nyuuzyou/wikis huggingface dataset:
Coset Product is Well-Defined

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Also see ==
* Coset Product of Normal Subgroup is Consistent with Subset Product Definition‎
* Congruence Modulo Normal Subgroup is Congruence Relation‎
== Sources ==
* : $\S 2.2$: Homomorphisms
* 
* 
Category:Coset Product
Category:Coset Product is Well-Defined
Coset Product is Well-Defined/Proof 1

== Theorem ==
== Proof ==
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is well-defined, we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
So:
By Cosets are Equal iff Product with Inverse in Subgroup:
:$\paren {a \circ b}^{-1} \circ \paren {a' \circ b'} \in N \implies \paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
and the proof is complete.
Category:Coset Product is Well-Defined
Coset Product is Well-Defined/Proof 2

== Theorem ==
== Proof ==
Let $N \lhd G$ where $G$ is a group.
Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a subset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$
This is justified by Coset Product of Normal Subgroup is Consistent with Subset Product Definition.
Since $N$ is normal, each conjugate $b^{-1} \circ N \circ b$ is contained in $N$.
So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.
So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:
That is:
:$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
Let $n \in N$ be arbitrary.
Then:
So:
:$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
and
:$\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
The result follows by definition of set equality.
== Sources ==
* : Chapter $\text{II}$: Groups: Quotient Groups: Theorem $3$
Category:Coset Product is Well-Defined
Coset Product is Well-Defined/Proof 3

== Theorem ==
== Proof ==
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$N \circ a = N \circ a'$
and:
:$N \circ b = N \circ b'$
We need to show that:
:$N \circ \paren {a \circ b} = N \circ \paren {a' \circ b'}$
So:
== Sources ==
* : $\S 6.7$. Quotient groups
Category:Coset Product is Well-Defined
Coset Product is Well-Defined/Proof 4

== Theorem ==
== Proof ==
Let $N \lhd G$ where $G$ is a group.
By Left Congruence Class Modulo Subgroup is Left Coset, it can be shown that the left congruence modulo $N$ is an equivalence relation.
Let $\RR^l_N$ denote the equivalence relation on $G$
:$\RR^l_N := \set {\tuple {x, y} \in G \times G: x^{-1} y \in N}$
Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$, where $\circ_N$ denotes the operation induced on $G / N$ by $\circ$.
By Left Congruence Class Modulo Subgroup is Left Coset again, the equivalence class $\eqclass g {\RR^l_N} \in \struct {G / N, \circ_N}$ of an element $g \in G$ is the left coset $g N$.
Let $a, a', b, b' \in G$, such that:
:$\eqclass a {\RR^l_N} = \eqclass {a'} {\RR^l_N}$
and:
:$\eqclass b {\RR^l_N} = \eqclass {b'} {\RR^l_N}$
It follows that if:
:$a, a', b, b' \in G$ 
then:
:$ a \circ N = a' \circ N$
and:
:$ b \circ N = b' \circ N$.
We need to show that:
:$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.
That is:
:$ \paren { a \circ b } \circ N = \paren {a' \circ b'} \circ N$
We have:
Hence:
:$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.
== Sources ==
* : $\S 50.1$ Quotient groups
Category:Coset Product is Well-Defined
Coset Product is Well-Defined/Proof 5

== Theorem ==
== Proof ==
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
To show that the coset product is well-defined, we need to demonstrate that:
:$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
So:
== Sources ==
* : $\S 1.10$
Category:Coset Product is Well-Defined
Coset Product of Normal Subgroup is Consistent with Subset Product Definition

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is consistent with the definition of the coset product as the subset product of $a \circ N$ and $b \circ N$:
:$\paren {a \circ N} \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
== Proof ==
Consider the set:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
As $e \in N$, have:
:$\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
by of $\circ$.
Hence $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$.
Now let $x \in a \circ N$ and $y \in b \circ N$.
Then by the definition of subset product:
:$\exists n_1 \in N: x = a \circ n_1$
:$\exists n_2 \in N: y = b \circ n_2$
It follows that:
So the definition by subset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
leads to the definition of coset product as:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
Category:Coset Product
Coset Product on Non-Normal Subgroup is not Well-Defined

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ which is not normal.
Let $a, b \in G$.
Then it is not necessarily the case that the coset product:
:$\paren {a \circ H} \circ \paren {b \circ H} = \paren {a \circ b} \circ H$
is well-defined.
== Proof ==
Proof by Counterexample:
Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:
Consider the subgroups of $S_3$:
Let $H = \set {e, \tuple {12} }$.
From Normal Subgroups in Symmetric Group on 3 Letters, $H$ is not normal.
Let $A = \set {\tuple {123}, \tuple {23} }$.
We have:
and so $A$ is the left coset of $H$ by both $\tuple {123}$ and $\tuple {23}$.
Let $B = \set {\tuple {132}, \tuple {13} }$.
and so $B$ is the left coset of $H$ by both $\tuple {132}$ and $\tuple {13}$.
Now consider:
But:
So two different evaluations of the coset product give two different result.
Hence by definition the coset product is not well-defined on $H$.
The result follows.
== Sources ==
* : $\S 6.7$. Quotient groups: Example $125$
Category:Coset Product
Coset Space forms Partition/Examples/Dihedral Group D3/Cosets of Subgroup Generated by b

== Examples of Use of Coset Space forms Partition ==
Consider the dihedral group $D_3$.
From Dihedral Group $D_3$: Cosets of $\gen b$, the left cosets of of the subgroup $\gen b$ generated by $b$ are:
It follows from Coset Space forms Partition that these are consequences of:
== Sources ==
* : Chapter $5$: Cosets and Lagrange's Theorem: Example $5.6$
Category:Coset Space forms Partition
Category:Dihedral Group D3
Coset Space wrt Subring forms Ring iff Subring is Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a subring of $R$.
For $a \in R$, let $\paren {a + S}$ denote the coset of $S$ by $a$.
Then:
:The coset space of $R$ $S$ forms a ring
:
:$S$ is an ideal of $R$.
== Proof ==
== Sources ==
* 
* 
Category:Coset Spaces of Rings
Category:Ideals of Rings
Coset by Identity

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
=== Left Coset by Identity ===
=== Right Coset by Identity ===
== Also see ==
This is consistent with the definition of the concept of coset by means of the subset product:
: $e H = \left\{{e}\right\} H$
: $H e = H \left\{{e}\right\}$
Category:Coset Product
Category:Cosets
Category:Coset by Identity
Coset of Subgroup of Subgroup

== Theorem ==
Let $G$ be a group.
Let $H, K \le G$ be subgroups of $G$.
Let $K \subseteq H$.
Let $x \in G$.
Then either:
:$x K \subseteq H$
or:
:$x K \cap H = \O$
where $x K$ denotes the left coset of $K$ by $x$.
== Proof ==
Suppose $x K \cap H \ne \O$.
Then:
We have that:
:$y \in H$
and:
:$K \subseteq H$
As $H$ is a group, it is closed.
Hence:
:$\forall x \in K: y x \in H$
which means, by definition of subset product:
:$y K \subseteq H$
Hence the result.
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $8$
Category:Cosets
Coset of Trivial Subgroup is Singleton

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $E := \struct {\set e, \circ}$ denote the trivial subgroup of $\struct {G, \circ}$.
Let $g \in G$.
Then the left coset and right coset of $E$ by $g$ is $\set g$.
== Proof ==
Similarly:
== Sources ==
* : Chapter $6$: Cosets: Exercise $2$
Category:Cosets
Category:Trivial Group
Cosets are Equal iff Element in Other Coset

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
==== Left Cosets are Equal iff Element in Other Left Coset ====
==== Right Cosets are Equal iff Element in Other Right Coset ====
Category:Cosets
Cosets are Equivalent

== Theorem ==
All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.
That is, any two left cosets are in one-to-one correspondence.
The same applies to right cosets.
As a special case of this:
:$\forall x \in G: \order {x H} = \order H = \order {H x}$
where $H$ is a subgroup of $G$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Problem $\text{GG}$
* : Chapter $2$: Subgroups and Cosets: $\S 37 \alpha$
* : $\S 41.3$ Multiplication of subsets of a group
* : $\S 3.3$: Group actions and coset decompositions
* : Chapter $5$: Cosets and Lagrange's Theorem
Category:Cosets
Category:Cosets are Equivalent
Cosets are Equivalent/Proof 1

== Theorem ==
== Proof ==
Let us set up mappings $\theta: H \to H x$ and $\phi: H x \to H$ as follows:
:$\forall u \in H: \map \theta u = u x$
:$\forall v \in H x: \map \phi v = v x^{-1}$
From Element in Right Coset iff Product with Inverse in Subgroup:
:$v \in H x \implies v x^{-1} \in H$
Now:
:$\forall v \in H x: \theta \circ \map \phi v = v x^{-1} x = v$
:$\forall u \in H: \phi \circ \map \theta u = u x x^{-1} = u$
Thus $\theta \circ \phi = I_{H x}$ and $\phi \circ \theta = I_H$ are identity mappings.
So $\theta = \phi^{-1}$: both are bijections and one is the inverse of the other.
This establishes, for each $x \in G$, the set equivalence between $H$ and $H x$:
:$H \simeq H x$
In particular, for any $x, y \in G$, it follows from $H x \simeq H$ and $H y \simeq H$ that:
:$H x \simeq H y$
by Set Equivalence behaves like Equivalence Relation.
Similarly, we can set up mappings $\alpha: H \to x H$ and $\beta: x H \to H$ as follows:
:$\forall u \in H: \map \alpha u = x u$
:$\forall v \in x H: \map \beta v = x^{-1} v$
Analogous to above reasoning gives $\alpha = \beta^{-1}$ which establishes $H \simeq x H$. 
Also similarly, $x H \simeq y H$ for all $x, y \in G$.
Hence the result.
== Sources ==
* : $\S 6.4$. Lagrange's theorem
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.8$
Category:Cosets are Equivalent
Cosets in Abelian Group

== Theorem ==
Let $G$ be an abelian group.
Then every right coset modulo $H$ is a left coset modulo $H$.
That is:
:$\forall x \in G: x H = H x$
In an abelian group, therefore, we can talk about congruence modulo $H$ and not worry about whether it is left or right.
== Proof ==
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $113$
* : $\S 1.9$
* : Chapter $\text{II}$: Groups: Subgroups
* : $\S 2.2$: Homomorphisms
* : Chapter $2$: Subgroups and Cosets: $\S 37 \gamma$
* : $\S 42$. Another approach to cosets
Category:Cosets
Category:Abelian Groups
Cosets of Positive Reals in Multiplicative Group of Complex Numbers

== Theorem ==
Let $S$ be the positive real axis of the complex plane:
:$S = \set {z \in \C: z = x + 0 i, x \in \R_{>0} }$
Consider the algebraic structure $\struct {S, \times}$ as a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
The cosets of $\struct {S, \times}$ are the sets of the form:
:$\set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ for some $\theta \in \hointr 0 {2 \pi}$
That is, the sets of all complex numbers with a constant argument.
== Proof ==
Let $z_0 \in \C_{\ne 0}$.
Write $z_0 = r_0 e^{i \theta}$, where $r_0 > 0$ and $\theta \in \hointr 0 {2 \pi}$.
We will show that:
:$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$
Pick any $w \in z_0 S$.
Then there exists some $x \in S$ such that $w = z_0 x$.
Note that $x \in \R_{>0}$ and $r_0 x \in \R_{>0}$.
Thus:
:$w = z_0 x = r_0 x e^{i \theta} \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$
This shows that:
:$z_0 S \subseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$
Pick any $w \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$.
Write $w = r e^{i \theta}$, where $r \in \R_{>0}$.
Note that $\dfrac r {r_0} \in \R_{>0}$.
Then:
:$w = \paren {\dfrac r {r_0}} r_0 e^{i \theta} = \paren {\dfrac r {r_0}} z_0 \in z_0 S$
This shows that:
:$z_0 S \supseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$
By definition of set equality:
:$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{M}$
Category:Real Multiplication
Category:Multiplicative Group of Complex Numbers
Category:Examples of Cosets
Cosine Formula for Dot Product

== Theorem ==
Let $\mathbf v, \mathbf w$ be two non-zero vectors in $\R^n$.
The dot product of $\mathbf v$ and $\mathbf w$ can be calculated by:
:$\mathbf v \cdot \mathbf w = \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
where:
:$\norm {\, \cdot \,}$ denotes vector length and
:$\theta$ is the angle between $\mathbf v$ and $\mathbf w$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Angle Between Vectors in Terms of Dot Product
Category:Cosine Formula for Dot Product
Category:Dot Product
Cosine Formula for Dot Product/Proof 1

== Theorem ==
== Proof ==
There are two cases:
:$(1): \quad \mathbf v$ and $\mathbf w$ are not scalar multiples of each other
:$(2): \quad \mathbf v$ and $\mathbf w$ are scalar multiples of each other.
=== Case 1 ===
:
Let the two vectors $\mathbf v$ and $\mathbf w$ not be scalar multiples of each other. 
Then by the definition of angle between vectors, we have $\theta$ defined as in the triangle as shown above.
(Note that from Angle Between Non-Zero Vectors Always Defined, such a triangle is guaranteed to exist).
By the Law of Cosines:
:$\norm {\mathbf v - \mathbf w}^2 = \norm {\mathbf w}^2 + \norm {\mathbf v}^2 - 2 \norm {\mathbf v} \norm {\mathbf w} \cos \theta$
Now, observe that:
Equating these two expressions for $\norm {\mathbf v - \mathbf w}^2$ gives:
which is exactly the desired result.
=== Case 2 ===
Let $\mathbf v = \paren {v_1, v_2, \ldots, v_n}$ and $\mathbf w = \paren {w_1, w_2, \ldots, w_n}$.
, let $\mathbf v = c \mathbf w$, where $c$ is some scalar. 
If $c > 0$, then by the definition of angle between vectors:
:$\theta = 0 \implies \cos \theta = 1$
If $c < 0$, then by the definition of angle between vectors:
:$\theta = \pi \implies \cos \theta = -1$
(Note that $c$ cannot be $0$ because we have stipulated $\mathbf v$ and $\mathbf w$ to be non-zero).
Then:
== Sources ==
* 
Category:Cosine Formula for Dot Product
Cosine Formula for Dot Product/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf v$ and $\mathbf w$ be considered to be embedded in a Cartesian plane $\CC$.
By Dot Product is Invariant under Coordinate Rotation, we may rotate $\CC$ arbitrarily, and $\mathbf v \cdot \mathbf w$ will not change.
So, let us rotate $\CC$ to $\CC'$ such that the $x$-axis is parallel to $\mathbf v$.
Hence $\mathbf v$ can be expressed as:
Hence: by definition of dot product:
Hence the result.
== Sources ==
* : Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product: $(1.23)$
Category:Cosine Formula for Dot Product
Cosine Function in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\map \cos x = \map F {\dfrac 1 2, -\dfrac 1 2; \dfrac 1 2; \sin^2 x}$
where:
:$x$ is a real number with $\size x < 1$
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Arcsine Function in terms of Gaussian Hypergeometric Function
* Arctangent Function in terms of Gaussian Hypergeometric Function
* Cosecant Function in terms of Gaussian Hypergeometric Function
* Cosine Function in terms of Hypergeometric Function
* Secant Function in terms of Gaussian Hypergeometric Function
== Sources ==
* : $\S 32$: Hypergeometric Functions: Special Cases: $31.6$
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Cosine Function in terms of Hypergeometric Function

== Theorem ==
:$\cos x = \map { {}_0 \operatorname F_1} {\dfrac 1 2; \dfrac {-x^2} 4}$
where:
:$x$ is a real number with $x \in \R$
:$\cos$ denotes the cosine function
:$\map { {}_0 \operatorname F_1} {a; x}$ is the generalized hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac 1 { a^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Sine Function in terms of Hypergeometric Function
* Hyperbolic Sine Function in terms of Hypergeometric Function
* Hyperbolic Cosine Function in terms of Hypergeometric Function
Category:Hypergeometric Functions
Category:Cosine Function
Cosine Function is Absolutely Convergent

== Theorem ==
Let $\cos$ be the cosine function.
Then:
:$\cos x$ is absolutely convergent for all $x \in \R$.
=== Complex Case ===
== Proof ==
Recall the definition of the cosine function:
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
For:
:$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ 
to be absolutely convergent, we want:
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } = \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!}$ 
to be convergent.
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!}$ 
is just the terms of:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n} {n!}$
for even $n$.
Thus:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^{2 n} } {\paren {2 n}!} < \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!}$
But:
:$\ds \sum_{n \mathop = 0}^\infty \frac {\size x^n}{n!} = \exp \size x$ 
from the Taylor Series Expansion for Exponential Function of $\size x$, which converges for all $x \in \R$.
Also, the sequence of partial sums:
:$\ds \sum_{n \mathop = 0}^k \frac {\size x^{2 n} } {\paren {2 n}!}$
is increasing.
The result follows from an application of the Monotone Convergence Theorem (Real Analysis).
== Sources ==
* : $\S 16.2$
Category:Cosine Function
Cosine Function is Absolutely Convergent/Complex Case

== Theorem ==
The complex cosine function $\cos: \C \to \C$ is absolutely convergent.
== Proof 1 ==
== Proof 2 ==
== Sources == 
* : $\S 1.5$
Category:Complex Cosine Function is Absolutely Convergent
Category:Complex Cosine Function
Category:Absolute Convergence
Cosine Function is Absolutely Convergent/Complex Case/Proof 1

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\cos z$ be the cosine of $z$.
Then:
:$\cos z$ is absolutely convergent for all $z \in \C$.
== Proof ==
The definition of the complex cosine function is:
:$\ds \cos z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$
By definition of absolutely convergent complex series, we must show that the power series
:$\ds \sum_{n \mathop = 0}^\infty \size {\paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} }$
is convergent.
We have
The result follows from Squeeze Theorem for Complex Sequences.
== Sources ==
* $\S 1.5$
Category:Cosine Function
Cosine Function is Absolutely Convergent/Complex Case/Proof 2

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\cos z$ be the cosine of $z$.
Then:
:$\cos z$ is absolutely convergent for all $z \in \C$.
== Proof ==
Radius of Convergence of Power Series Expansion for Cosine Function shows that the radius of convergence of the complex cosine function is infinite.
Then Existence of Radius of Convergence of Complex Power Series shows that the complex cosine function is absolutely convergent.
Category:Cosine Function
Cosine Function is Continuous

== Theorem ==
Let $x \in \R$ be a real number.
Let $\cos x$ be the cosine of $x$.
Then:
:$\cos x$ is continuous on $\R$.
== Proof ==
Recall the definition of the cosine function:
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
Thus $\cos x$ is expressible in the form of a power series.
From Cosine Function is Absolutely Convergent, we have that the interval of convergence of $\cos x$ is the whole of $\R$. 
From Power Series is Differentiable on Interval of Convergence, it follows that $\cos x$ is continuous on the whole of $\R$.
 
Category:Cosine Function
Cosine Function is Even

== Theorem ==
For all $z \in \C$:
:$\map \cos {-z} = \cos z$
That is, the cosine function is even.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine Function is Odd
* Tangent Function is Odd
* Cotangent Function is Odd
* Secant Function is Even
* Cosecant Function is Odd
* Hyperbolic Cosine Function is Even
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.29$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* 
* : Appendix $12$: Trigonometric formulae: Symmetry
* : Appendix $14$: Trigonometric formulae: Symmetry
Category:Cosine Function is Even
Category:Cosine Function
Category:Examples of Even Functions
Cosine Function is Even/Proof 1

== Theorem ==
== Proof ==
Recall the definition of the cosine function:
From Even Power is Non-Negative:
:$\forall n \in \N: z^{2 n} = \paren {-z}^{2 n}$
The result follows.
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$: $(4.15)$
* : $\S 16.3 \ (1) \ \text{(iii)}$
Category:Cosine Function is Even
Cosine Inequality

== Theorem ==
:$1 - \dfrac {x^2} 2 \le \cos x$
for all $x \in \R$.
== Proof ==
Let $\map f x = \cos x - \paren {1 - \dfrac {x^2} 2}$.
By Derivative of Cosine Function:
:$\map {f'} x = x - \sin x$
From Sine Inequality, we know $\sin x \le x$ for $x \ge 0$.
Hence $\map {f'} x \ge 0$ for $x \ge 0$.
From Derivative of Monotone Function, $\map f x$ is increasing for $x \ge 0$.
By Cosine of Zero is One, $\map f 0 = 0$.
It follows that $\map f x \ge 0$ for $x \ge 0$.
By Cosine Function is Even, $\map f x$ is an even function.
This implies $\map f x \ge 0$ for all $x \in \R$.
Category:Cosine Function
Cosine in Terms of Haversine

== Theorem ==
:$\cos \theta = 1 - 2 \hav \theta$
where $\cos$ denotes cosine and $\hav$ denotes haversine.
== Proof ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $13$. The haversine formula.
Category:Cosine Function
Category:Haversines