avewright/proofwiki-math huggingface dataset:
Theorem: Equivalence of Definitions of Cosecant of Angle

Let $\theta$ be an angle.
{{TFAE|def = Cosecant of Angle|view = cosecant}}

Proof:
=== Definition from Triangle implies Definition from Circle ===
Let $\csc \theta$ be defined as $\dfrac {\text{Hypotenuse}} {\text{Opposite}}$ in a right triangle.
Consider the triangle $\triangle OAB$.
By construction, $\angle OAB$ is a right angle.
From Parallelism implies Equal Alternate Angles:
:$\angle OBA = \theta$
Thus:
{{begin-eqn}}
{{eqn | l = \csc \theta
      | r = \frac {OB} {OA}
      | c = 
}}
{{eqn | r = \frac {OB} 1
      | c = as $OA$ is the radius of the unit circle
}}
{{eqn | r = OB
      | c = 
}}
{{end-eqn}}
That is:
:$\csc \theta = OB$
{{qed|lemma}}
Theorem: Cosine of Angle in Cartesian Plane

Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\cos \theta = \dfrac x r$
where $\cos$ denotes the cosine of $\theta$.

Proof:
:500px
Let a unit circle $C$ be drawn with its center at the origin $O$.
Let $Q$ be the point on $C$ which intersects $OP$.
From Parallelism implies Equal Alternate Angles, $\angle OQR = \theta$.
Thus:
:$(1): \quad \cos \theta = RQ$
by definition of cosine
$\angle OSP = \angle ORQ$, as both are right angles.
Both $\triangle OSP$ and $\triangle ORQ$ share angle $\angle SOP$.
By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.
By definition of similarity:
Then:
{{begin-eqn}}
{{eqn | l = \frac x r
      | r = \frac {SP} {OP}
      | c = 
}}
{{eqn | r = \frac {RQ} {OQ}
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | r = RQ
      | c = as $OP$ is the radius of the unit circle
}}
{{eqn | r = \cos \theta
      | c = from $(1)$ above
}}
{{end-eqn}}
{{qed}}
:500px
When $\theta$ is obtuse, the same argument holds, except that this time both $x$ and $\cos \theta$ are negative.
When $\theta = \dfrac \pi 2$ we have that $x = 0$ and $\cos \theta = 0 = \dfrac x r$.
Thus $y = r$ and $\sin \theta = 1 \dfrac y r$.
Thus the relation holds for $\theta = \dfrac \pi 2$.
When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.
Thus the relation continues to hold.
When $\theta = 0$ we have that $y = 0$ and $\cos \theta = 1 = \dfrac x r$.
When $\theta = \pi$ we have that $y = 0$ and $x = -r$, while $\cos \theta = -1 = \dfrac x r$.
Hence the result.
{{qed}}
Theorem: Tangent of Angle in Cartesian Plane

Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\tan \theta = \dfrac y x$
where $\tan$ denotes the tangent of $\theta$.

Proof:
:500px
Let a unit circle $C$ be drawn with its center at the origin $O$.
Let a tangent line be drawn to $C$ parallel to $PS$ meeting $C$ at $R$.
Let $Q$ be the point on $OP$ which intersects this tangent line.
$\angle OSP = \angle ORQ$, as both are right angles.
Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$.
By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.
Thus:
Then:
{{begin-eqn}}
{{eqn | l = \frac y x
      | r = \frac {SP} {OS}
      | c = 
}}
{{eqn | r = \frac {RQ} {OR}
      | c = {{Defof|Similar Triangles}}
}}
{{eqn | r = RQ
      | c = $OP$ is Radius of the Unit Circle
}}
{{eqn | r = \tan \theta
      | c = {{Defof|Tangent Function|subdef = Definition from Circle}}
}}
{{end-eqn}}
When $\theta$ is obtuse, the same argument holds, but both $x$ and $\tan \theta$ are negative.
When $\theta = \dfrac \pi 2$ we have that $x = 0$.
Then $OP$ is parallel to the tangent line at $R$ which it therefore does not meet.
Thus when $\theta = \dfrac \pi 2$, it follows that $\tan \theta$ is not defined.
Likewise $\dfrac y x$ is not defined when $x = 0$.
Thus the relation holds for $\theta = \dfrac \pi 2$.
When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.
In this case, $y$ is negative.
Thus the relation continues to hold.
When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\tan \theta = 0 = \dfrac y x$.
Hence the result.
{{qed}}
Theorem: Cotangent of Angle in Cartesian Plane

Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\cot \theta = \dfrac x y$
where $\cot$ denotes the cotangent of $\theta$.

Proof:
{{ProofWanted|Anybody want to take this on? I seem to have lost interest.}}
Theorem: Secant of Angle in Cartesian Plane

Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\sec \theta = \dfrac r x$
where $\sec$ denotes the secant of $\theta$.

Proof:
{{ProofWanted|Anybody want to take this on? I seem to have lost interest.}}
Theorem: Cosecant of Angle in Cartesian Plane

Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\csc \theta = \dfrac r x$
where $\csc$ denotes the secant of $\theta$.

Proof:
{{ProofWanted|Anybody want to take this on? I seem to have lost interest.}}
Theorem: Tangent is Sine divided by Cosine

Let $\theta$ be an angle such that $\cos \theta \ne 0$.
Then:
:$\tan \theta = \dfrac {\sin \theta} {\cos \theta}$
where $\tan$, $\sin$ and $\cos$ mean tangent, sine and cosine respectively.

Proof:
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \frac {\sin \theta} {\cos \theta}
      | r = \frac {y / r} {x / r}
      | c = Sine of Angle in Cartesian Plane and Cosine of Angle in Cartesian Plane
}}
{{eqn | r = \frac y r \frac r x
      | c = 
}}
{{eqn | r = \frac y x
      | c = 
}}
{{eqn | r = \tan \theta
      | c = Tangent of Angle in Cartesian Plane
}}
{{end-eqn}}
When $\cos \theta = 0$ the expression $\dfrac {\sin \theta} {\cos \theta}$ is not defined.
{{qed}}
Theorem: Cotangent is Cosine divided by Sine

Let $\theta$ be an angle such that $\sin \theta \ne 0$.
Then:
:$\cot \theta = \dfrac {\cos \theta} {\sin \theta}$
where $\cot$, $\sin$ and $\cos$ mean cotangent, sine and cosine respectively.

Proof:
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \frac {\cos \theta} {\sin \theta}
      | r = \frac {x / r} {y / r}
      | c = Cosine of Angle in Cartesian Plane and Sine of Angle in Cartesian Plane
}}
{{eqn | r = \frac x r \frac r y
      | c = 
}}
{{eqn | r = \frac x y
      | c = 
}}
{{eqn | r = \cot \theta
      | c = Cotangent of Angle in Cartesian Plane
}}
{{end-eqn}}
When $\sin \theta = 0$ the expression $\dfrac {\cos \theta} {\sin \theta}$ is not defined.
{{qed}}
Theorem: Cotangent is Reciprocal of Tangent

:$\cot \theta = \dfrac 1 {\tan \theta}$

Proof:
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \cot \theta
      | r = \frac x y
      | c = Cotangent of Angle in Cartesian Plane
}}
{{eqn | r = \frac 1 {y / x}
      | c = 
}}
{{eqn | r = \frac 1 {\tan \theta}
      | c = Tangent of Angle in Cartesian Plane
}}
{{end-eqn}}
$\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$.
{{qed}}
Theorem: Secant is Reciprocal of Cosine

:$\sec \theta = \dfrac 1 {\cos \theta}$

Proof:
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \sec \theta
      | r = \frac r x
      | c = Secant of Angle in Cartesian Plane
}}
{{eqn | r = \frac 1 {x / r}
      | c = 
}}
{{eqn | r = \frac 1 {\cos \theta}
      | c = Cosine of Angle in Cartesian Plane
}}
{{end-eqn}}
When $\cos \theta = 0$, $\dfrac 1 {\cos \theta}$ is not defined.
{{qed}}
Theorem: Cosecant is Reciprocal of Sine

:$\csc \theta = \dfrac 1 {\sin \theta}$

Proof:
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \csc \theta
      | r = \frac r y
      | c = Cosecant of Angle in Cartesian Plane
}}
{{eqn | r = \frac 1 {y / r}
      | c = 
}}
{{eqn | r = \frac 1 {\sin \theta}
      | c = Sine of Angle in Cartesian Plane
}}
{{end-eqn}}
When $\sin \theta = 0$, $\dfrac 1 {\sin \theta}$ is not defined.
{{qed}}
Theorem: Positive-Term Generalized Sum Converges iff Supremum

Let $\struct {G, \circ, \le}$ be an abelian totally ordered group, considered under the order topology.
Let $\set {x_i: i \in I}$ be an indexed set of positive elements of $G$.
{{explain|What actually does "positive" mean in this context?}}
Then:
:the generalized sum $\ds \sum \set {x_i: i \in I}$ converges to a point $x \in G$
{{iff}}
:$x$ is the supremum of:
::$P := \ds \set {\sum_{i \mathop \in F} x_i: \text{$F \subseteq I$ and $F$ is finite} }$

Proof:
{{explain|There are plenty of details to fill in.}}
Theorem: Half Angle Formulas/Sine

{{begin-eqn}}
{{eqn | l = \sin \frac \theta 2
      | r = +\sqrt {\frac {1 - \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {II}$
}}
{{eqn | l = \sin \frac \theta 2
      | r = -\sqrt {\dfrac {1 - \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text {III}$ or quadrant $\text {IV}$
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \cos \theta
      | r = 1 - 2 \ \sin^2 \frac \theta 2
      | c = {{Corollary|Double Angle Formula for Cosine|2}}
}}
{{eqn | ll= \leadsto
      | l = 2 \ \sin^2 \frac \theta 2
      | r = 1 - \cos \theta
}}
{{eqn | ll= \leadsto
      | l = \sin \frac \theta 2
      | r = \pm \sqrt {\frac {1 - \cos \theta} 2}
}}
{{end-eqn}}
We also have that:
:In quadrant $\text I$ and quadrant $\text {II}$, $\sin \theta > 0$
:In quadrant $\text {III}$ and quadrant $\text {IV}$, $\sin \theta < 0$.
{{qed}}
Theorem: Half Angle Formulas/Sine

{{begin-eqn}}
{{eqn | l = \sin \frac \theta 2
      | r = +\sqrt {\frac {1 - \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {II}$
}}
{{eqn | l = \sin \frac \theta 2
      | r = -\sqrt {\dfrac {1 - \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text {III}$ or quadrant $\text {IV}$
}}
{{end-eqn}}

Proof:
Define:
:$u = \dfrac \theta 2$
Then:
{{begin-eqn}}
{{eqn | l = \sin^2 u
      | r = \frac {1 - \cos 2 u} 2
      | c = Power Reduction Formulas
}}
{{eqn | ll= \leadsto
      | l = \sin \frac \theta 2
      | r = \pm \sqrt {\frac {1 - \cos \theta} 2}
}}
{{end-eqn}}
We also have that:
:In quadrant $\text I$, and quadrant $\text {II}$, $\sin \theta > 0$
:In quadrant $\text {III}$, and quadrant $\text {IV}$, $\sin \theta < 0$.
{{qed}}
Theorem: Half Angle Formulas/Cosine

{{begin-eqn}}
{{eqn | l = \cos \frac \theta 2
      | r = +\sqrt {\frac {1 + \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {IV}$
}}
{{eqn | l = \cos \frac \theta 2
      | r = -\sqrt {\frac {1 + \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {III}$
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \cos \theta
      | r = 2 \cos^2 \frac \theta 2 - 1
      | c = {{Corollary|Double Angle Formula for Cosine|1}}
}}
{{eqn | ll= \leadsto
      | l = 2 \cos^2 \frac \theta 2
      | r = 1 + \cos \theta
}}
{{eqn | ll= \leadsto
      | l = \cos \frac \theta 2
      | r = \pm \sqrt {\frac {1 + \cos \theta} 2}
}}
{{end-eqn}}
We also have that:
:In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
:In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.
{{qed}}
Theorem: Half Angle Formulas/Cosine

{{begin-eqn}}
{{eqn | l = \cos \frac \theta 2
      | r = +\sqrt {\frac {1 + \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {IV}$
}}
{{eqn | l = \cos \frac \theta 2
      | r = -\sqrt {\frac {1 + \cos \theta} 2}
      | c = for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {III}$
}}
{{end-eqn}}

Proof:
Define:
:$u = \dfrac \theta 2$
Then:
{{begin-eqn}}
{{eqn | l = \cos^2 u
      | r =  \frac {1 + \cos 2 u} 2
      | c = Power Reduction Formulas
}}
{{eqn | ll= \leadsto
      | l = \cos \frac \theta 2
      | r = \pm \sqrt {\frac {1 + \cos \theta} 2}
}}
{{end-eqn}}
We also have that:
:In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
:In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.
{{qed}}
Theorem: Half Angle Formulas/Tangent

{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = +\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }
      | c = for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {III}$
}}
{{eqn | l = \tan \frac \theta 2
      | r = -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }
      | c = for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {IV}$
}}
{{end-eqn}}
where $\tan$ denotes tangent and $\cos$ denotes cosine.
When $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.

Proof:
{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = \frac {\sin \frac \theta 2} {\cos \frac \theta 2}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\pm \sqrt {\frac {1 - \cos \theta} 2} } {\pm \sqrt {\frac {1 + \cos \theta} 2} }
      | c = Half Angle Formula for Sine and Half Angle Formula for Cosine
}}
{{eqn | r = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }
}}
{{end-eqn}}
Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$.
When $\cos \theta = -1$ it follows that $\cos \theta + 1 = 0$ and then $\tan \dfrac \theta 2$ is undefined.
This happens when $\theta = \paren {2 k + 1} \pi$.
We have that:
:$\tan \dfrac \theta 2 = \dfrac {\sin \dfrac \theta 2} {\cos \dfrac \theta 2}$
Theorem: Kernel of Normal Operator is Kernel of Adjoint

Let $H$ be a Hilbert space.
Let $A \in \map B H$ be a normal operator.
Then:
:$\ker A = \ker A^*$
where:
:$\ker$ denotes kernel
:$A^*$ denotes the adjoint of $A$.

Proof:
Let $x \in H$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \ker A
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = A x
      | r = \mathbf 0_H
      | c = {{Defof|Kernel of Linear Transformation}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \innerprod {A x} {A x}
      | r = 0
      | c = {{Defof|Inner Product}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \innerprod {A^* A x} x
      | r = 0
      | c = {{Defof|Adjoint Linear Transformation}} and Adjoint is Involutive
}}
{{eqn | ll= \leadstoandfrom
      | l = \innerprod {A A^* x} x
      | r = 0
      | c = {{Defof|Normal Operator}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \innerprod {A^*x} {A^* x}
      | r = 0
      | c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | ll= \leadstoandfrom
      | l = A^*x
      | r = \mathbf 0_H
      | c = Inner product is positive
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \in
      | r = \ker A^*
      | c = {{Defof|Kernel of Linear Transformation}}
}}
{{end-eqn}}
Hence, by definition of set equality:
:$\ker A = \ker A^*$
{{qed}}
Theorem: Half Angle Formula for Tangent/Corollary 1

:$\tan \dfrac \theta 2 = \dfrac {\sin \theta} {1 + \cos \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }
      | c = Half Angle Formula for Tangent
}}
{{eqn | r = \pm \sqrt {\frac {\paren {1 - \cos \theta} \paren {1 + \cos \theta} } {\paren {1 + \cos \theta}^2} }
      | c = multiplying top and bottom by $\sqrt {1 + \cos \theta}$
}}
{{eqn | r = \pm \sqrt {\frac {1 - \cos^2 \theta} {\paren {1 + \cos \theta}^2} }
      | c = Difference of Two Squares
}}
{{eqn | r = \pm \sqrt {\frac {\sin^2 \theta} {\paren {1 + \cos \theta}^2} }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \pm \frac {\sin \theta} {1 + \cos \theta}
}}
{{end-eqn}}
Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$.
When $\cos \theta = -1$ it follows that $1 + \cos \theta = 0$.
This happens when $\theta = \paren {2 k + 1} \pi$, for $k \in \Z$.
At these points, $\sin \theta = 0$ as well.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\sin \theta} {1 + \cos \theta}
      | o = =
      | r = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\dfrac \d {\d \theta} \sin \theta} {\map {\dfrac \d {\d \theta} } {1 + \cos \theta} }
      | c = L'Hôpital's Rule
}}
{{eqn | r = \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\cos \theta} {\sin \theta}
      | c = 
}}
{{eqn | r = \frac 1 0
      | c = 
}}
{{end-eqn}}
So it follows that at $\theta = \paren {2 k + 1} \pi$, $\dfrac {\sin \theta} {1 + \cos \theta}$ is undefined.
{{qed|lemma}}
At all other values of $\theta$, $\cos \theta + 1 > 0$.
Therefore the sign of $\dfrac {\sin \theta} {1 + \cos \theta}$ is equal to the sign of $\sin \theta$.
We recall:
:In quadrant $\text I$ and quadrant $\text {II}$: $\sin \theta > 0$
:In quadrant $\text {III}$ and quadrant $\text {IV}$: $\sin \theta < 0$
Thus it follows that the same applies to $\dfrac {\sin \theta} {1 + \cos \theta}$.
Let $\dfrac \theta 2$ be in quadrant I or quadrant $\text {III}$.
Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text I$ or quadrant $\text {II}$.
Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is positive.
Let $\dfrac \theta 2$ be in quadrant $\text {II}$ or quadrant $\text {IV}$.
Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text {III}$ or quadrant $\text {IV}$.
Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is negative.
{{qed}}
Theorem: Half Angle Formula for Tangent/Corollary 2

:$\tan \dfrac \theta 2 = \dfrac {1 - \cos \theta} {\sin \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }
      | c = Half Angle Formula for Tangent
}}
{{eqn | r = \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {\paren {1 + \cos \theta} \paren {1 - \cos \theta} } }
      | c = multiplying both numerator and denominator by $\sqrt {1 - \cos \theta}$
}}
{{eqn | r = \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {1 - \cos^2 \theta} }
      | c = Difference of Two Squares
}}
{{eqn | r = \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {\sin^2 \theta} }
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \pm \frac {1 - \cos \theta} {\sin \theta}
}}
{{end-eqn}}
When $\sin \theta = 0$, the above is undefined.
This happens when $\theta = k \pi$ for $k \in \Z$.
When $\theta = \paren {2 k + 1} \pi$, the value of $1 - \cos \theta$ is $2$.
Thus at $\theta = \paren {2 k + 1} \pi$, the value of $\tan \dfrac \theta 2$ is undefined.
{{qed|lemma}}
When $\theta = 2 k \pi$, the value of $\cos \theta = 1$ and so $1 - \cos \theta$ is $0$.
Then:
{{begin-eqn}}
{{eqn | l = \lim_{x \mathop \to 0^+} \frac {1 - \cos \theta} {\sin \theta}
      | r = \lim_{x \mathop \to 0^+} \frac {\map {\dfrac \d {\d \theta} } {1 - \cos \theta} } {\dfrac \d {\d \theta} \sin \theta}
      | c = L'Hôpital's Rule
}}
{{eqn | r = \lim_{x \mathop \to 0^+} \frac {\sin \theta} {\cos \theta}
      | c = 
}}
{{eqn | r = \frac 0 1
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
Thus $\tan \dfrac \theta 2$ is defined at $\theta = 2 k \pi$, and equals $0$.
{{qed|lemma}}
At all other values of $\theta$, $1 - \cos \theta > 0$.
Therefore the sign of $\dfrac {1 - \cos \theta} {\sin \theta}$ is equal to the sign of $\sin \theta$.
We recall:
:In quadrant $\text I$ and quadrant $\text {II}$: $\sin \theta > 0$
:In quadrant $\text {III}$ and quadrant $\text {IV}$: $\sin \theta < 0$
Thus it follows that the same applies to $\dfrac {1 - \cos \theta} {\sin \theta}$.
Let $\dfrac \theta 2$ be in quadrant $\text I$ or quadrant $\text {III}$.
Then from {{Corollary|Bisection of Angle in Cartesian Plane}}, $\theta$ is in quadrant $\text I$ or quadrant $\text {II}$.
Therefore $\dfrac {1 - \cos \theta} {\sin \theta}$ is positive.
Let $\dfrac \theta 2$ be in quadrant $\text {II}$ or quadrant $\text {IV}$.
Then from {{Corollary|Bisection of Angle in Cartesian Plane}}, $\theta$ is in quadrant $\text {III}$ or quadrant $\text {IV}$.
Therefore $\dfrac {1 - \cos \theta} {\sin \theta}$ is negative.
{{qed}}
Theorem: Half Angle Formula for Tangent/Corollary 2

:$\tan \dfrac \theta 2 = \dfrac {1 - \cos \theta} {\sin \theta}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = \frac {\sin \frac \theta 2} {\cos \frac \theta 2}
      | c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \frac {\sin \frac \theta 2} {\cos \frac \theta 2} \frac {2 \sin \frac \theta 2} {2 \sin \frac \theta 2}
      | c = multiplying both numerator and denominator by $\dfrac {2 \sin \frac \theta 2} {2 \sin \frac \theta 2}$
}}
{{eqn | r = \frac {2 \sin^2 \frac \theta 2} {2 \sin \frac \theta 2 \cos \frac \theta 2}
      | c = 
}}
{{eqn | r = \frac {1 - \cos \theta} {2 \sin \frac \theta 2 \cos \frac \theta 2}
      | c = {{Corollary|Double Angle Formula for Cosine|5}}
}}
{{eqn | r = \frac {1 - \cos \theta} {\sin \theta}
      | c = Double Angle Formula for Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Half Angle Formula for Tangent/Corollary 3

:$\tan \dfrac \theta 2 = \csc \theta - \cot \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \frac \theta 2
      | r = \frac {1 - \cos \theta} {\sin \theta}
      | c = {{Corollary|Half Angle Formula for Tangent|2}}
}}
{{eqn | r = \frac 1 {\sin \theta} - \frac {\cos \theta} {\sin \theta}
      | c = 
}}
{{eqn | r = \csc \theta - \cot \theta
      | c = Cosecant is Reciprocal of Sine and Cotangent is Cosine divided by Sine
}}
{{end-eqn}}
When $\theta = k \pi$, both $\cot \theta$ and $\csc \theta$ are undefined.
{{qed}}
Theorem: Bisection of Angle in Cartesian Plane

Let $\theta$ be the angular coordinate of a point $P$ in a polar coordinate plane.
Let $QOR$ be a straight line that bisects the angle $\theta$.
Then the angular coordinates of $Q$ and $R$ are $\dfrac \theta 2$ and $\pi + \dfrac \theta 2$.

Proof:
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Let $A$ be a point on the polar axis.
By definition of bisection, $\angle AOQ = \dfrac \theta 2$.
This is the angular coordinate of $Q$.
{{qed|lemma}}
Consider the conjugate angle $\map \complement {\angle AOP}$ of $\angle AOP$.
By definition of conjugate angle:
:$\map \complement {\angle AOP} = -2 \pi - \theta$
where the negative sign arises from the fact that it is measured clockwise.
Then the angle $\angle AOR$ is half of $\map \complement {\angle AOP}$:
{{begin-eqn}}
{{eqn | l = \angle AOR
      | r = -\paren {\frac {2 \pi - \theta} 2}
      | c = 
}}
{{eqn | r = -\pi + \frac \theta 2
      | c = 
}}
{{end-eqn}}
The angular coordinate of point $R$ is the conjugate angle $\map \complement {\angle AOR}$ of $\angle AOR$:
{{begin-eqn}}
{{eqn | l = \map \complement {\angle AOR}
      | r = 2 \pi - \paren {-\pi + \frac \theta 2}
      | c = 
}}
{{eqn | r = 2 \pi - \pi + \frac \theta 2
      | c = 
}}
{{eqn | r = \pi + \frac \theta 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Analytic Geometry
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Theorem: Sine of 15 Degrees

:$\sin 15 \degrees = \sin \dfrac \pi {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 15 \degrees
      | r = \sin \frac {30 \degrees} 2
      | c = 
}}
{{eqn | r = \sqrt {\frac {1 - \cos 30 \degrees} 2}
      | c = Half Angle Formula for Sine: $\theta$ is in Quadrant I
}}
{{eqn | r = \sqrt {\frac {1 - \frac {\sqrt 3} 2} 2}
      | c = {{cos|30}}
}}
{{eqn | r = \sqrt {\frac {2 - \sqrt 3} 4}
      | c = 
}}
{{eqn | r = \sqrt {\frac {8 - 4 \sqrt 3} {16} }
      | c = 
}}
{{eqn | r = \sqrt {\frac {6 + 2 - 2 \sqrt 2 \sqrt 6} {16} }
      | c = 
}}
{{eqn | r = \sqrt {\frac {\paren {\sqrt 6 - \sqrt 2}^2} {16} }
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} 4
      | c = positive because $\theta$ is in the first quadrant
}}
{{end-eqn}}
{{qed}}