nyuuzyou/wikis huggingface dataset:
Dot Product Associates with Scalar Multiplication

== Theorem ==
Let $\mathbf u, \mathbf v$ be vectors in a vector space $\mathbf V$ of $n$ dimensions.
Let $c$ be a real scalar.
Then:
:$\paren {c \mathbf u} \cdot \mathbf v = c \paren {\mathbf u \cdot \mathbf v}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 2$.
* 
* 
Category:Dot Product Associates with Scalar Multiplication
Category:Dot Product
Dot Product Associates with Scalar Multiplication/Proof 3

== Theorem ==
== Proof ==
From Dot Product Operator is Bilinear:
:$\left({c \mathbf u + \mathbf v}\right) \cdot \mathbf w = c \left({\mathbf u \cdot \mathbf w}\right) + \left({\mathbf v \cdot \mathbf w}\right)$
Setting $\mathbf v = 0$ and renaming $\mathbf w$ yields the result.
Category:Dot Product Associates with Scalar Multiplication
Dot Product Distributes over Addition

== Theorem ==
Let $\mathbf u, \mathbf v, \mathbf w$ be vectors in a vector space $\mathbf V$ of $n$ dimensions.
Then:
:$\paren {\mathbf u + \mathbf v} \cdot \mathbf w = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w$
== Proof 1 ==
== Proof 2 ==
The following proof is valid in the context of a Cartesian $3$-space:
== Proof 3 ==
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
* : $\S 22$: Dot or Scalar Product: $22.9$
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $113 \ \text{(a)}$
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product
* 
* 
Category:Dot Product Distributes over Addition
Category:Dot Product
Category:Vector Addition
Category:Examples of Distributive Operations
Dot Product Distributes over Addition/Proof 2

== Theorem ==
== Proof ==
Let the vectors $\mathbf u$, $\mathbf v$ and $\mathbf w$ be embedded in a Cartesian $3$-space.
It is noted that $\mathbf u$, $\mathbf v$ and $\mathbf w$ are not necessarily coplanar.
:
Let instances of $\mathbf u$ and $\mathbf w$ be selected so their initial points are at some point $O$.
Let an instance of $\mathbf v$ be selected so its initial point is positioned at the terminal point $U$ of $\mathbf u$.
Let the terminal point $\mathbf v$ be $V$.
Let $UA$ be dropped perpendicular to $\mathbf w$.
Let $VB$ be dropped perpendicular to $\mathbf w$.
Let another instance of $\mathbf w$ be selected so that its initial point is at $U$.
Let $VC$ be dropped perpendicular to this second instance of $\mathbf w$.
Let $CB$ be dropped from $C$ to the first instance of $\mathbf w$.
We have that:
:$UA \parallel CB$
and:
:$UC \parallel AB$
Thus $\Box ABCU$ is a parallelogram.
Hence:
:$AB = UC$
Then we have that:
Hence the result.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.7)$
* : Chapter $\text I$: Vector Algebra: $\S 2$.
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product
Category:Dot Product Distributes over Addition
Dot Product Distributes over Addition/Proof 3

== Theorem ==
== Proof ==
From Dot Product Operator is Bilinear:
:$\paren {c \mathbf u + \mathbf v} \cdot \mathbf w = c \paren {\mathbf u \cdot \mathbf w} + \paren {\mathbf v \cdot \mathbf w}$
Setting $c = 1$ yields the result.
Category:Dot Product Distributes over Addition
Dot Product Operator is Bilinear

== Theorem ==
Let $\mathbf u, \mathbf v, \mathbf w$ be vectors in the real Euclidean space $\R^n$.
Let $c$ be a real scalar.
Then:
:$\paren {c \mathbf u + \mathbf v} \cdot \mathbf w = c \paren {\mathbf u \cdot \mathbf w} + \paren {\mathbf v \cdot \mathbf w}$
== Proof ==
Category:Dot Product
Dot Product Operator is Commutative

== Theorem ==
Let $\mathbf u, \mathbf v$ be vectors in a vector space $\mathbf V$ of $n$ dimensions.
Then:
:$\mathbf u \cdot \mathbf v = \mathbf v \cdot \mathbf u$
== Proof 1 ==
== Proof 2 ==
The following proof is valid in the context of a Cartesian $3$-space:
== Examples ==
== Sources ==
* : $\S 22$: Dot or Scalar Product: $22.8$
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $111 \ \text{(a)}$
* 
* 
Category:Dot Product Operator is Commutative
Category:Dot Product
Category:Examples of Commutative Operations
Dot Product Operator is Commutative/Examples

== Examples of Use of Dot Product Operator is Commutative ==
=== Example: $\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$ ===
Category:Examples of Use of Dot Product Operator is Commutative
Dot Product Operator is Commutative/Examples/2+5i dot 3-i

== Example of Use of Dot Product Operator is Commutative ==
==== Example: $\paren {2 + 5 i} \circ \paren {3 - i}$ ====
==== Example: $\paren {3 - i} \circ \paren {2 + 5 i}$ ====
As can be seen:
:$\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$
Category:Examples of Use of Dot Product Operator is Commutative
Dot Product is Inner Product

== Theorem ==
The dot product is an inner product.
== Proof ==
Let $\mathbf u, \mathbf v \in \R^n$.
We will check the four defining properties of an inner product in turn.
=== Conjugate Symmetry ===
Thus the dot product possesses conjugate symmetry.
=== Bilinearity ===
From Dot Product Operator is Bilinear, the dot product possesses bilinearity.
=== Non-Negative Definiteness ===
From Dot Product with Self is Non-Negative, the dot product possesses non-negative definiteness.
=== Positiveness ===
From Dot Product with Self is Zero iff Zero Vector, the dot product possesses positiveness.
Hence the dot product satisfies the definition of an inner product.
== Sources ==
* : 1. Functions on Euclidean Space: Norm and Inner Product
Category:Dot Product
Category:Inner Products
Dot Product of Antiparallel Vectors is Negative

== Theorem ==
Let $\mathbf u$ and $\mathbf v$ be non-zero antiparallel vectors.
Then:
:$\mathbf u \cdot \mathbf v < 0$
where $\cdot$ denotes dot product.
== Proof ==
We have that $\mathbf u$ and $\mathbf v$ are antiparallel.
Hence we can define $\mathbf u$ and $\mathbf v$ as:
where $x$, $y$ and $z$ are arbitrary real numbers such that at least one of $x$, $y$ and $z$ is not zero.
By definition of dot product:
== Sources ==
* 
* 
Category:Antiparallel Vectors
Category:Dot Product
Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero

== Theorem ==
Let:
:$\map {\mathbf f} x = \ds \sum_{k \mathop = 1}^n \map {f_k} x \mathbf e_k$
be a differentiable vector-valued function.
Let $\map {\mathbf f} x$ be such that its magnitude is constant:
:$\size {\map {\mathbf f} x} = c$
for some $c \in \R$.
Then the dot product of $\mathbf f$ with its derivative is zero:
:$\map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x} = 0$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products
* : $\S 22$: Formulas involving Derivatives: $22.27$
Category:Differential Calculus
Category:Vector Calculus
Category:Dot Product
Category:Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero
Dot Product of Elements of Standard Ordered Basis

== Theorem ==
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
:$\mathbf i \cdot \mathbf j = \mathbf j \cdot \mathbf k = \mathbf k \cdot \mathbf i = 0$
where $\cdot$ denotes the dot product.
== Proof ==
By definition, the Cartesian $3$-space is a frame of reference consisting of a rectangular coordinate system.
By definition of rectangular coordinate system, the coordinate axes are perpendicular to each other.
By definition of Component of Vector in $3$-space, the vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ are the unit vectors in the direction of the $x$-axis, $y$-axis and $z$-axis respectively.
Hence:
:$\mathbf i$ is perpendicular to $\mathbf j$
:$\mathbf j$ is perpendicular to $\mathbf k$
:$\mathbf k$ is perpendicular to $\mathbf i$
The result follows from Dot Product of Perpendicular Vectors.
== Also see ==
* Self-Product of Standard Ordered Basis Element equals 1
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.5)$
* : Chapter $\text I$: Vector Algebra: $\S 2$. $(2)$
* : Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product: $(1.22 b)$
Category:Dot Product
Category:Standard Ordered Bases
Dot Product of Like Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a$ and $\mathbf b$ are like.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b}$
where $\norm {\, \cdot \,}$ denotes the magnitude of a vector.
== Proof ==
By definition of dot product:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.
When $\mathbf a$ and $\mathbf b$ are like, by definition $\theta = 0$.
The result follows by Cosine of Zero is One, which gives that $\cos 0 \degrees = 1$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.3)$
Category:Dot Product
Dot Product of Like Vectors/Mistake

== Source Work ==
:
:Chapter $\text {II}$: The Products of Vectors
::$2$. The Scalar Product: $(2.3)$
== Mistake ==
:When two vectors are perpendicular, therefore,
::$\mathbf A \cdot \mathbf B = 0$, $\qquad (2.2)$
:and when they are parallel,
::$\mathbf A \cdot \mathbf B = A B$. $\qquad (2.3)$
== Correction ==
Being parallel is insufficient.
Parallel vector quantities may have opposite directions, that is, such that $\theta = 180 \degrees$, in which case $\mathbf A \cdot \mathbf B = -A B$.
For $\mathbf A \cdot \mathbf B = AB$, it is necessary for $\mathbf A$ and $\mathbf B$ to be like, that is, for their directions to be equal.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.3)$
Category:Mistakes/An Introduction to Vector Analysis
Dot Product of Orthonormal Basis Vectors

== Theorem ==
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be an orthonormal basis of a vector space $V$.
Then:
:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \delta_{i j}$
where:
:$\mathbf e_i \cdot \mathbf e_j$ denotes the dot product of $\mathbf e_i$ and $\mathbf e_j$
:$\delta_{i j}$ denotes the Kronecker delta.
== Proof ==
By definition of orthonormal basis:
:$(1): \quad \tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is an orthogonal basis of $V$
:$(2): \quad \norm {\mathbf e_1} = \norm {\mathbf e_2} = \cdots = \norm {\mathbf e_1} = 1$
From $(1)$ we have by definition that $\mathbf e_i$ and $\mathbf e_j$ are perpendicular whenever $i \ne j$
Thus, from Dot Product of Perpendicular Vectors, for all $i$ and $j$, when $i \ne j$:
:$\mathbf e_i \cdot \mathbf e_j = 0$
From Dot Product of Vector with Itself, for all $i$ we have:
:$\mathbf e_i \cdot \mathbf e_i = \norm {\mathbf e_i}^2$
From $(2)$, we have that:
:$\norm {\mathbf e_i} = 1$
and so:
:$\mathbf e_i \cdot \mathbf e_i = 1$
Putting this together, we have:
:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$
The result follows by definition of the Kronecker delta.
== Sources ==
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product: $(1.1)$
Category:Dot Product
Dot Product of Perpendicular Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities such that $\mathbf a \ne \bszero$ and $\mathbf b \ne \bszero$.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ and $\mathbf b$.
Then:
:$\mathbf a \cdot \mathbf b = 0$
:
:$\mathbf a$ and $\mathbf b$ are perpendicular.
== Proof ==
By definition of dot product:
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.
When $\mathbf a$ and $\mathbf b$ be perpendicular, by definition $\theta = 90 \degrees$.
The result follows by Cosine of Right Angle, which gives that $\cos 90 \degrees = 0$.
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.2)$
* : Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product
Category:Dot Product
Dot Product of Sum with Difference of Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then:
:$\paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a - \mathbf b} = \mathbf a^2 - \mathbf b^2$
where:
:$\cdot$ denotes dot product
:$\mathbf a^2$ denotes the square of $\mathbf a$, that is: $\mathbf a \cdot \mathbf a$.
== Proof ==
== Also see ==
* Square of Sum of Vectors
* Difference of Two Squares
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
Category:Vector Addition
Category:Dot Product
Dot Product of Unit Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be unit vectors.
Then their dot product $\mathbf a \cdot \mathbf b$ is:
:$\mathbf a \cdot \mathbf b = \cos \theta$
where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$.
== Proof ==
We have by definition of dot product :
:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \, \norm {\mathbf b} \cos \theta$
where $\norm {\mathbf a}$ denotes the length of $\mathbf a$.
From Length of Unit Vector is 1:
:$\norm {\mathbf a} = \norm {\mathbf b} = 1$
from which the result follows.
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
Category:Dot Product
Category:Unit Vectors
Dot Product of Vector-Valued Function with its Derivative

== Theorem ==
Let:
:$\map {\mathbf f} x = \ds \sum_{k \mathop = 1}^n \map {f_k} x \mathbf e_k$
be a differentiable vector-valued function.
The dot product of $\mathbf f$ with its derivative is given by:
:$\map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x} = \size {\map {\mathbf f} x} \dfrac {\d \size {\map {\mathbf f} x} } {\d x}$
where $\size {\map {\mathbf f} x} \ne 0$.
== Proof ==
Then:
Hence the result.
== Sources ==
* : $\S 22$: Formulas involving Derivatives: $22.26$
Category:Differential Calculus
Category:Vector Calculus
Category:Dot Product
Dot Product of Vector Cross Products

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ with $\mathbf b$.
Then:
:$\paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d} = \paren {\mathbf a \cdot \mathbf c} \paren {\mathbf b \cdot \mathbf d} - \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.20$
Category:Dot Product of Vector Cross Products
Category:Vector Cross Product
Category:Dot Product
Dot Product of Vector Cross Products/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be embedded in a Cartesian space:
where $\tuple {\mathbf e_1, \mathbf e_2, \mathbf e_3}$ denotes the standard ordered basis of $\mathbf V$.
Then:
Hence the result.
Category:Dot Product of Vector Cross Products
Dot Product of Vector with Itself

== Theorem ==
Let $\mathbf u$ be a vector in the real Euclidean space $\R^n$.
Then:
:$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$
where $\norm {\mathbf u}$ is the length of $\mathbf u$.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This can also be seen presented as:
:$\norm {\mathbf u} = \paren {\mathbf u \cdot \mathbf u}^{1/2}$
or:
:$\norm {\mathbf u} = u = \sqrt {\mathbf u \cdot \mathbf u}$
and so on.
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product: $(2.4)$
* : Volume One: Chapter $1$ Vectors in Classical Physics: $1.3$ The Scalar Product
Category:Dot Product
Category:Dot Product of Vector with Itself
Dot Product of Vector with Itself/Proof 2

== Theorem ==
== Proof ==
=== Note ===
Because this theorem is used to prove the general ($n$-dimensional) case of Cosine Formula for Dot Product, this proof is circular the way we have defined the dot product.
However, some texts use the cosine formula as the definition of the dot product and derive the sum of products form as a consequence.
The two definitions are equivalent, so we have included this proof to show how the statement would be proved from the cosine definition.
Category:Dot Product of Vector with Itself
Dot Product with Self is Non-Negative

== Theorem ==
Let $\mathbf u$ be a vector in the real Euclidean space $\R^n$.
Then:
:$\mathbf u \cdot \mathbf u \ge 0$
where $\cdot$ denotes the dot product operator.
== Proof 1 ==
== Proof 2 ==
Category:Dot Product with Self is Non-Negative
Category:Dot Product
Dot Product with Self is Zero iff Zero Vector

== Theorem ==
Let $\mathbf u$ be a vector in the real Euclidean space $\R^n$.
Then:
:$\mathbf u \cdot \mathbf u = 0 \iff \mathbf u = \mathbf 0$
== Proof 1 ==
== Proof 2 ==
Category:Dot Product with Self is Zero iff Zero Vector
Category:Dot Product
Dot Product with Self is Zero iff Zero Vector/Proof 2

== Theorem ==
== Proof==
Let $\mathbf u \cdot \mathbf u = 0$.
Then:
The only way for this to happen is if:
:$\norm {\mathbf u} = 0$
which implies:
:$\mathbf u = \mathbf 0$
Now suppose $\mathbf u = \mathbf 0$.
Then:
Category:Dot Product with Self is Zero iff Zero Vector
Dot Product with Zero Vector is Zero

== Theorem ==
Let $\mathbf u$ be a vector quantity.
Let $\cdot$ denote the dot product operator.
Then:
:$\mathbf u \cdot \mathbf 0 = 0$
where $\mathbf 0$ denotes the zero vector.
== Proof ==
By definition of dot product:
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $2$. The Scalar Product
Category:Dot Product
Double Angle Formula for Cosine/Corollary 1

== Corollary to Double Angle Formula for Cosine ==
:$\cos 2 \theta = 2 \cos^2 \theta - 1$
where $\cos$ denotes cosine.
== Proof ==
== Also known as ==
This identity and Corollary $2$ are sometimes known as Carnot's Formulas, for .
== Also see ==
* Square of Cosine: $\cos^2 \theta = \dfrac {1 + \cos 2 \theta} 2$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(14)$
* : $\S 5$: Trigonometric Functions: $5.39$
* 
* : Appendix $12$: Trigonometric formulae: Double-angle formulae
* : Appendix $14$: Trigonometric formulae: Double-angle formulae
Category:Double Angle Formula for Cosine
Double Angle Formula for Cosine/Corollary 2

== Corollary to Double Angle Formula for Cosine ==
:$\cos 2 \theta = 1 - 2 \sin^2 \theta$
where $\cos$ and $\sin$ denote cosine and sine respectively.
== Proof ==
== Also known as ==
This identity and Corollary $1$ are sometimes known as Carnot's Formulas, for .
== Also see ==
* Square of Sine: $\sin^2 \theta = \dfrac {1 - \cos 2 \theta} 2$
== Sources ==
* : $\text V$. Trigonometry: Formulae $(15)$
* : $\S 5$: Trigonometric Functions: $5.39$
* : Appendix $12$: Trigonometric formulae: Double-angle formulae
* : Appendix $14$: Trigonometric formulae: Double-angle formulae
Category:Double Angle Formula for Cosine
Double Angle Formula for Cosine/Corollary 4/Proof

== Corollary to Double Angle Formula for Cosine ==
== Proof ==
From :
:$\cos 2 \theta = 2 \cos^2 \theta - 1$
The result follows from substituting $\dfrac \theta 2$ for $\theta$ and rearranging.
== Sources ==
* : $\text V$. Trigonometry: Tangents of sum and difference
Category:Double Angle Formula for Cosine
Double Angle Formula for Cosine/Corollary 5/Proof

== Corollary to Double Angle Formula for Cosine ==
== Proof ==
From :
:$\cos 2 \theta = 1 - 2 \sin^2 \theta$
The result follows from substituting $\dfrac \theta 2$ for $\theta$ and rearranging.
== Sources ==
* : $\text V$. Trigonometry: Tangents of sum and difference
Category:Double Angle Formula for Cosine
Double Angle Formula for Hyperbolic Cosine/Corollary 1

== Corollary to Double Angle Formula for Hyperbolic Cosine ==
:$\cosh 2 x = 2 \cosh^2 x - 1$
where $\cosh$ denotes hyperbolic cosine.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.25$: Double Angle Formulas
Category:Hyperbolic Cosine Function
Double Angle Formula for Hyperbolic Cosine/Corollary 2

== Corollary to Double Angle Formula for Hyperbolic Cosine ==
:$\cosh 2 x = 1 + 2 \sinh^2 x$
where $\cosh$ and $\sinh$ denote hyperbolic cosine and hyperbolic sine respectively.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.25$: Double Angle Formulas
Category:Hyperbolic Cosine Function
Double Angle Formula for Tangent/Corollary

== Theorem ==
Let $u = \tan \dfrac \theta 2$.
Then:
:$\tan \theta = \dfrac {2 u} {1 - u^2}$
where $\tan$ denotes tangent.
== Proof ==
From Double Angle Formula for Tangent:
:$\tan 2 \theta = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$
The result follows by substituting $\dfrac \theta 2$ for $\theta$.
== Sources ==
* 
* 
* 
* : Appendix $12$: Trigonometric formulae: Tangent-of-half-angle formulae
* : Appendix $14$: Trigonometric formulae: Tangent-of-half-angle formulae
Category:Double Angle Formula for Tangent
Double Angle Formulas

== Theorem ==
=== Double Angle Formula for Sine ===
=== Double Angle Formula for Cosine ===
=== Double Angle Formula for Tangent ===
where $\sin, \cos, \tan$ denote sine, cosine and tangent respectively.
== Hyperbolic Functions ==
== Also known as ==
== Also see ==
* Triple Angle Formulas
== Sources ==
* 
* 
* 
* 
Category:Double Angle Formulas
Category:Multiple Angle Formulas
Category:Trigonometric Identities
Double Angle Formulas/Cosine

== Theorem ==
:$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$
where $\cos$ and $\sin$ denote cosine and sine respectively.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
=== Corollary 5 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also known as ==
Corollary $1$ and Corollary $2$ are sometimes known as Carnot's Formulas, for .
== Also see ==
* Double Angle Formula for Sine
== Sources ==
* : $\text V$. Trigonometry: Formulae $(13)$
* : $\S 5$: Trigonometric Functions: $5.39$
* : $\S 16.3 \ (3) \ \text{(iv)}$
* 
* 
* 
* 
* : Appendix $12$: Trigonometric formulae: Double-angle formulae
* : Appendix $14$: Trigonometric formulae: Double-angle formulae
Category:Double Angle Formula for Cosine
Category:Double Angle Formulas
Category:Cosine Function
Double Angle Formulas/Cosine/Proof 4

== Theorem ==
== Proof ==
Consider an isosceles triangle $\triangle ABC$ with base $BC$, and apex $\angle BAC = 2 \alpha$.
Draw an angle bisector to $\angle BAC$ and name it $AH$.
:$\angle BAH = \angle CAH = \alpha$
From Angle Bisector and Altitude Coincide iff Triangle is Isosceles:
:$AH \perp BC$
From Law of Cosines:
From Pythagoras's Theorem:
By definition of sine:
By definition of cosine:
:$AH = AB \cos \alpha = AC \cos \alpha$
So:
Now:
Hence we get the equation:
Category:Double Angle Formula for Cosine
Double Angle Formulas/Hyperbolic Cosine

== Theorem ==
:$\cosh 2 x = \cosh^2 x + \sinh^2 x$
where $\cosh$ and $\sinh$ denote hyperbolic cosine and hyperbolic sine respectively.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.25$: Double Angle Formulas
* 
* 
Category:Hyperbolic Cosine Function
Double Angle Formulas/Hyperbolic Functions

== Theorem ==
=== Double Angle Formula for Hyperbolic Sine ===
=== Double Angle Formula for Hyperbolic Cosine ===
=== Double Angle Formula for Hyperbolic Tangent ===
where $\sinh, \cosh, \tanh$ denote hyperbolic sine, hyperbolic cosine and hyperbolic tangent respectively.
== Also known as ==
Some sources hyphenate: double-angle formulas.
== Sources ==
* 
Category:Hyperbolic Functions
Double Angle Formulas/Hyperbolic Sine

== Theorem ==
:$\sinh 2 x = 2 \sinh x \cosh x$
where $\sinh$ and $\cosh$ denote hyperbolic sine and hyperbolic cosine respectively.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.24$: Double Angle Formulas
* 
* 
Category:Hyperbolic Sine Function
Category:Double Angle Formula for Hyperbolic Sine