hcju/proofwikiqa huggingface dataset:
Let $S$ be a sphere of radius $R$ whose center is located for convenience at the origin. Let $P$ be a plane which intersects $S$ but is not a tangent plane to $S$. It is to be shown that $S \cap P$ is a circle. Let $S$ and $P$ be embedded in a (real) cartesian space of $3$ dimensions. Let this space be rotated until $P$ is parallel to the plane $z = 0$. Thus from the Equation of Plane we have that $P$ can be described as: :$z = c$ where $\size c < R$ (or $P$ would not intersect $S$). Let $A = \tuple {x, y, z}$ be an arbitrary point on $S \cap P$. We have that: {{begin-eqn}} {{eqn | l = x^2 + y^2 + z^2       | r = R^2       | c = Equation of Sphere }} {{eqn | ll= \leadsto       | l = x^2 + y^2 + c^2       | r = R^2       | c = Equation of Plane: $z - c$ }} {{eqn | ll= \leadsto       | l = x^2 + y^2       | r = R^2 - c^2       | c =  }} {{eqn | o = >       | r = 0       | c = as $-R < c < R$ }} {{end-eqn}} The result follows from Equation of Circle. {{qed}}
By definition, the $p$-adic numbers are the unique (up to isometric isomorphism) non-Archimedean valued field that completes $\struct {\Q, \norm {\,\cdot\,}_p}$. By definition of a non-Archimedean valued field, $\norm {\,\cdot\,}_p$ is a non-Archimedean norm. By definition of a non-Archimedean norm, $\norm {\,\cdot\,}_p$ satisfies the axioms: $\text N 1$, $\text N 2$ and $\text N 4$. {{qed}} Category:P-adic Number Theory o44h5n17xfi18x9tdhtyj6m3d4131su
Let the poles of $C$ and $D$ be $A$ and $B$. $C$ and $D$ each lie embedded in a plane. Both of these planes by definition are perpendicular to $AB$. The result follows from Planes Perpendicular to same Straight Line are Parallel. {{qed}}
Because $A$, $B$ and $C$ do not lie on the same great circle, no two of these points are the endpoints of the same diameter of $S$. Otherwise it would be possible to construct a great circle passing through all $3$ points $A$, $B$ and $C$. Let a great circle $E$ be constructed through $A$ and $B$. Then as $C$ is not on $E$, it is on either one side or the other. Hence there is a finite spherical angle $\phi$ between $E$ and $C$. Let a diameter $PQ$ of $S$ be constructed whose endpoints are on $E$ such that: :neither $P$ nor $Q$ coincides with $A$ or $B$ :both $A$ and $B$ are on the same semicircle of $E$ into which $PQ$ divides $E$. Let two great circles $F$ and $G$ be constructed through $PQ$ which are at a spherical angle $\dfrac \phi 2$ from $E$, one in one direction and one in the other. Then $F$ and $G$ both divide $S$ into two hemispheres: :one such division has a hemisphere which contains $A$ and $B$ but does not contain $E$ :the other such division has a hemisphere which contains $A$, $B$ and $E$. Hence the result. {{qed}}
$A$ and $B$ are two points on a great circle $E$ of $S$ which are not both on the same diameter. So $AB$ is not equal to $2$ right angles. Then it is noted that both $A$ and $B$ are in the same hemisphere, from Three Points on Sphere in Same Hemisphere. That means the distance along $E$ is less than one semicircle of $E$. The result follows by definition of spherical angle and length of side of $AB$. {{qed}}
By metric axiom $(\text M 1)$: :$\map d {a, a} = 0$ By assumption: :$\epsilon > 0$ Hence: :$\map d {a, a} < \epsilon$ By definition of the closed $\epsilon$-ball of $a$ $\map {{B_\epsilon}^-} a$ in $M$: :$a \in \map { {B_\epsilon}^-} a$ {{qed}} Category:Metric Spaces 8qmx694flnod8zy7dm2tndr6ifttdir
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. From Closed Ball in Normed Division Ring is Closed Ball in Induced Metric, $\map { {B_\epsilon}^-} a$ is the closed $\epsilon$-ball of $a$ in the metric space $\struct{R,d}$. From Center is Element of Closed Ball: :$a \in \map { {B_\epsilon}^-} a$ {{qed}} Category:Normed Division Rings njelynv9t36p3joi480cdvqbcsryu6v
By definition, $\map { {B_\epsilon}^-} a$ is the closed $\epsilon$-ball of $a$ in the normed division ring $\struct {\Q_p, \norm {\,\cdot\,}_p}$. From Center is Element of Closed Ball in Normed Division Ring: $a \in \map { {B_\epsilon}^-} a$ {{qed}} Category:Topology of P-adic Numbers rq0scft6ipnmp4fwnx6xdkpws6dwjx1
:500px Let $R$ denote the center of $\bigcirc FCD$. Let $O$ denote the center of $S$, which is also the center of $\bigcirc EAB$. We have: :$CD = RC \times \angle CRD$ Similarly: :$AB = OA \times \angle AOB$ By Circles with Same Poles are Parallel: :$\bigcirc FCD \parallel \bigcirc EAB$ Hence $RC$ and $RD$ are parallel to $OA$ and $OB$ respectively. Thus: {{begin-eqn}} {{eqn | l = \angle CRD       | r = \angle AOB       | c =  }} {{eqn | ll= \leadsto       | l = CD       | r = \dfrac {RC} {OA} AB       | c =  }} {{eqn | r = \dfrac {RC} {OC} AB       | c = as $OA = OC$ are both radii of $S$ }} {{end-eqn}} We also have that: {{begin-eqn}} {{eqn | l = RC       | o = \perp       | r = OR       | c =  }} {{eqn | ll= \leadsto       | l = RC       | r = OC \cos \angle RCO       | c =  }} {{end-eqn}} and that: {{begin-eqn}} {{eqn | l = RC       | o = \parallel       | r = OA       | c =  }} {{eqn | ll= \leadsto       | l = \angle RCO       | r = \angle AOC       | c =  }} {{end-eqn}} We have that $\angle AOC$ is the (plane) angle subtended at $O$ by the arc $AC$ of $\bigcirc EAB$. Thus: {{begin-eqn}} {{eqn | l = CD       | r = AB \cos AC       | c =  }} {{eqn | r = AB \, \map \cos {PA - PC}       | c =  }} {{eqn | r = AB \sin PC       | c = as $PA$ is a right angle, and Cosine of Complement equals Sine }} {{end-eqn}} Hence the result. {{qed}}
We have that $\size {\operatorname {Long}_J - \operatorname {Long}_K}$ is the magnitude of the spherical angle between the meridians on which $J$ and $K$ lie. Thus from Length of Arc of Small Circle: :$\size {\operatorname {Long}_J - \operatorname {Long}_K} \cos \operatorname {Lat}_J$ degrees Then by definition of nautical mile: :$1$ nautical mile $= \dfrac 1 {60}$ degree of arc of latitude along a meridian measured along Earth's surface. Thus, approximately, $1$ degree of arc along the equator equals $60$ nautical miles. The result follows. {{qed}}
By metric space axiom $(\text M 1)$: :$\map d {a, a} = 0$ By assumption: :$\epsilon > 0$ Hence: :$\map d {a, a} < \epsilon$ By definition of the open $\epsilon$-ball of $a$ $\map {B_\epsilon} a$ in $M$: :$a \in \map {B_\epsilon} a$ {{qed}} Category:Open Balls Category:Center is Element of Open Ball 7iep1hfualttjot1dz00ffgnuv0n89z
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. From Open Ball in Normed Division Ring is Open Ball in Induced Metric, $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in the metric space $\struct{R,d}$. From Center is Element of Open Ball: :$a \in \map {B_\epsilon} a$ {{qed}} Category:Normed Division Rings Category:Center is Element of Open Ball hbfz7z7mx8al9c4afrud8yyvslxss3i
By definition, $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in the normed division ring $\struct {\Q_p, \norm {\,\cdot\,}_p}$. From Center is Element of Open Ball in Normed Division Ring: $a \in \map {B_\epsilon} a$ {{qed}} Category:Topology of P-adic Numbers Category:Center is Element of Open Ball h7csuqfstpfwo5dbvlgrky6l26y4g3m
=== Definition 1 implies Definition 2 ===   Let $\norm {\,\cdot\,} : R \to \R_{\ge 0}$ be a norm on a division ring satisfying: {{begin-axiom}} {{axiom | n = \text N 4         | lc= Ultrametric Inequality:         | q = \forall x, y \in R         | ml= \norm {x + y}         | mo= \le         | mr= \max \set {\norm x, \norm y} }} {{end-axiom}} It remains only to show that $\norm {\,\cdot\,}$ satisfies $(\text N 1)$ and $(\text N 2)$. This follows from the definition of a norm on a division ring. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\norm{\,\cdot\,} : R \to \R_{\ge 0}$ satisfy the non-Archimedean norm axioms: $(\text N 1)$, $(\text N 2)$ and $(\text N 4)$. To show that $\norm{\,\cdot\,}$ is a norm on a division ring satisfying $(\text N 4)$, it remains to show that $\norm{\,\cdot\,}$ satisfies: {{begin-axiom}} {{axiom | n = \text N 3         | lc= Triangle Inequality:         | q = \forall x, y \in R         | ml= \norm {x + y}         | mo= \le         | mr= \norm x + \norm y }} {{end-axiom}} Let $x, y \in R$. {{WLOG}}, suppose $\norm x \le \norm y$. From non-Archimedean norm axiom $(\text N 1)$ : Positive Definiteness: :$0 \le \norm x$ Then: {{begin-eqn}} {{eqn | l = \norm{x + y}       | o = \le        | r = \max \set {\norm x, \norm y}       | c = Non-Archimedean Norm Axiom $(\text N 4)$ : Ultrametric Inequality }} {{eqn | r = \norm y       | c = as $\norm x \le \norm y$ by assumption }} {{eqn | o = \le       | r = \norm x + \norm y       | c = as $0 \le \norm x$ }} {{end-eqn}} The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \cos b \cos c + \sin b \sin c \cos A       | r = \cos a       | c = Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \sin b \sin c \cos A       | r = \cos a - \cos b \cos c       | c =  }} {{eqn | ll= \leadsto       | l = \cos A       | r = \dfrac {\cos a - \cos b \cos c} {\sin b \sin c}       | c =  }} {{eqn | r = \cosec b \cosec c \paren {\cos a - \cos b \cos c}       | c = {{Defof|Real Cosecant Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos a       | r = \cos b \cos c + \sin b \sin c \cos A       | c = Spherical Law of Cosines }} {{eqn | r = \cos b \cos c + \sin b \sin c \paren {1 - 2 \sin^2 \dfrac A 2}       | c = Double Angle Formula for Cosine: Corollary 2 }} {{eqn | r = \map \cos {b - c} - 2 \sin b \sin c \sin^2 \dfrac A 2       | c = Cosine of Difference }} {{eqn | ll= \leadsto       | l = \map \cos {b - c} - \cos a       | r = 2 \sin b \sin c \sin^2 \dfrac A 2       | c = rearranging }} {{eqn | ll= \leadsto       | l = 2 \sin \dfrac {a + \paren {b - c} } 2 \sin \dfrac {a - \paren {b - c} } 2       | r = 2 \sin b \sin c \sin^2 \dfrac A 2       | c = Prosthaphaeresis Formula for Cosine minus Cosine }} {{eqn | ll= \leadsto       | l = 2 \, \map \sin {\dfrac {a + b + c} 2 - c}  \, \map \sin {\dfrac {a + b + c} 2 - b}       | r = 2 \sin b \sin c \sin^2 \dfrac A 2       | c =  }} {{eqn | ll= \leadsto       | l = \map \sin {s - c} \, \map \sin {s - b}       | r = \sin b \sin c \sin^2 \dfrac A 2       | c = setting $s = \dfrac {a + b + c} 2$ and simplifying }} {{end-eqn}} The result follows. {{qed}}
Let $n \in \Z$. Then: {{begin-eqn}} {{eqn | l = x \in \map { B_{p^{-n} } } a       | o = \leadstoandfrom       | r = \norm {x - a}_p < p^{-n}       | c = {{Defof|Open Ball in P-adic Numbers|Open Ball in $p$-adic Numbers}} }} {{eqn | o = \leadstoandfrom       | r = \norm {x - a}_p \le p^{-\paren {n + 1} }       | c = $p$-adic Norm of $p$-adic Number is Power of $p$ }} {{eqn | o = \leadstoandfrom       | r = x \in \map {B^-_{p^{-\paren {n + 1} } } } a       | c = {{Defof|Closed Ball in P-adic Numbers|Closed Ball in $p$-adic Numbers}} }} {{end-eqn}} By set equality: :$\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$ {{qed}} Category:Topology of P-adic Numbers 1vz77urn3ktbwn01ub8lcq64kxw667p
{{begin-eqn}} {{eqn | l = \tan \dfrac A 2       | r = \dfrac {\sqrt {\dfrac {\sin \paren {s - b} \sin \paren {s - c} } {\sin b \sin c} } } {\sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} } }       | c = Sine of Half Angle for Spherical Triangles, Cosine of Half Angle for Spherical Triangles }} {{eqn | r = \dfrac {\sqrt {\sin \paren {s - b} \sin \paren {s - c} } } {\sqrt {\sin s \, \map \sin {s - a} } }       | c = simplification }} {{end-eqn}} The result follows. {{qed}}
From Sequence of Powers of Number less than One, $\sequence{p^{-n}}$ is a real null sequence. From Null Sequence induces Local Basis in Metric Space, for all $a \in \Q_p$ the set $\set {\map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of $a$. From Union of Local Bases is Basis, the set: :$\BB' = \displaystyle \bigcup_{a \in \Q_p} \set {\map {B_{p^{-n} } } a : n \in \Z} = \set {\map {B_{p^{-n} } } a : a \in Q_p, n \in \Z}$ is a basis for $\tau_p$. Let $a \in \Q_p$ and $n \in \Z$. By definition of the $p$-adic numbers, $\Q$ is everywhere dense in $\Q_p$. Hence: :$\exists q \in \Q: q \in \map {B_{p^{-n} } } a$ From Characterization of Open Ball in P-adic Numbers, :$\map {B_{p^{-n} } } q = \map {B_{p^{-n} } } a$ Hence: :$\map {B_{p^{-n} } } a \in \BB_p$ Since $q$ and $n$ were arbitrary, then: :$\BB' \subseteq \BB_p$ It follows that $\BB_p$ is a basis for $\tau_p$. It remains to show that $\BB_p$ is countable. For all $q \in \Q$, let $\map {\BB_p} q = \set {\map {B_{p^{-n} } } q : n \in \Z}$. Then:  :$\BB_p = \displaystyle \bigcup_{q \mathop \in \Q} \map {\BB_p} q$ From Integers are Countably Infinite and Surjection from Countably Infinite Set iff Countable: :$\forall q \in \Q: \map {\BB_p} q$ is a countable set. From Rational Numbers are Countably Infinite: :$\BB_p$ is a countable union of countable sets. From Countable Union of Countable Sets is Countable: :$\BB_p$ is countable. {{qed}}
From Countable Basis for P-adic Numbers, the topological space $\struct {\Q_p, \tau_p}$ has a countable basis. By definition, $\struct {\Q_p, \tau_p}$ is second-countable. {{qed}}
By definition of the $p$-adic numbers, $\norm {\,\cdot\,}_p$ is a non-Archimedean norm. From Non-Archimedean Division Ring is Totally Disconnected, $\struct {\Q_p, \tau_p}$ is totally disconnected. {{qed}}
Let $d_p$ be the metric induced by the norm $\norm {\,\cdot\,}_p$. By definition of the topology induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$, $\tau_p$ is the topology induced by the metric $d_p$. From Metric Space is Hausdorff, it follows that $\struct{\Q_p, \tau_p}$ is Hausdorff. {{qed}}
From Local Basis of P-adic Number: :$\forall a \in \Q_p$: the set $\set {\map {B_{p^{-n} } } a: n \in Z}$ is a local basis of $a$. From Open and Closed Balls in P-adic Numbers are Compact Subspaces; :$\forall a \in \Q_p$: the set $\set {\map {B_{p^{-n} } } a: n \in Z}$ is a local basis of compact sets of $a$. Hence $\struct {\Q_p, \norm {\,\cdot\,}_p}$ is locally compact by definition. {{qed}}
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$. From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved. Let $d$ denote the subspace metric induced on $\map {B^-_{p^{-n}}} a$ by the $p$-adic Metric. From Open and Closed Balls in P-adic Numbers are Totally Bounded, the closed ball $\map {B^-_{p^{-n}}} a$ is totally bounded in $d$. Recall that by definition, the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ are complete. {{disambiguate|Definition:Complete}} From Open and Closed Balls in P-adic Numbers are Clopen in P-adic Metric, the closed ball $\map {B^-_{p^{-n}}} a$ is closed in the $p$-adic metric. {{disambiguate|Definition:Closed}} From Subspace of Complete Metric Space is Closed iff Complete, the closed ball $\map {B^-_{p^{-n}}} a$ is complete in $d$. {{disambiguate|Definition:Complete}} From Complete and Totally Bounded Metric Space is Sequentially Compact, the closed ball $\map {B^-_{p^{-n}}} a$ is sequentially compact in $d$. From Sequentially Compact Metric Space is Compact, the closed ball $\map {B^-_{p^{-n} } } a$ is compact in $d$. Hence the closed ball $\map {B^-_{p^{-n} } } a$ is a compact subspace in the $p$-adic metric by definition. The result follows. {{qed}} Category:Topology of P-adic Numbers tecnqtzfnasm0wj0ebbetr8rdidprs1
From the Spherical Law of Sines, we have: :$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$ from which: :$\sin A = \dfrac {\sin a \sin A} {\sin b}$ We find that $0 < \sin A \le 1$. We have that: :$\sin A = \map \sin {\pi - A}$ and so unless $\sin A = 1$ and so $A = \dfrac \pi 2$, it is not possible to tell which of $A$ or $\pi - A$ provides the correct solution. {{qed}}
From the Law of Sines, we have: :$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$ from which: :$\sin A = \dfrac {\sin a \sin B} {\sin b}$ We find that $0 < \sin A \le 1$. We have that: :$\sin A = \map \sin {\pi - A}$ and so unless $\sin A = 1$ and so $A = \dfrac \pi 2$, it is not possible to tell which of $A$ or $\pi - A$ provides the correct solution. :400px {{qed}}
:400px {{begin-eqn}} {{eqn | l = \cos b       | r = \cos a \cos c + \sin a \sin c \cos B       | c = Spherical Law of Cosines }} {{eqn | l = \cos c       | r = \cos b \cos a + \sin b \sin a \cos C       | c = Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \cos b       | r = \cos a \paren {\cos b \cos a + \sin b \sin a \cos C} + \sin a \sin c \cos B       | c = substituting for $\cos c$ }} {{eqn | ll= \leadsto       | l = \cos b       | r = \cos^2 a \cos b + \cos a \sin b \sin a \cos C + \sin a \sin c \cos B       | c =  }} {{eqn | ll= \leadsto       | l = \cos b \paren {1 - \cos^2 a}       | r = \cos a \sin b \sin a \cos C + \sin a \sin c \cos B       | c =  }} {{eqn | ll= \leadsto       | l = \cos b \sin^2 a       | r = \cos a \sin b \sin a \cos C + \sin a \sin c \cos B       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \cot b \sin a       | r = \cos a \cos C + \dfrac {\sin c} {\sin b} \cos B       | c = dividing both sides by $\sin a \sin b$ }} {{eqn | ll= \leadsto       | l = \cot b \sin a       | r = \cos a \cos C + \dfrac {\sin C} {\sin B} \cos B       | c = Spherical Law of Sines }} {{eqn | ll= \leadsto       | l = \cos a \cos C       | r = \sin a \cot b - \sin C \cot B       | c = simplification }} {{end-eqn}} {{qed}}
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$.   From Open Ball in P-adic Numbers is Closed Ball: :$\BB_p = \set {\map {B^-_{p^{-n} } } q : q \in \Q, n \in \Z} = \set {\map {B_{p^{-n + 1} } } q : q \in \Q, n \in \Z} = \set {\map {B_{p^{-n} } } q : q \in \Q, n \in \Z}$ From Countable Basis for P-adic Numbers: :$\BB_p$ is a countable basis for $\struct{\Q_p, \tau_p}$. {{qed}}
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon^-} a$ denote the closed $\epsilon$-ball of $a$.   From Closed Balls of P-adic Number: :$\BB_p = \set {q + p^n \Z_p : q \in \Q, n \in \Z} = \set {\map {B^-_{p^{-n} } } q : q \in \Q, n \in \Z}$ From Countable Closed Ball Basis for P-adic Numbers: :$\BB_p$ is a countable basis for $\struct{\Q_p, \tau_p}$. {{qed}}
Follows from: :P-adic Numbers is Hausdorff Topological Space :P-adic Numbers is Second Countable Topological Space :P-adic Numbers is Totally Disconnected Topological Space :P-adic Numbers is Locally Compact Topological Space {{qed}}
Let $\triangle ABC$ be a right spherical triangle such that the angle $\sphericalangle C$ is a right angle. :500px Let the remaining parts of $\triangle ABC$ be arranged according to the '''interior''' of the circle above, where the symbol $\Box$ denotes a right angle. ==== $\sin a$ ==== {{begin-eqn}} {{eqn | l = \dfrac {\sin a} {\sin A}       | r = \dfrac {\sin c} {\sin C}       | c = Spherical Law of Sines for side $a$ }} {{eqn | ll= \leadsto       | l = \dfrac {\sin a} {\sin A}       | r = \dfrac {\sin c} 1       | c = Sine of Right Angle as $C = \Box$ }} {{eqn | ll= \leadsto       | l = \sin a       | r = \sin A \sin c       | c =  }} {{eqn | ll= \leadsto       | l = \sin a       | r = \map \cos {\Box - A} \map \cos {\Box - c}       | c = Cosine of Complement equals Sine }} {{end-eqn}} {{qed|lemma}} ==== $\sin b$ ==== {{begin-eqn}} {{eqn | l = \dfrac {\sin b} {\sin B}       | r = \dfrac {\sin c} {\sin C}       | c = Spherical Law of Sines for side $b$ }} {{eqn | ll= \leadsto       | l = \dfrac {\sin b} {\sin B}       | r = \dfrac {\sin c} 1       | c = Sine of Right Angle as $C = \Box$ }} {{eqn | ll= \leadsto       | l = \sin b       | r = \sin B \sin c       | c =  }} {{eqn | ll= \leadsto       | l = \sin b       | r = \map \cos {\Box - B} \map \cos {\Box - c}       | c = Cosine of Complement equals Sine }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {\Box - A}$ ==== {{begin-eqn}} {{eqn | l = \cos A       | r = -\cos B \cos C + \sin B \sin C \cos a       | c = Spherical Law of Cosines for angle $A$ }} {{eqn | r = -\cos B \times 0 + \sin B \times 1 \times \cos a       | c = Cosine of Right Angle and Sine of Right Angle as $C = \Box$ }} {{eqn | r = \sin B \cos a       | c =  }} {{eqn | ll= \leadsto       | l = \map \sin {\Box - A}       | r = \map \cos {\Box - B} \cos a       | c = Sine of Complement equals Cosine, Cosine of Complement equals Sine }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {\Box - c}$ ==== {{begin-eqn}} {{eqn | l = \cos c       | r = \cos a \cos b + \sin a \sin b \cos C       | c = Spherical Law of Cosines for side $c$ }} {{eqn | r = \cos a \cos b       | c = Cosine of Right Angle as $C = \Box$ }} {{eqn | ll= \leadsto       | l = \map \sin {\Box - c}       | r = \cos a \cos b       | c = Sine of Complement equals Cosine }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {\Box - B}$ ==== {{begin-eqn}} {{eqn | l = \cos B       | r = -\cos A \cos C + \sin A \sin C \cos b       | c = Spherical Law of Cosines for angle $B$ }} {{eqn | r = -\cos A \times 0 + \sin A \times 1 \cos b       | c = Cosine of Right Angle as $C = \Box$ }} {{eqn | r = \sin A \cos b       | c =  }} {{eqn | ll= \leadsto       | l = \map \sin {\Box - B}       | r = \map \cos {\Box - A} \cos c       | c = Sine of Complement equals Cosine, Cosine of Complement equals Sine }} {{end-eqn}} {{qed}}
Let $\triangle ABC$ be a quadrantal triangle on the surface of a sphere whose center is $O$ such that side $c$ is a right angle.. :500px Let the remaining parts of $\triangle ABC$ be arranged according to the '''exterior''' of the circle above, where the symbol $\Box$ denotes a right angle. ==== $\sin A$ ==== {{begin-eqn}} {{eqn | l = \dfrac {\sin A} {\sin a}       | r = \dfrac {\sin C} {\sin c}       | c = Spherical Law of Sines for angle $A$ }} {{eqn | ll= \leadsto       | l = \dfrac {\sin A} {\sin a}       | r = \dfrac {\sin C} 1       | c = Sine of Right Angle as $c = \Box$ }} {{eqn | ll= \leadsto       | l = \sin A       | r = \sin a \sin C       | c =  }} {{eqn | ll= \leadsto       | l = \sin A       | r = \map \cos {\Box - a} \, \map \cos {\Box - C}       | c = Cosine of Complement equals Sine }} {{eqn | r = \map \cos {\Box - a} \, \map \cos {C - \Box}       | c = Cosine Function is Even }} {{end-eqn}} {{qed|lemma}} ==== $\sin B$ ==== {{begin-eqn}} {{eqn | l = \dfrac {\sin B} {\sin b}       | r = \dfrac {\sin C} {\sin c}       | c = Spherical Law of Sines for angle $B$ }} {{eqn | ll= \leadsto       | l = \dfrac {\sin B} {\sin b}       | r = \dfrac {\sin C} 1       | c = Sine of Right Angle as $c = \Box$ }} {{eqn | ll= \leadsto       | l = \sin B       | r = \sin b \sin C       | c =  }} {{eqn | ll= \leadsto       | l = \sin B       | r = \map \cos {\Box - b} \, \map \cos {\Box - C}       | c = Cosine of Complement equals Sine }} {{eqn | r = \map \cos {\Box - b} \, \map \cos {C - \Box}       | c = Cosine Function is Even }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {\Box - a}$ ==== {{begin-eqn}} {{eqn | l = \cos a       | r = \cos b \cos c + \sin b \sin c \cos A       | c = Spherical Law of Cosines for side $a$ }} {{eqn | r = \cos b \times 0 + \sin b \times 1 \times \cos A       | c = Cosine of Right Angle and Sine of Right Angle as $c = \Box$ }} {{eqn | r = \sin b \cos A       | c =  }} {{eqn | ll= \leadsto       | l = \map \sin {\Box - a}       | r = \map \cos {\Box - b} \cos A       | c = Sine of Complement equals Cosine, Cosine of Complement equals Sine }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {C - \Box}$ ==== {{begin-eqn}} {{eqn | l = \cos C       | r = -\cos A \cos B + \sin A \sin B \cos c       | c = Spherical Law of Cosines for angle $C$ }} {{eqn | ll= \leadsto       | l = -\cos C       | r = \cos A \cos B       | c = Cosine of Right Angle as $c = \Box$ }} {{eqn | ll= \leadsto       | l = \map \sin {C - \Box}       | r = \cos A \cos B       | c = Sine of Complement equals Cosine and Sine Function is Odd }} {{end-eqn}} {{qed|lemma}} ==== $\map \sin {\Box - b}$ ==== {{begin-eqn}} {{eqn | l = \cos b       | r = \cos a \cos c + \sin a \sin c \cos B       | c = Spherical Law of Cosines for side $b$ }} {{eqn | r = \cos a \times 0 + \sin a \times 1 \cos B       | c = Cosine of Right Angle as $c = \Box$ }} {{eqn | r = \sin a \cos B       | c =  }} {{eqn | ll= \leadsto       | l = \map \sin {\Box - b}       | r = \map \cos {\Box - a} \cos B       | c = Sine of Complement equals Cosine, Cosine of Complement equals Sine }} {{end-eqn}} {{qed}}