nyuuzyou/wikis huggingface dataset:
Cosine in terms of Cotangent

== Theorem ==
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
where $\cos$ denotes the real cosine function and $\cot$ denotes the real cotangent function.
== Proof ==
It remains to determine the sign.
From Sign of Cosine:
From Sign of Cotangent:
This means:
Thus on $(1)$, $\cos x$ and $\cot x$ are the same sign:
:$\cos x > 1$ and $\cot x > 1$
and so:
:$\cos x = +\dfrac {\cot x} {\sqrt {1 + \cot^2 x} }$
On $(2)$, $\cos x$ and $\cot x$ are also the same sign:
:$\cos x < 1$ and $\cot x < 1$
and so:
:$\cos x = +\dfrac {\cot x} {\sqrt {1 + \cot^2 x} }$
On $(3)$, $\cos x$ and $\cot x$ are of opposite sign:
:$\cos x < 1$ and $\cot x > 1$
and so:
:$\cos x = -\dfrac {\cot x} {\sqrt {1 + \cot^2 x} }$
On $(4)$, $\cos x$ and $\cot x$ are also of opposite sign:
:$\cos x > 1$ and $\cot x < 1$
and so:
:$\cos x = -\dfrac {\cot x} {\sqrt {1 + \cot^2 x} }$
When $x = \paren {2 n + \dfrac 1 2} \pi$ and $x = \paren {2 n + \dfrac 3 2} \pi$, both $\cos x = 0$ and $\cot x = 0$
When $x$ is an integer $\sin x = 0$ and so $\cot x$ is undefined.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Cosine Function
Category:Cotangent Function
Cosine in terms of Hyperbolic Cosine

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\cos z = \map \cosh {i z}$
where:
:$\cos$ denotes the complex cosine
:$\cosh$ denotes the hyperbolic cosine
:$i$ is the imaginary unit: $i^2 = -1$.
== Proof ==
 
== Also see ==
* Sine in terms of Hyperbolic Sine
* Tangent in terms of Hyperbolic Tangent
* Cotangent in terms of Hyperbolic Cotangent
* Secant in terms of Hyperbolic Secant
* Cosecant in terms of Hyperbolic Cosecant
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $(4.22)$
* : $\S 8$: Hyperbolic Functions: $8.81$: Relationship between Hyperbolic and Trigonometric Functions
* : $4.$ Mathematics: $4.3$ Trigonometric identities and hyperbolic functions
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $5$
* 
* 
Category:Cosine Function
Category:Hyperbolic Cosine Function
Cosine in terms of Tangent

== Theorem ==
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
where $\sin$ denotes the real sine function and $\tan$ denotes the real tangent function.
== Proof ==
Then from Sign of Cosine:
When $\cos x = 0$, $\tan x$ is undefined.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Cosine Function
Category:Tangent Function
Cosine is Reciprocal of Secant

== Theorem ==
Let $\theta$ be an angle such that $\cos \theta \ne 0$.
Then:
:$\cos \theta = \dfrac 1 {\sec \theta}$
where $\cos$ denotes the cosine function and $\sec$ denotes the secant function.
== Proof ==
$\sec \theta$ and $\dfrac 1 {\cos \theta}$ are not defined when $\cos \theta = 0$.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Cosine Function
Category:Secant Function
Cosine is of Exponential Order Zero

== Theorem ==
Let $\cos t$ be the cosine of $t$, where $t \in \R$.
Then $\cos t$ is of exponential order $0$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Functions of Exponential Order
Category:Exponential Order
Category:Cosine Function
Category:Cosine is of Exponential Order Zero
Cosine minus Cosine/Examples/Cosine 20 minus Cosine 50

== Example of Use of Cosine minus Cosine ==
:$\cos 20 \degrees - \cos 50 \degrees = 2 \sin 50 \degrees \cos 10 \degrees$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $2$.
Category:Cosine minus Cosine
Cosine minus Cosine/Examples/Cosine 2 A minus Cosine 4 A

== Example of Use of Cosine minus Cosine ==
:$\cos 2 A - \cos 4 A = 2 \sin 3 A \sin A$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $6$.
Category:Cosine minus Cosine
Cosine of 105 Degrees

== Theorem ==
:$\cos 105 \degrees = \cos \dfrac {7 \pi} {12} = - \dfrac {\sqrt 6 - \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 120 Degrees

== Theorem ==
:$\cos 120 \degrees = \cos \dfrac {2 \pi} 3 = -\dfrac 1 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 135 Degrees

== Theorem ==
:$\cos 135 \degrees = \cos \dfrac {3 \pi} 4 = -\dfrac {\sqrt 2} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 150 Degrees

== Theorem ==
:$\cos 150 \degrees = \cos \dfrac {5 \pi} 6 = -\dfrac {\sqrt 3} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 15 Degrees

== Theorem ==
:$\cos 15 \degrees = \cos \dfrac \pi {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$
where $\cos$ denotes the cosine.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Category:Cosine of 15 Degrees
Cosine of 165 Degrees

== Theorem ==
:$\cos 165 \degrees = \cos \dfrac {11 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 195 Degrees

== Theorem ==
:$\cos 195 \degrees = \cos \dfrac {13 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 210 Degrees

== Theorem ==
:$\cos 210 \degrees = \cos \dfrac {7 \pi} 6 = -\dfrac {\sqrt 3} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 22.5 Degrees

== Theorem ==
:$\cos 22.5 \degrees = \cos \dfrac \pi 8 = \dfrac 1 2 \sqrt {2 + \sqrt 2}$
where $\cos$ denotes cosine.
== Proof ==
== Also see ==
* Sine of $22.5 \degrees$
* Tangent of $22.5 \degrees$
Category:Cosine Function
Cosine of 225 Degrees

== Theorem ==
:$\cos 225 \degrees = \cos \dfrac {5 \pi} 4 = -\dfrac {\sqrt 2} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 240 Degrees

== Theorem ==
:$\cos 240 \degrees = \cos \dfrac {4 \pi} 3 = -\dfrac 1 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 255 Degrees

== Theorem ==
:$\cos 255^\circ = \cos \dfrac {17 \pi} {12} = - \dfrac {\sqrt 6 - \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 285 Degrees

== Theorem ==
:$\cos 285^\circ = \cos \dfrac {19 \pi} {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 300 Degrees

== Theorem ==
:$\cos 300 \degrees = \cos \dfrac {5 \pi} 3 = \dfrac 1 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 30 Degrees

== Theorem ==
:$\cos 30 \degrees = \cos \dfrac \pi 6 = \dfrac {\sqrt 3} 2$
where $\cos$ denotes the cosine.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* 
* 
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Cosine Function
Cosine of 315 Degrees

== Theorem ==
:$\cos 315 \degrees = \cos \dfrac {7 \pi} 4 = \dfrac {\sqrt 2} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 330 Degrees

== Theorem ==
:$\cos 330 \degrees = \cos \dfrac {11 \pi} 6 = \dfrac {\sqrt 3} 2$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 345 Degrees

== Theorem ==
:$\cos 345 \degrees = \cos \dfrac {23 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$
where $\cos$ denotes cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of 36 Degrees

== Theorem ==
:$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the golden mean.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Cosine Function
Category:Golden Mean
Category:Cosine of 36 Degrees
Cosine of 36 Degrees/Proof 1

== Theorem ==
== Proof ==
Let $u = \cos 72 \degrees, v = \cos 36 \degrees$.
Recall from Cosine of Complement equals Sine that $\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
Therefore:
Therefore:
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $19$
Category:Cosine of 36 Degrees
Cosine of 36 Degrees/Proof 2

== Theorem ==
== Proof ==
From Complex Algebra: $z^4 - 3z^2 + 1 = 0$, the roots of $z^4 - 3z^2 + 1 = 0$ are:
:$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$
Then:
By Shape of Cosine Function:
:$\paren {2 \cos 36^\circ}^2 > \paren {2 \cos 45^\circ}^2 = 2 > \dfrac 3 2 - \dfrac {\sqrt 5} 2$
so $\paren {2 \cos 36^\circ}^2 = \dfrac 3 2 + \dfrac {\sqrt 5} 2$.
Let $z = k \paren {a + \sqrt 5}$, where $k \in \Q$ and $a \in \Z$.
Then for $z = 2 \cos 36^\circ$:
By comparing coefficients:
We have, from the expansion above:
:$4 a k^2 = 1$
which leads to:
:$\sqrt a = \dfrac 1 {2 k}$
so $a$ must be square.
Thus $a = 1$ and hence:
:$2 \cos 36^\circ = k \paren {a + \sqrt 5} = \dfrac {1 + \sqrt 5} 2$
:$\cos 36^\circ = \dfrac {1 + \sqrt 5} 4$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $92 \ \text {(a)}$
Category:Cosine of 36 Degrees
Cosine of 3 Degrees

== Theorem ==
:$\cos 3 \degrees = \cos \dfrac {\pi} {60} = \dfrac {\sqrt {30} - \sqrt {10} - \sqrt 6 + \sqrt 2 + 2 \sqrt {15 + 3 \sqrt 5} + 2 \sqrt {5 + \sqrt 5} } {16}$
where $\cos$ denotes the cosine function.
== Proof ==
Category:Cosine Function
Cosine of 45 Degrees

== Theorem ==
:$\cos 45 \degrees = \cos \dfrac \pi 4 = \dfrac {\sqrt 2} 2$
where $\cos$ denotes the cosine.
== Proof ==
== Also presented as ==
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* 
* 
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Cosine of 45 Degrees
Category:Cosine Function
Cosine of 45 Degrees/Also presented as

== Cosine of 45 Degrees: Also presented as == 
Some sources present the cosine of $45 \degrees$ as:
:$\cos 45 \degrees = \dfrac 1 {\sqrt 2}$
== Sources ==
* : $\text V$. Trigonometry: Special angles
* 
* 
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Cosine of 45 Degrees
Cosine of 60 Degrees

== Theorem ==
:$\cos 60 \degrees = \cos \dfrac \pi 3 = \dfrac 1 2$
where $\cos$ denotes the cosine.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* 
* 
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Cosine Function
Cosine of 72 Degrees

== Theorem ==
:$\cos 72 \degrees = \cos \dfrac {2 \pi} 5 = \dfrac {\sqrt 5 - 1} 4$
where $\cos$ denotes the cosine function.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $92 \ \text {(b)}$
Category:Cosine Function
Category:Cosine of 72 Degrees
Cosine of 75 Degrees

== Theorem ==
:$\cos 75^\circ = \cos \dfrac {5 \pi}{12} = \dfrac {\sqrt 6 - \sqrt 2} 4$
where $\cos$ denotes the cosine.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of Angle in Cartesian Plane

== Theorem ==
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 ==
:
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:
:
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.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.8$
Category:Cosine Function
Cosine of Angle plus Full Angle

== Theorem ==
:$\map \cos {x + 2 \pi} = \cos x$
=== Corollary ===
== Proof ==
== Also see ==
* Sine of Angle plus Full Angle
* Tangent of Angle plus Full Angle
* Cotangent of Angle plus Full Angle
* Secant of Angle plus Full Angle
* Cosecant of Angle plus Full Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cosine Function
Cosine of Angle plus Full Angle/Corollary

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$\map \cos {x + 2 n \pi} = \cos x$
== Proof ==
From Cosine of Angle plus Full Angle:
:$\map \cos {x + 2 \pi} = \cos x$
The result follows from the General Periodicity Property:
If:
:$\forall x \in X: \map f x = \map f {x + L}$
then:
:$\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$
== Sources ==
* : Appendix $12$: Trigonometric formulae: Shifts and periodicity
* : Appendix $14$: Trigonometric formulae: Shifts and periodicity
Category:Cosine Function
Cosine of Angle plus Right Angle

== Theorem ==
:$\map \cos {x + \dfrac \pi 2} = -\sin x$
== Proof ==
== Also see ==
* Sine of Angle plus Right Angle
* Tangent of Angle plus Right Angle
* Cotangent of Angle plus Right Angle
* Secant of Angle plus Right Angle
* Cosecant of Angle plus Right Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : Appendix $12$: Trigonometric formulae: Shifts and periodicity
* : Appendix $14$: Trigonometric formulae: Shifts and periodicity
Category:Cosine Function
Category:Reduction Formulae (Trigonometry)
Cosine of Angle plus Straight Angle

== Theorem ==
:$\map \cos {x + \pi} = -\cos x$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Sine of Angle plus Straight Angle
* Tangent of Angle plus Straight Angle
* Cotangent of Angle plus Straight Angle
* Secant of Angle plus Straight Angle
* Cosecant of Angle plus Straight Angle
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$: Examples
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : Appendix $12$: Trigonometric formulae: Shifts and periodicity
* : Appendix $14$: Trigonometric formulae: Shifts and periodicity
Category:Cosine Function
Category:Cosine of Angle plus Straight Angle
Cosine of Angle plus Straight Angle/Proof 4

== Theorem ==
== Proof ==
From the discussion in the proof of Real Cosine Function is Periodic:
:$\map \sin {x + \eta} = \cos x$
:$\map \cos {x + \eta} = -\sin x$
for $\eta \in \R_{>0}$.
From Sine and Cosine are Periodic on Reals: Pi, we define $\pi \in \R$ as $\pi := 2 \eta$.
It follows that $\eta = \dfrac \pi 2$, thus:
:$\map \cos {x + \pi} = -\map \sin {x + \dfrac \pi 2} = -\cos x$
Category:Cosine of Angle plus Straight Angle
Cosine of Angle plus Three Right Angles

== Theorem ==
:$\map \cos {x + \dfrac {3 \pi} 2} = \sin x$
== Proof ==
== Also see ==
* Sine of Angle plus Three Right Angles
* Tangent of Angle plus Three Right Angles
* Cotangent of Angle plus Three Right Angles
* Secant of Angle plus Three Right Angles
* Cosecant of Angle plus Three Right Angles
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cosine Function
Cosine of Complement equals Sine

== Theorem ==
:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
where $\cos$ and $\sin$ are cosine and sine respectively.
That is, the sine of an angle is the cosine of its complement.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Sine of Complement equals Cosine
* Tangent of Complement equals Cotangent
* Cotangent of Complement equals Tangent
* Secant of Complement equals Cosecant
* Cosecant of Complement equals Secant
== Historical Note ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(4)$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : Appendix $12$: Trigonometric formulae: Symmetry
* : Appendix $14$: Trigonometric formulae: Symmetry
Category:Cosine of Complement equals Sine
Category:Sine Function
Category:Cosine Function
Category:Complementary Angles
Cosine of Complement equals Sine/Proof 4

== Theorem ==
== Proof ==
:
Let $\angle xOP$ and $\angle QOy$ be complementary.
Then:
:$\angle xOP = \angle QOy$
Hence:
:the projection of $OP$ on the $x$-axis
equals:
:the projection of $OQ$ on the $y$-axis.
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: Complementary angles
Category:Cosine of Complement equals Sine
Cosine of Complex Number

== Theorem ==
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\cos \left({a + b i}\right) = \cos a \cosh b - i \sin a \sinh b$
where:
:$\cos$ denotes the cosine function (real and complex)
:$\sin$ denotes the real sine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine of Complex Number
* Tangent of Complex Number
* Cosecant of Complex Number
* Secant of Complex Number
* Cotangent of Complex Number
Category:Cosine Function
Category:Complex Numbers
Category:Cosine of Complex Number
Cosine of Conjugate Angle

== Theorem ==
:$\map \cos {2 \pi - \theta} = \cos \theta$
where $\cos$ denotes cosine.
That is, the cosine of an angle equals its conjugate.
== Proof ==
== Also see ==
* Sine of Conjugate Angle
* Tangent of Conjugate Angle
* Cotangent of Conjugate Angle
* Secant of Conjugate Angle
* Cosecant of Conjugate Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cosine Function
Category:Conjugate Angles
Cosine of Difference

== Theorem ==
:$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$
where $\sin$ denotes the sine and $\cos$ denotes the cosine.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine of Difference
== Historical Note ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(9)$
* : $\S 5$: Trigonometric Functions: $5.35$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* 
* : Appendix $12$: Trigonometric formulae: Addition formulae
* : Appendix $14$: Trigonometric formulae: Addition formulae
Category:Cosine of Difference
Category:Cosine Function
Category:Trigonometric Subtraction Formulas
Cosine of Difference/Proof 2

== Theorem ==
== Proof ==
:
Consider two radii $OP$ and $OQ$ of a unit circle whose center is at the origin of a Cartesian plane.
Let:
Then the coordinates of $P$ and $Q$ are given by:
Hence:
== Sources ==
* : $\text V$. Trigonometry: The addition formulae
Category:Cosine of Difference
Cosine of Full Angle

== Theorem ==
:$\cos 360 \degrees = \cos 2 \pi = 1$
where $\cos$ denotes cosine and $360 \degrees = 2 \pi$ is the full angle.
== Proof ==
A direct implementation of Cosine of Multiple of Pi:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
In this case, $n = 2$ and so:
:$\cos 2 \pi = \paren {-1}^2 = 1$
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of Half-Integer Multiple of Pi

== Theorem ==
Let $x \in \R$ be a real number.
Let $\cos x$ denote the cosine of $x$.
Then:
:$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$
== Proof ==
This is established in Zeroes of Sine and Cosine.
Category:Cosine Function
Cosine of Half Angle for Spherical Triangles

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos \dfrac A 2 = \sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} }$
where $s = \dfrac {a + b + c} 2$.
== Proof ==
The result follows.
== Also see ==
* The other Half Angle Formulas for Spherical Triangles:
** Sine of Half Angle for Spherical Triangles
** Tangent of Half Angle for Spherical Triangles
* Half Side Formulas for Spherical Triangles:
** Sine of Half Side for Spherical Triangles
** Cosine of Half Side for Spherical Triangles
** Tangent of Half Side for Spherical Triangles
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.99$
* : Chapter $\text I$. Spherical Trigonometry: $5$. The cosine-formula.
* : $\S 12$: Trigonometric Functions: $12.99.$
Category:Half Angle Formulas for Spherical Triangles
Cosine of Half Angle for Spherical Triangles/Mistake

== Source Work ==
:Chapter $5$: Trigonometrical Functions:
::Relationships between Sides and Angles of a Spherical Triangle: $5.99$
:
:Chapter $12$: Trigonometrical Functions:
::Relationships between Sides and Angles of a Spherical Triangle: $12.99.$
== Mistake ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos \dfrac A 2 = \sqrt {\dfrac {\sin s \, \map \sin {s - c} } {\sin b \sin c} }$
where $s = \dfrac {a + b + c} 2$.
== Correction ==
The correct formula is:
:$\cos \dfrac A 2 = \sqrt {\dfrac {\sin s \, \map \sin {s - a} } {\sin b \sin c} }$
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.99$
* : $\S 12$: Trigonometric Functions: $12.99.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables