nyuuzyou/wikis huggingface dataset:
Half-Open Rectangles Closed under Intersection

== Theorem ==
Let $\horectr {\mathbf a} {\mathbf b}$ and $\horectr {\mathbf c} {\mathbf d}$ be half-open $n$-rectangles.
Then $\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$ is also a half-open $n$-rectangle.
== Proof ==
As $\O$ is trivially a half-open $n$-rectangle, let us assume that:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} \ne \O$
By Cartesian Product of Intersections: General Case, it follows that:
:$\ds \horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} = \prod_{i \mathop = 1}^n \hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$
which leaves only to prove that the $\hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$ are half-open intervals.
Now let $x \in \hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$.
Then $x$ is subject to:
* $x \ge a_i$ and $x \ge c_i$, i.e., $x \ge \max \set {a_i, c_i}$
* $x < b_i$ and $x < d_i$, i.e., $x < \min \set {b_i, d_i}$
and we see that these conditions are satisfied :
:$x \in \hointr {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
Thus, we conclude:
:$\hointr {a_i} {b_i} \cap \hointr {c_i} {d_i} = \hointr {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
showing that indeed the intersection is a half-open interval.
Combining this with the above reasoning, it follows that indeed:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$
is again a half-open $n$-rectangle.
== Sources ==
* : $\S 5$: Problem $7$
Category:Analysis
Half-Open Rectangles form Semiring of Sets

== Theorem ==
The half-open $n$-rectangles $\JJ_{ho}^n$ form a semiring of sets.
== Proof ==
By definition, $\O$ is considered to be a half-open $n$-rectangle.
That $\JJ_{ho}^n$ is $\cap$-stable follows from Half-Open Rectangles Closed under Intersection.
Thus, it remains to show condition $(3')$ for a semiring of sets:
:$(3'):\quad$ If $A, B \in \JJ_{ho}^n$, then there exists a finite sequence of pairwise disjoint half-open $n$-rectangles $A_1, A_2, \ldots, A_m \in \JJ_{ho}^n$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^m A_k$.
To prove this, proceed by induction on $n$:
=== Basis for the Induction ===
Suppose that $n = 1$, and let $I := \hointr a b$ and $J := \hointr c d$ be half-open intervals.
It is to be demonstrated that $I \setminus J$ is a finite union of pairwise disjoint half-open intervals.
Equivalently, by the above verification of the other axioms, that $\JJ_{ho}^1$ is a semiring of sets.
By swapping the roles of $I$ and $J$ if necessary, it may be arranged that $a \le c$.
Now if $b \le c$ as well, then $I \cap J = \O$.
Subsequently, this implies $I \setminus J = I$, and the statement trivially holds.
Next, consider the other case, that is, $c < b$.
If also $d < b$, then we may write $I$ as the following disjoint union:
:$I = \hointr a b = \hointr a c \cup \hointr c d \cup \hointr d b$
The middle term equals $J$, and we immediately obtain:
:$I \setminus J = \hointr a c \cup \hointr d b$
verifying the statement in this case.
Lastly, suppose that $b \le d$.
Then $I \cap J = \hointr a b \cap \hointr c d = \hointr c b$.
Therefore, by Set Difference with Intersection is Difference:
:$I \setminus J = I \setminus \paren {I \cap J} = \hointr a b \setminus \hointr c b = \hointr a c$
and, having verified this last case, the result follows from Proof by Cases.
This constitutes the induction basis.
=== Induction Hypothesis ===
Now assume the induction hypothesis.
That is, for some fixed $n \ge 1$, assume that $\JJ_{ho}^n$ is a semiring of sets.
Next, it is to be shown that $\JJ_{ho}^{n + 1}$ is also a semiring of sets.
=== Induction Step ===
The induction step goes as follows.
By definition of half-open rectangle, it holds that:
:$\JJ_{ho}^{n + 1} = \JJ_{ho}^n \times \JJ_{ho}^1$
Further, we have that $\JJ_{ho}^n$ and $\JJ_{ho}^1$ are semirings of sets.
Hence by Cartesian Product of Semirings of Sets, $\JJ_{ho}^{n + 1}$ is a semiring of sets, too.
== Sources ==
* : $6.4$
Category:Semirings of Sets
Half-Range Fourier Cosine Series/Cosine of Non-Integer Multiple of x over 0 to Pi

== Theorem ==
Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
:
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \cos \lambda x$
Then its half-range Fourier cosine series can be expressed as:
== Proof ==
By definition of half-range Fourier cosine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x$
Thus by definition of $f$:
Because $\lambda \notin \Z$ we have that $\lambda \ne n$ for all $n$.
Thus for $n > 0$:
Finally:
== Also presented as ==
Some sources present this result as:
== Sources ==
* : Exercises on Chapter $\text I$: $5$.
* : $\S 23$: Miscellanous Fourier Series: $23.20$
Category:Examples of Half-Range Fourier Series
Half-Range Fourier Cosine Series/Identity Function/0 to Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:
:$\map f x = x$
Then its half-range Fourier cosine series can be expressed as:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Exercises on Chapter $\text I$: $6 \, \text{(ii)}$.
Category:Fourier Series for Identity Function
Half-Range Fourier Cosine Series/Identity Function/0 to Pi/Proof 1

== Theorem ==
== Proof ==
By definition of half-range Fourier cosine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos n x$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \map f x \cos n x \rd x$
Thus by definition of $f$:
For $n > 0$:
When $n$ is even, $\paren {-1}^n = 1$.
We can express $n = 2 r$ for $r \ge 1$.
Hence in that case:
When $n$ is odd, $\paren {-1}^n = -1$.
We can express $n = 2 r - 1$ for $r \ge 1$.
Hence in that case:
Finally:
Category:Fourier Series for Identity Function
Half-Range Fourier Cosine Series/Identity Function/0 to Pi/Proof 2

== Theorem ==
== Proof ==
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:
:$\forall x \in \openint 0 \lambda: \map f x = x$
From Half-Range Fourier Cosine Series for Identity Function, the half-range Fourier cosine series for $\map f x$ can be expressed as:
The result follows by setting $\lambda = \pi$.
Category:Fourier Series for Identity Function
Half-Range Fourier Cosine Series over Negative Range

== Theorem ==
Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.
Let $f$ be expressed using the half-range Fourier cosine series over $\openint 0 \lambda$:
:$\ds \map C x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \frac {n \pi x} \lambda$
where:
:$a_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \cos \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
Then over the closed real interval $\openint {-\lambda} 0$, $\map C x$ takes the values:
:$\map C x = \map f {-x}$
That is, the real function expressed by the half-range Fourier cosine series over $\openint 0 \lambda$ is an even function over $\openint {-\lambda} \lambda$.
== Proof ==
From Fourier Series for Even Function over Symmetric Range, $\map C x$ is the Fourier series of an even real function over the interval $\openint 0 \lambda$.
We have that $\map C x \sim \map f x$ over $\openint 0 \lambda$.
Thus over $\openint {-\lambda} 0$ it follows that $\map C x = \map f {-x}$.
== Sources ==
* : Chapter One: $\S 6$. Half-Range Cosine Series
Category:Half-Range Fourier Series
Half-Range Fourier Series/Identity Function

== Theorem ==
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:
:$\forall x \in \openint 0 \lambda: \map f x = x$
The half-range Fourier series of $f$ over $\openint 0 \lambda$ can be given in the following forms:
=== Cosine Series for Identity Function ===
=== Sine Series for Identity Function ===
Category:Examples of Half-Range Fourier Series
Category:Identity Mappings
Category:Fourier Series for Identity Function
Category:Half-Range Fourier Series for Identity Function
Half-Range Fourier Series/Identity Function/Cosine

== Theorem ==
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:
:$\forall x \in \openint 0 \lambda: \map f x = x$
The half-range Fourier cosine series for $\map f x$ can be expressed as:
== Proof ==
By definition of half-range Fourier cosine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos \dfrac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \cos \dfrac {n \pi x} \lambda \rd x$
Thus by definition of $f$:
For $n > 0$:
When $n$ is even, $\paren {-1}^n = 1$.
We can express $n = 2 r$ for $r \ge 1$.
Hence in that case:
When $n$ is odd, $\paren {-1}^n = -1$.
We can express $n = 2 r + 1$ for $r \ge 0$.
Hence in that case:
Finally:
Category:Half-Range Fourier Series for Identity Function
Half-Range Fourier Series/Identity Function/Sine

== Theorem ==
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:
:$\forall x \in \openint 0 \lambda: \map f x = x$
The half-range Fourier sine series for $\map f x$ can be expressed as:
== Proof ==
By definition of half-range Fourier sine series:
:$(1): \quad \map f x \sim \ds \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
Substituting for $b_n$ in $(1)$: 
:$\ds \map f x = \dfrac {2 \lambda} \pi \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \sin \frac {n \pi x} \lambda$
as required.
Category:Half-Range Fourier Series for Identity Function
Half-Range Fourier Sine Series/Cosine over 0 to Pi

== Theorem ==
:
On the interval $\openint 0 \pi$:
== Proof ==
Let $\map f x$ be the function defined as:
:$\forall x \in \openint 0 \pi: \map f x = \cos x$
Let $f$ be expressed by a half-range Fourier sine series:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \cos x \sin \frac {n \pi x} \lambda \rd x$
In this context, $\lambda = \pi$ and so this can be expressed more simply as:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \pi \int_0^\pi \cos x \sin n x \rd x$
First the case when $n = 1$:
When $n \ne 1$:
Thus for $n = 2 m$ for $m \in \Z$:
and for $n = 2 m + 1$ for $m \in \Z$:
Thus we have:
and so over the given interval:
== Sources ==
* : Chapter One: $\S 5$. Half-Range Sine Series: Example $4$
* : $\S 23$: Special Fourier Series and their Graphs: $23.13$
Category:Examples of Half-Range Fourier Series‎
Half-Range Fourier Sine Series/Sine of Half x over 0 to Pi, Minus Sine of Half x over Pi to 2 Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:
:$\map f x = \begin {cases}
\sin \dfrac x 2 & : 0 \le x < \pi \\
-\sin \dfrac x 2 & : \pi < x \le 2 \pi \end {cases}$
Then its Fourier series can be expressed as:
== Proof ==
By definition of half-range Fourier sine series:
:$\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin \dfrac {n x} 2$
where:
for all $n \in \Z_{>0}$.
Thus:
When $\dfrac {n x} 2 \ne \dfrac x 2$, that is, when $n \ne 1$, we have:
and so for $n \ne 1$:
When $\dfrac {n x} 2 = \dfrac x 2$, that is, when $n = 1$, we have:
and so for $n = 1$:
Hence:
When $n$ is odd, we have $n = 2 r + 1$ for $r \ge 1$, and so:
When $n$ is even, we have $n = 2 r$ for $r \ge 1$, and so:
== Sources ==
* : Exercises on Chapter $\text I$: $2$.
Category:Examples of Half-Range Fourier Series
Half-Range Fourier Sine Series/Sine of Non-Integer Multiple of x over 0 to Pi

== Theorem ==
Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = \sin \lambda x$
Then its half-range Fourier sine series can be expressed as:
== Proof ==
By definition of half-range Fourier sine series:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \sin n x$
where for all $n \in \Z_{> 0}$:
:$a_n = \ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$
Because $\lambda \notin \Z$ we have that $\lambda \ne n$ for all $n$.
Thus for $n > 0$:
Finally:
== Also presented as ==
Some sources present this result as:
== Sources ==
* : $\S 23$: Miscellanous Fourier Series: $23.19$
Category:Examples of Half-Range Fourier Series
Half-Range Fourier Sine Series/x by Pi minus x over 0 to Pi

== Theorem ==
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:$\map f x = x \paren {\pi - x}$
Then its half-range Fourier sine series can be expressed as:
:$\ds \map f x \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 r + 1} x} {\paren {2 r + 1}^3}$
== Proof ==
By definition of half-range Fourier sine series:
:$\ds \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$
where for all $n \in \Z_{> 0}$:
:$b_n = \ds \frac 2 \pi \int_0^\pi \map f x \sin n x \rd x$
Thus by definition of $f$:
Splitting it up into two:
and:
Thus:
When $n$ is even, $\paren {-1}^n = 1$ and so $\dfrac {4 \paren {1 - \paren {-1}^n} } {\pi n^3} = 0$.
When $n$ is odd, $n$ can be expressed as $n = 2 r + 1$ for $r \ge 0$.
Thus:
Hence:
:$\ds \map f x \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \paren {2 r + 1} x} {\paren {2 r + 1}^3}$
== Sources ==
* : $\S 23$: Special Fourier Series and their Graphs: $23.18$
Category:Examples of Half-Range Fourier Series
Half-Range Fourier Sine Series over Negative Range

== Theorem ==
Let $\map f x$ be a real function defined on the open real interval $\openint 0 \lambda$.
Let $f$ be expressed using the half-range Fourier sine series over $\openint 0 \lambda$:
:$\ds \map S x \sim \sum_{n \mathop = 1}^\infty b_n \sin \frac {n \pi x} \lambda$
where:
:$b_n = \ds \frac 2 \lambda \int_0^\lambda \map f x \sin \frac {n \pi x} \lambda \rd x$
for all $n \in \Z_{\ge 0}$.
Then over the interval $\openint {-\lambda} 0$, $\map S x$ takes the values:
:$\map S x = -\map f {-x}$
That is, the real function expressed by the half-range Fourier sine series over $\openint 0 \lambda$ is an odd function over $\openint {-\lambda} \lambda$.
== Proof ==
From Fourier Series for Odd Function over Symmetric Range, $\map S x$ is the Fourier series of an odd real function over the interval $\openint 0 \lambda$.
We have that $\map S x \sim \map f x$ over $\openint 0 \lambda$.
Thus over $\openint {-\lambda} 0$ it follows that:
:$\map S x = -\map f {-x}$
== Sources ==
* : Chapter One: $\S 5$. Half-Range Sine Series
Category:Half-Range Fourier Series
Half Angle Formula for Hyperbolic Tangent/Corollary 1

== Theorem ==
Let $x \in \R$.
Then:
:$\tanh \dfrac x 2 = \dfrac {\sinh x} {\cosh x + 1}$
where $\tanh$ denotes hyperbolic tangent, $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
== Proof ==
Since $\cosh x > 0$, it follows that $\cosh x + 1 > 0$.
We also have that:
:when $x \ge 0$, $\tanh \dfrac x 2 \ge 0$ and $\sinh x \ge 0$
:when $x \le 0$, $\tanh \dfrac x 2 \le 0$ and $\sinh x \le 0$.
Thus:
:$\tanh \dfrac x 2 = \dfrac {\sinh x} {\cosh x + 1}$
== Also see ==
* Half Angle Formula for Hyperbolic Sine
* Half Angle Formula for Hyperbolic Cosine
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.29$: Double Angle Formulas
Category:Hyperbolic Tangent Function
Half Angle Formula for Hyperbolic Tangent/Corollary 2

== Theorem ==
For $x \ne 0$:
:$\tanh \dfrac x 2 = \dfrac {\cosh x - 1} {\sinh x}$
where $\tanh$ denotes hyperbolic tangent, $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
== Proof ==
Since $\cosh x \ge 1$, it follows that $\cosh x - 1 \ge 0$, with equality happening at $x = 0$.
We also have that:
:when $x > 0$, $\tanh \dfrac x 2 > 0$ and $\sinh x > 0$
:when $x < 0$, $\tanh \dfrac x 2 < 0$ and $\sinh x < 0$.
Thus:
:$\tanh \dfrac x 2 = \dfrac {\cosh x - 1} {\sinh x}$
== Also see ==
* Half Angle Formula for Hyperbolic Sine
* Half Angle Formula for Hyperbolic Cosine
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.29$: Double Angle Formulas
Category:Hyperbolic Tangent Function
Half Angle Formula for Tangent/Corollary 1

== Theorem ==
:$\tan \dfrac \theta 2 = \dfrac {\sin \theta} {1 + \cos \theta}$
where $\tan$ denotes tangent, $\sin$ denotes sine and $\cos$ denotes cosine.
Where $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.
== Proof ==
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:
So it follows that at $\theta = \paren {2 k + 1} \pi$, $\dfrac {\sin \theta} {1 + \cos \theta}$ is undefined.
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.
== Also see ==
* Half Angle Formula for Sine
* Half Angle Formula for Cosine
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.43$
Category:Half Angle Formula for Tangent
Half Angle Formula for Tangent/Corollary 2

== Theorem ==
:$\tan \dfrac \theta 2 = \dfrac {1 - \cos \theta} {\sin \theta}$
where $\tan$ denotes tangent, $\sin$ denotes sine and $\cos$ denotes cosine.
When $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Half Angle Formula for Sine
* Half Angle Formula for Cosine
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.43$
Category:Half Angle Formula for Tangent
Half Angle Formula for Tangent/Corollary 2/Proof 1

== Theorem ==
== Proof ==
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.
When $\theta = 2 k \pi$, the value of $\cos \theta = 1$ and so $1 - \cos \theta$ is $0$.
Then:
Thus $\tan \dfrac \theta 2$ is defined at $\theta = 2 k \pi$, and equals $0$.
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 Bisection of Angle in Cartesian Plane: Corollary, $\theta$ 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 Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text {III}$ or quadrant $\text {IV}$.
Therefore $\dfrac {1 - \cos \theta} {\sin \theta}$ is negative.
Category:Half Angle Formula for Tangent
Half Angle Formula for Tangent/Corollary 3

== Theorem ==
:$\tan \dfrac \theta 2 = \csc \theta - \cot \theta$
where $\tan$ denotes tangent, $\csc$ denotes cosecant and $\cot$ denotes cotangent.
When $\theta = k \pi$, the of this formula is undefined.
== Proof ==
When $\theta = k \pi$, both $\cot \theta$ and $\csc \theta$ are undefined.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.43$
Category:Half Angle Formula for Tangent
Half Angle Formulas

== Theorem ==
=== Half Angle Formula for Sine ===
=== Half Angle Formula for Cosine ===
=== Half Angle Formula for Tangent ===
=== Half Angle Formula for Tangent: Corollary 1 ===
=== Half Angle Formula for Tangent: Corollary 2 ===
=== Half Angle Formula for Tangent: Corollary 3 ===
=== One Plus Tangent Half Angle over One Minus Tangent Half Angle ===
Let $x \in \R$.
Then:
=== Half Angle Formula for Hyperbolic Sine ===
=== Half Angle Formula for Hyperbolic Cosine ===
=== Half Angle Formula for Hyperbolic Tangent ===
=== Half Angle Formula for Hyperbolic Tangent: Corollary 1 ===
=== Half Angle Formula for Hyperbolic Tangent: Corollary 2 ===
== Also known as ==
Some sources hyphenate: half-angle formulas.
The British English plural is formulae.
== Sources ==
* 
* 
Category:Trigonometric Identities
Category:Hyperbolic Functions
Category:Half Angle Formulas
Half Angle Formulas/Cosine

== Theorem ==
where $\cos$ denotes cosine.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Half Angle Formula for Sine
* Half Angle Formula for Tangent
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.42$
* 
* 
Category:Cosine Function
Category:Half Angle Formula for Cosine
Half Angle Formulas/Cosine/Proof 1

== Theorem ==
== Proof ==
We also have that:
:In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
:In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.
Category:Half Angle Formula for Cosine
Half Angle Formulas/Cosine/Proof 2

== Theorem ==
== Proof ==
Define:
:$u = \dfrac \theta 2$
Then:
We also have that:
:In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
:In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.
Category:Half Angle Formula for Cosine
Half Angle Formulas/Hyperbolic Cosine

== Theorem ==
Let $x \in \R$.
Then:
:$\cosh \dfrac x 2 = +\sqrt {\dfrac {\cosh x + 1} 2}$
where $\cosh$ denotes hyperbolic cosine.
== Proof ==
As $\forall x \in \R: \cosh x > 0$, the result follows.
== Also see ==
* Half Angle Formula for Hyperbolic Sine
* Half Angle Formula for Hyperbolic Tangent
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.28$: Double Angle Formulas
Category:Hyperbolic Cosine Function
Half Angle Formulas/Hyperbolic Sine

== Theorem ==
Let $x \in \R$.
Then:
where $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
== Proof ==
We also have that:
:when $x \ge 0$, $\sinh x \ge 0$
:when $x \le 0$, $\sinh x \le 0$.
== Also see ==
* Half Angle Formula for Hyperbolic Cosine
* Half Angle Formula for Hyperbolic Tangent
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.27$: Double Angle Formulas
Category:Hyperbolic Sine Function
Half Angle Formulas/Hyperbolic Tangent

== Theorem ==
Let $x \in \R$.
Then:
where $\tanh$ denotes hyperbolic tangent and $\cosh$ denotes hyperbolic cosine.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
We also have that:
:when $x \ge 0$, $\tanh \dfrac x 2 \ge 0$
:when $x \le 0$, $\tanh \dfrac x 2 \le 0$.
== Also see ==
* Half Angle Formula for Hyperbolic Sine
* Half Angle Formula for Hyperbolic Cosine
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.29$: Double Angle Formulas
Category:Hyperbolic Tangent Function
Half Angle Formulas/Sine

== Theorem ==
where $\sin$ denotes sine and $\cos$ denotes cosine.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Half Angle Formula for Cosine
* Half Angle Formula for Tangent
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.41$
* 
* : Chapter $5$: Eternal Triangles: Ptolemy
* 
Category:Sine Function
Category:Half Angle Formula for Sine
Half Angle Formulas/Sine/Proof 1

== Theorem ==
== Proof ==
We also have that:
:In quadrant $\text I$ and quadrant $\text {II}$, $\sin \theta > 0$
:In quadrant $\text {III}$ and quadrant $\text {IV}$, $\sin \theta < 0$.
Category:Half Angle Formula for Sine