nyuuzyou/wikis huggingface dataset:
Speed of Body under Free Fall from Height/Proof 2

== Theorem ==
== Proof ==
From Acceleration is Second Derivative of Displacement with respect to Time:
:$\mathbf g = \dfrac {\d^2 \mathbf s} {\d t^2}$
Integrating with respect to $t$, and by definition of velocity:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf c_1$
When $t = 0$, we have that $\mathrm c_1$ is the initial velocity $\mathbf v_0$, and so:
:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf v_0$
Integrating with respect to $t$ again:
:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf c_2$
When $t = 0$, we have that $\mathrm c_2$ is the initial displacement $\mathbf s_0$, and so:
:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf s_0$
We have that $B$ is released at rest starting at $\mathbf s = \mathbf 0$.
Thus $\mathbf v_0 = \mathbf s_0 = \mathbf 0$ and so:
:$\mathbf s = \dfrac {\mathbf g t^2} 2$
:$\mathbf v = \mathbf g t$
By eliminating $t$ and taking the scalar quantities:
:$v = \sqrt {2 g s}$
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Category:Speed of Body under Free Fall from Height
Speed of Body under Free Fall from Height/Proof 3

== Theorem ==
== Proof ==
From the Principle of Conservation of Energy:
:$K + P = C$
where:
:$K$ is the kinetic energy of $B$
:$P$ is the potential energy of $B$
:$C$ is a constant.
Let the mass of $B$ be $m$.
From Kinetic Energy of Motion:
:$K = \dfrac {m v^2} 2$
where $v$ is the speed of $B$.
From Potential Energy of Position:
:$P = m g s$
where $s$ is the distance fallen by $B$.
Since $B$ falls from rest, its initial kinetic energy is zero.
Having fallen a distance $s$, $B$ has lost potential energy $m g s$.
Therefore:
:$\dfrac {m v^2} 2 = m g s$
from which the result follows by dividing both sides by $m$ and extracting the square root.
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
Category:Speed of Body under Free Fall from Height
Speed of Hour Hand

== Theorem ==
Consider an analogue clock $C$.
The hour hand of $C$ rotates at $\dfrac 1 2$ of a degree per minute.
== Proof ==
It takes $12$ hours, for the hour hand to go round the dial one time.
That is, in $12$ minutes the hour hand travels $360 \degrees$.
So in $1$ hour, the hour hand travels $\dfrac {360} {12} \degrees$, that is, $30 \degrees$.
So in $1$ minute, the hour hand travels $\dfrac 1 {60} \times 30 \degrees$, that is, $\dfrac 1 2 \degrees$.
== Also see ==
* Speed of Minute Hand
Category:Clocks
Speed of Minute Hand

== Theorem ==
Consider an analogue clock $C$.
The minute hand of $C$ rotates at $6$ degrees per minute.
== Proof ==
It takes one hour, that is $60$ minutes, for the minute hand to go round the dial one time.
That is, in $60$ minutes the minute hand travels $360 \degrees$.
So in $1$ minute, the minute hand travels $\dfrac {360} {60} \degrees$, that is, $6 \degrees$.
== Also see ==
* Speed of Hour Hand
Category:Clocks
Sphere in Normed Division Ring is Sphere in Induced Metric

== Theorem ==
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {S_\epsilon} {a; \norm {\,\cdot\,} }$ denote the sphere in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {S_\epsilon} {a; d }$ denote the sphere in the metric space $\struct {R, d}$.
Then:
:$\map {S_\epsilon} {a; \norm {\,\cdot\,} }$ = $\map {S_\epsilon} {a; d }$
== Proof ==
The result follows from Equality of Sets.
== Also see ==
* Closed Ball in Normed Division Ring is Closed Ball in Induced Metric
* Open Ball in Normed Division Ring is Open Ball in Induced Metric
Category:Definitions/Normed Division Rings
Sphere is Disjoint Union of Open Balls in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$:
:let $\map {S_\epsilon} a$ denote the sphere of $a$ of radius $\epsilon$. 
:let $\map {B_\epsilon} a$ denote the open ball of $a$ of radius $\epsilon$. 
Then:
:$\ds \forall n \in Z: \map {S_{p^{-n} } } a = \bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n}$
== Proof ==
For all $\epsilon \in \R_{>0}$:
:let $\map {B^-_\epsilon} a$ denote the closed ball of $a$ of radius $\epsilon$. 
Let $n \in \Z$.
Then:
From Closed Ball is Disjoint Union of Open Balls in P-adic Numbers:
:$\set {\map {B_{p^{-n} } } {a + i p^n}: i = 0, \dots, p - 1}$ is a set of pairwise disjoint open balls.
Continuing from above:
== Also see ==
* Closed Ball is Disjoint Union of Open Balls in P-adic Numbers
Category:Topology of P-adic Numbers
Sphere is Set Difference of Closed Ball with Open Ball

== Theorem ==
Let $M = \struct{A, d}$ be a metric space or pseudometric space.
Let $a \in A$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} {a; d}$ denote the closed $\epsilon$-ball of $a$ in $M$.
Let $\map {B_\epsilon} {a; d}$ denote the open $\epsilon$-ball of $a$ in $M$.
Let $\map {S_\epsilon} {a; d}$ denote the $\epsilon$-sphere of $a$ in $M$.
Then:
:$\map {S_\epsilon} {a; d} = \map { {B_\epsilon}^-} {a; d} \setminus \map {B_\epsilon} {a; d}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Category:Metric Spaces
Category:Closed Balls
Category:Open Balls
Sphere is Set Difference of Closed Ball with Open Ball/Normed Division Ring

== Theorem ==
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-closed ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {B_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-open ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {S_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-sphere of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Then:
:$\map {S_\epsilon} {a; \norm {\,\cdot\,} } = \map { {B_\epsilon}^-} {a; \norm {\,\cdot\,} } \setminus \map {B_\epsilon} {a; \norm {\,\cdot\,} }$
=== Corollary ===
== Proof ==
The result follows directly from:
:Closed Ball in Normed Division Ring is Closed Ball in Induced Metric
:Open Ball in Normed Division Ring is Open Ball in Induced Metric
:Sphere in Normed Division Ring is Sphere in Induced Metric
:Sphere is Set Difference of Closed Ball with Open Ball
Category:Normed Division Rings
Sphere is Set Difference of Closed Ball with Open Ball/P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct{\Q_p,\norm{\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} a$ denote the $\epsilon$-closed ball of $a$ in $\Q_p$.
Let $\map {B_\epsilon} a$ denote the $\epsilon$-open ball of $a$ in $\Q_p$.
Let $\map {S_\epsilon} a$ denote the $\epsilon$-sphere of $a$ in $\Q_p$.
Then:
:$\map {S_\epsilon} a = \map { {B_\epsilon}^-} a \setminus \map {B_\epsilon} a$
== Proof ==
The result follows directly from:
:P-adic Closed Ball is Instance of Closed Ball of a Norm
:P-adic Open Ball is Instance of Open Ball of a Norm
:P-adic Sphere is Instance of Sphere of a Norm
:Sphere is Set Difference of Closed and Open Ball in Normed Division Ring
Category:P-adic Number Theory
Spheres in 24 Dimensions in Leech Lattice

== Theorem ==
Let a set of identical spheres in a $24$-dimensional space be arranged in a Leech lattice.
Then each sphere will touch $196 \, 560$ other spheres.
This is believed to be the densest possible sphere packing in $24$ dimensions.
== Proof ==
== Sources ==
* 
* : $24$
* : $196,560$
* : $24$
* : $196,560$
Category:Geometry
Category:24
Spherical Coordinate Form of Laplace's Equation

== Theorem ==
Laplace's equation can be expressed in spherical coordinates as:
:$\dfrac 1 {r^2} \map {\dfrac \partial {\partial r} } {r^2 \dfrac {\partial V} {\partial r} } + \dfrac 1 {r^2 \sin^2 \theta} \dfrac {\partial^2 V} {\partial \phi^2} + \dfrac 1 {r^2 \sin \theta} \map {\dfrac \partial {\partial \theta} } {\sin \theta \dfrac {\partial V} {\partial \theta} } = 0$
== Proof ==
== Sources ==
* 
* 
Category:Laplace's Equation
Spherical Law of Cosines

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos a = \cos b \cos c + \sin b \sin c \cos A$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Spherical Law of Sines
* Law of Cosines
* Law of Sines
== Historical Note ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.97$
* 
* 
* : Chapter $5$: Eternal Triangles: Early trigonometry
:::... where he misattributes it to 
Category:Spherical Law of Cosines
Category:Spherical Trigonometry
Category:Named Theorems
Spherical Law of Cosines/Also known as

== Spherical Law of Cosines: Also known as == 
Some sources refer to the Spherical Law of Cosines as just:
:the cosine-formula
:the cosine law
:the cosine rule
:the rule of cosines
:the law of cosines
but all of these are also used to refer to the plane version, so on this is not recommended.
Some sources carefully refer to it as the law of cosines for spherical triangles.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.97$
* 
* 
* : Chapter $5$: Eternal Triangles: Early trigonometry
:::... where he misattributes it to 
Category:Spherical Law of Cosines
Spherical Law of Cosines/Angles

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos A = -\cos B \cos C + \sin B \sin C \cos a$
== Proof ==
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.
From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:
:not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
:but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
We have:
== Historical Note ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.97$
* : Chapter $\text I$. Spherical Trigonometry: $11$. Polar formulae.
* 
* 
Category:Spherical Law of Cosines
Category:Spherical Trigonometry
Spherical Law of Cosines/Proof 1

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $AD$ be the tangent to the great circle $AB$.
Let $AE$ be the tangent to the great circle $AC$.
Thus the radius $OA$ of $S$ is perpendicular to $AD$ and $AE$.
By construction, $AD$ lies in the same plane as $AB$.
Thus when $OB$ is produced, it will intersect $AD$ at $D$, say.
Similarly, $OC$ can be produced to intersect $AE$ at $E$, say.
The spherical angle $\sphericalangle BAC$ is defined as the angle between the tangents $AD$ and $AE$.
Thus:
:$\sphericalangle BAC = \angle DAE$
or, denoting that spherical angle $\sphericalangle BAC$ as $A$:
:$A = \angle DAE$
In the (plane) triangle $OAD$, we have that $\angle OAD$ is a right angle.
We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the length of a side of a spherical triangle.
Thus:
and by similar analysis of $\triangle OAE$, we have:
From consideration of $\triangle DAE$:
From consideration of $\triangle DOE$:
Thus:
and the result follows.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.
Category:Spherical Law of Cosines
Spherical Law of Cosines/Proof 2

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $O$ be joined to each of $A$, $B$ and $C$.
Let $P$ be an arbitrary point on $OC$.
Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$.
Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$.
In the plane $OAB$:
:construct $QS$ perpendicular to $OA$
:construct $RS$ perpendicular to $OB$
where $S$ is the point where $QS$ and $RS$ intersect.
Let $OS$ and $PS$ be joined.
Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$.
These tangents contain the spherical angle $A$.
But by construction, $QS$ and $QP$ are parallel to these tangents.
Hence $\angle PQS = \sphericalangle A$.
Similarly, $\angle PRS = \sphericalangle B$.
Also we have:
It is to be proved that $PS$ is perpendicular to the plane $AOB$.
By construction, $OQ$ is perpendicular to both $PQ$ and $QS$.
Thus $OQ$ is perpendicular to the plane $PQS$.
Similarly, $OR$ is perpendicular to the plane $PRS$.
Thus $PS$ is perpendicular to both $OQ$ and $OR$.
Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$.
That is, $PS$ is perpendicular to the plane $OAB$.
In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$
It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles.
From the right triangles $\triangle OQP$ and $\triangle ORP$, we have:
Let us denote the angle $\angle SOQ$ by $x$.
Then:
:$\angle ROS = c - x$
We have that:
But we also have:
Hence the result.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $9$. Alternative proofs of the formulae $\mathbf A$, $\mathbf B$ and $\mathbf C$.
Category:Spherical Law of Cosines
Spherical Law of Haversines

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\hav a = \map \hav {b - c} + \sin b \sin c \hav A$
where $\hav$ denotes haversine.
== Proof ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $13$. The haversine formula.
Category:Haversines
Category:Spherical Trigonometry
Spherical Law of Sines

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Spherical Law of Cosines
* Law of Sines
* Law of Cosines
== Historical Note ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.96$
* 
* 
* 
* : Chapter $5$: Eternal Triangles: Early trigonometry
:::... where he misattributes it to 
Category:Spherical Trigonometry
Category:Named Theorems
Category:Spherical Law of Sines
Spherical Law of Sines/Also known as

== Spherical Law of Sines: Also known as == 
Some sources refer to the spherical law of sines as just:
:the sine-formula
:the sine law
:the sine rule
:the rule of sines
:the law of sines
but all of these are also used to refer to the plane version, so on this is not recommended.
Some sources carefully refer to it as the law of sines for spherical triangles.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.96$
* 
* 
* 
* : Chapter $5$: Eternal Triangles: Early trigonometry
Category:Spherical Law of Sines
Spherical Law of Sines/Historical Note/Mistake

== Source Work ==
:
:Chapter $5$: Eternal Triangles
::Early trigonometry
== Mistake ==
: calculated sines for a circle of radius $10^{15}$ -- effectively, tables accurate to $15$ decimal places, but multiplying all numbers by $10^{15}$ to get integers -- for all multiples of one second of arc. He stated the law of sines for spherical triangles
::$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
:and the law of cosines
::$\cos a = \cos b \cos c + \sin b \sin c \cos A$
:in his De Triangulis, written in $1462$-$3$ but not published until $1533$.
== Correction ==
The tables published by appeared in fact in his .
This was published in $1596$, some $20$ years after the death of its author.
The tables contained in it were computed in intervals of $10$ seconds, not $1$ second, and calculated to $10$ decimal places, not $15$.
The Spherical Law of Sines and Spherical Law of Cosines actually appeared in by , not , who was not born until $1514$.
As for the suggestion that was not published until $1533$, this is suspect, as no corroboration for this can be found online.
It is accepted that there exists a $1533$ edition of this work, as this can be found everywhere you look.
But while evidence of an actual $1464$ edition may be elusive, the fact that no online biographies mention the fact that its publication was delayed until $1533$, the suggestion is that this is false.
It is possible that has conflated the two works in question: by , and by .
== Sources ==
* : Chapter $5$: Eternal Triangles: Early trigonometry
Category:Mistakes/Curious and Interesting Numbers
Spherical Law of Sines/Proof 1

== Theorem ==
== Proof ==
Let $X \in \R_{>0}$ such that:
:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
Then from $(1)$:
In a spherical triangle, all of the sides are less than $\pi$ radians.
The same applies to the angles.
From Shape of Sine Function:
:$\sin \theta > 0$ for all $0 < \theta < \pi$
Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:
:$X = \dfrac {\sin A} {\sin a}$
Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:
we arrive at the same point:
where:
:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
as before.
Hence we have:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $6$. The sine-formula.
Category:Spherical Law of Sines
Spherical Law of Sines/Proof 2

== Theorem ==
== Proof ==
:
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $O$ be joined to each of $A$, $B$ and $C$.
Let $P$ be an arbitrary point on $OC$.
Construct $PQ$ perpendicular to $OA$ meeting $OA$ at $Q$.
Construct $PR$ perpendicular to $OB$ meeting $OB$ at $R$.
In the plane $OAB$:
:construct $QS$ perpendicular to $OA$
:construct $RS$ perpendicular to $OB$
where $S$ is the point where $QS$ and $RS$ intersect.
Let $OS$ and $PS$ be joined.
Let tangents be constructed at $A$ to the arcs of the great circles $AC$ and $AB$.
These tangents contain the spherical angle $A$.
But by construction, $QS$ and $QP$ are parallel to these tangents.
Hence $\angle PQS = \sphericalangle A$.
Similarly, $\angle PRS = \sphericalangle B$.
Also we have:
It is to be proved that $PS$ is perpendicular to the plane $AOB$.
By construction, $OQ$ is perpendicular to both $PQ$ and $QS$.
Thus $OQ$ is perpendicular to the plane $PQS$.
Similarly, $OR$ is perpendicular to the plane $PRS$.
Thus $PS$ is perpendicular to both $OQ$ and $OR$.
Thus $PS$ is perpendicular to every line in the plane of $OQ$ and $OR$.
That is, $PS$ is perpendicular to the plane $OAB$.
In particular, $PS$ is perpendicular to $OS$, $SQ$ and $SR$
It follows that $\triangle PQS$ and $\triangle PRS$ are right triangles.
From the right triangles $\triangle OQP$ and $\triangle ORP$, we have:
From the right triangles $\triangle PQS$ and $\triangle PRS$, we have:
The result follows by applying this technique to the other angles of $ABC$.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $9$. Alternative proofs of the formulae $\mathbf A$, $\mathbf B$ and $\mathbf C$.
Category:Spherical Law of Sines
Spherical Law of Tangents

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} } = \dfrac {\tan \frac 1 2 \paren {a + b} } {\tan \frac 1 2 \paren {a - b} }$
== Proof ==
Hence we have:
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.98$
* : $\S 12$: Trigonometric Functions: $12.98.$
Category:Spherical Trigonometry
Category:Named Theorems
Spherical Laws of Sines and Cosines/Mistake

== Source Work ==
:
: Chapter $5$: Eternal Triangles: Trigonometry and logarithms
:: Early trigonometry
This mistake can be seen in the $2008$ edition as published by Quercus: ISBN 978-1-84724-768-1
== Mistake ==
:"He [that is, ] stated the law of sines for spherical triangles ... and the law of cosines ... in his De Triangulis, written in 1462-3 but not published until 1533."
The book De Triangulis (Omnimodus) was in fact by and was actually published in $1464$.
== Also see ==
* Spherical Law of Sines
* Spherical Law of Cosines
Category:Mistakes/Taming the Infinite
Spherical Triangle is Polar Triangle of its Polar Triangle

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Then $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
== Proof ==
:
Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively.
Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $ALM$.
By construction we have that $B'$ is the pole of $AC$.
Thus the length of the arc of the great circle from $B$ to any point on $AC$ is a right angle.
Similarly, the length of the arc of the great circle from $A'$ to any point on $BC$ is also a right angle.
Hence:
:the length of the great circle arc $CA'$ is a right angle
:the length of the great circle arc $CB'$ is a right angle
and it follows by definition that $C$ is a pole of $A'B'$.
In the same way:
:$A$ is a pole of $B'C'$
:$B$ is a pole of $A'C'$.
Hence, by definition, $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $11$. Polar formulae.
* 
* 
Category:Polar Triangles
Split Epimorphism is Epic

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a split epimorphism.
Then $f: C \twoheadrightarrow D$ is epic.
== Proof ==
Let $g: D \to C$ be a morphism such that:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of a split epimorphism.
Suppose $x, y: D \to E$ are morphisms, such that:
:$x \circ f = y \circ f$
Then:
:$x \circ f \circ g = y \circ f \circ g$
Hence:
:$f \circ g = \operatorname{id}_D$
It follows that:
:$x \circ \operatorname{id}_D = y \circ \operatorname{id}_D$
which yields the result, by the definition of an identity morphism.
The situation is illustrated by the following commutative diagram:
$\quad\quad \begin{xy}
<0em,0em> *+{C} = "C",
<0em,-4em>*+{D} = "D",
<4em,0em> *+{D} = "D2",
<8em,0em> *+{E} = "E",
"D2"+/r.5em/+/^.25em/;"E"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x},
"D2"+/r.5em/+/_.25em/;"E"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y},
"C";"D2" **@{-} ?>*@{>} ?*!/_.6em/{f},
"D";"D2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C},
"D";"C" **@{-} ?>*@{>} ?*!/_.6em/{g},
\end{xy}$
== Sources ==
* : $\S 2.1.1$
Category:Category Theory