:$\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$
where:
:$\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
:$r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
:$k$ is some predetermined constant.
Then:
:$\grad S = -\dfrac {k \mathbf r} {r^3} = -\dfrac {k \mathbf {\hat r} } {r^2}$
where:
:$\grad$ denotes the gradient operator
:$\mathbf {\hat r}$ denotes the unit vector in the direction of $\mathbf r$.
The fact that the gradient of $S$ is negative indicates that direction of the vector quantities that compose the vector field that is $\grad S$ all point towards the origin.

Proof:
From the geometry of the sphere, the equal surfaces of $S$ are concentric spheres whose centers are at the origin.
As the origin is approached, the scalar potential is unbounded above.
We have:
{{begin-eqn}}
{{eqn | l = \grad S
      | r = \map \grad {\dfrac k r}
      | c = Definition of $S$
}}
{{eqn | r = \map \nabla {\dfrac k r}
      | c = {{Defof|Gradient Operator}}
}}
{{eqn | r = k \map {\paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } } {\dfrac 1 {\paren {x^2 + y^2 + z^2}^{1/2} } }
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = -k \paren {\dfrac {x \mathbf i + y \mathbf j + z \mathbf k} {\paren {x^2 + y^2 + z^2}^{3/2} } }
      | c = Power Rule for Derivatives
}}
{{eqn | r = -\dfrac {k \mathbf r} {r^3}
      | c = Definition of $\mathbf r$ and simplifying
}}
{{eqn | r = -\dfrac {k \mathbf {\hat r} } {r^2}
      | c = as $\mathbf r = r \mathbf {\hat r}$
}}
{{end-eqn}}
{{qed}}
Theorem: Newtonian Potential satisfies Laplace's Equation

Let $R$ be a region of space.
Let $S$ be a Newtonian potential over $R$ defined as:
:$\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$
where:
:$\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
:$r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
:$k$ is some predetermined constant.
Then $S$ satisfies Laplace's equation:
:$\nabla^2 S = 0$
where $\nabla^2$ denotes the Laplacian.

Proof:
From Gradient of Newtonian Potential:
:$\grad S = -\dfrac {k \mathbf r} {r^3}$
where $\grad$ denotes the gradient operator.
Then:
{{begin-eqn}}
{{eqn | l = \operatorname {div} \grad S
      | r = \map {\operatorname {div} } {-\dfrac {k \mathbf r} {r^3} }
      | c = where $\operatorname {div}$ denotes divergence
}}
{{eqn | r = \nabla \cdot \paren {-\dfrac {k \mathbf r} {r^3} }
      | c = Divergence Operator on Vector Space is Dot Product of Del Operator
}}
{{eqn | r = -k \paren {\mathbf i \dfrac \partial {\partial x} + \mathbf j \dfrac \partial {\partial y} + \mathbf k \dfrac \partial {\partial z} } \cdot {\dfrac {x \mathbf i + y \mathbf j + z \mathbf k} {\paren {x^2 + y^2 + z^2}^{3/2} } }
      | c = {{Defof|Del Operator}}
}}
{{eqn | r = 0
      | c = Power Rule for Derivatives and tedious messy algebra
}}
{{eqn | ll= \leadsto
      | l = \nabla^2 S
      | r = 0
      | c = {{Defof|Laplacian on Scalar Field}}
}}
{{end-eqn}}
{{qed}}
Theorem: Riemann Integral Operator is Linear Mapping

Let $C \closedint a b$ be the space of continuous Riemann integrable functions.
Let $\R$ be the set of real numbers.
Let $I : C \closedint a b \to \R$ be the Riemann integral operator.
Then $I$ is a linear mapping.

Proof:
Let $x, y \in C \closedint a b$ be Riemann integrable.
Let $\alpha \in \R$.
Theorem: Total Solid Angle Subtended by Spherical Surface

The total solid angle subtended by a spherical surface is $4 \pi$.

Proof:
:400px
Let $\d S$ be an element of a surface $S$.
Let $\mathbf n$ be a unit normal on $\d S$ positive outwards.
From a point $O$, let a conical pencil touch the boundary of $S$.
Let $\mathbf r_1$ be a unit vector in the direction of the position vector $\mathbf r = r \mathbf r_1$ with respect to $O$.
Let spheres be drawn with centers at $O$ of radii $1$ and $r$.
Let $\d \omega$ be the area cut from the sphere of radius $1$.
We have:
:$\dfrac {\d \omega} {1^2} = \dfrac {\d S \cos \theta} {r^2}$
where $\theta$ is the angle between $\mathbf n$ and $\mathbf r$.
Thus $\d \omega$ is the solid angle subtended by $\d S$ at $O$.
Hence by definition of solid angle subtended:
:$\ds \Omega = \iint_S \frac {\mathbf {\hat r} \cdot \mathbf {\hat n} \rd S} {r^2}$
{{WIP|I've lost the thread of the source work. I will restart vector analysis from a different viewpoint, as the coverage of vector integration by Hague is not easy to interpret.}}
Theorem: Operator of Integrated Weighted Derivatives is Linear Mapping

Let $n \in \N$.
Let $I := \closedint a b$ be a closed real interval.
Let $\map {a_i} x : I \to \R$ be Riemann integrable functions.
Let $f, g \in \map {C^n} I$ be Riemann integrable real-valued functions of differentiability class $k$.
Let $L : \map {C^n} I \to \R$ be the operator of integrated weighted derivatives.
Then $L$ is a linear mapping.

Proof:
=== Distributivity ===
{{begin-eqn}}
{{eqn | l = \map L {f + g} 
      | r = \int_a^b \sum_{i \mathop = 0}^n \map {a_i} x \dfrac {\d^i}{ {\d x}^i} \paren{\map f x + \map g x} \rd x 
      | c = {{Defof|Operator of Integrated Weighted Derivatives}}
}}
{{eqn | r = \int_a^b \sum_{i \mathop = 0}^n \map {a_i} x \paren{\map {f^{\paren i} } x + \map {g^{\paren i} } x} \rd x 
      | c = Linear Combination of Derivatives
}}
{{eqn | r = \int_a^b \sum_{i \mathop = 0}^n \map {a_i} x \map {f^{\paren i} } x \rd x + \int_a^b \sum_{i \mathop = 0}^n \map {a_i} x \map {g^{\paren i} } x \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \map L f + \map L g
      | c = {{Defof|Operator of Integrated Weighted Derivatives}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Component of Vector is Scalar Projection on Standard Ordered Basis Element

Let $\tuple {\mathbf e_1, \mathbf e_2, \mathbf e_3}$ be the standard ordered basis of Cartesian $3$-space $S$.
{{explain|It needs to be emphasised that the the standard ordered basis of Cartesian $3$-space is actually an orthonormal basis.}}
Let $\mathbf a = a_1 \mathbf e_1 + a_2 \mathbf e_2 + a_3 \mathbf e_3$ be a vector quantity in $S$.
Then:
:$\mathbf a \cdot \mathbf e_i = a_i$

Proof:
Using the Einstein summation convention
{{begin-eqn}}
{{eqn | l = \mathbf a \cdot \mathbf e_i
      | r = a_j \cdot \mathbf e_j \cdot \mathbf e_i
      | c = 
}}
{{eqn | r = a_j \delta_{i j}
      | c = Dot Product of Orthonormal Basis Vectors
}}
{{eqn | r = a_i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sextuple Angle Formulas/Sine/Corollary

:$\sin 6 \theta = 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos \theta + 32 \sin^5 \theta \cos \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 6 \theta
      | r = \paren {2 \cos \theta } \sin 5 \theta - \sin 4 \theta
      | c = Sine of Integer Multiple of Argument/Formulation 4
}}
{{eqn | r = \paren {2 \cos \theta } \paren { 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta } - \paren {4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta }
      | c = 
}}
{{eqn | r = \paren {10 - 4 } \sin \theta \cos \theta + \paren {-40 + 8} \sin^3 \theta \cos \theta + 32 \sin^5 \theta \cos \theta
      | c = Gathering terms
}}
{{eqn | r = 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos \theta + 32 \sin^5 \theta \cos \theta
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Sextuple Angle Formula for Sine
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Theorem: Complex Conjugation is not Linear Mapping

Let $\overline \cdot: \C \to \C: z \mapsto \overline z$ be the complex conjugation over the field of complex numbers.
Then complex conjugation is not a linear mapping.

Proof:
{{begin-eqn}}
{{eqn | l = \overline {i \cdot 1}
      | r = \overline i \cdot \overline 1
      | c = Product of Complex Conjugates
}}
{{eqn | r = - i \cdot \overline 1
      | c = {{defof|Complex Conjugate}}
}}
{{eqn | o = \ne
      | r = i \cdot \overline 1
}}
{{end-eqn}}
By definition, it is not a linear mapping.
{{qed}}