nyuuzyou/wikis huggingface dataset:
Cosine of Sum/Proof 7

== Theorem ==
== Proof ==
Let two triangles $\triangle ABC$ and $\triangle ABD$ be inscribed in a circle in the same semicircle:.
By Thales' Theorem, these are both right triangles with:
:$ \angle ACB = \angle ADB = 90 \degrees$
Let $AB = 1$.
Join $DC$.
By construction, $\Box ABCD$ is a cyclic quadrilateral.
Let:
:$\angle CAB = \alpha$
:$\angle DBA = \beta$
:$\angle DAC = \gamma$
From the construction above, we have the following:
:$\cos \alpha = AC$
:$\cos \beta = BD$
:$\sin \alpha = CB$
:$\sin \beta = AD$
:$\gamma = \dfrac \pi 2 - \alpha - \beta$
By Length of Chord of Circle:
:$CD = 2 r \map \sin {\gamma}$
Since $2r \mathop = 1$:
:$CD = \map \sin {\gamma}$
Therefore:
By Ptolemy's Theorem:
:$AC \times BD = AD \times BC + AB \times CD$
Substituting:
:$\cos \alpha \times \cos \beta = \sin \beta \times \sin \alpha + 1 \times \map \cos {\alpha + \beta}$
Rearranging:
:$\cos \tuple {\alpha + \beta} = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
By Equivalence of Definitions of Cosine of Angle, the definition of cosine from the circle, from the triangle and as a real function are equivalent.
It follows that all real numbers $x$ and $y$ correspond to values of $\alpha$ and $\beta$ for which the proof above applies, with one exception.
The exception occurs when both $\alpha$ and $\beta$ are equal to $\dfrac {\pi} 2$.
But then the result is simply:
:$ \cos {\pi} = \cos \dfrac {\pi} 2 \cos \dfrac {\pi} 2 - \sin \dfrac {\pi} 2 \sin \dfrac {\pi} 2$
:$ -1 = 0 \cdot 0 - 1 \cdot 1$
The result follows.
Category:Cosine of Sum
Cosine of Supplementary Angle

== Theorem ==
:$\map \cos {\pi - \theta} = -\cos \theta$
where $\cos$ denotes cosine.
That is, the cosine of an angle is the negative of its supplement.
== Proof ==
== Examples ==
== Also see ==
* Sine of Supplementary Angle
* Tangent of Supplementary Angle
* Cotangent of Supplementary Angle
* Secant of Supplementary Angle
* Cosecant of Supplementary Angle
== Sources ==
* : $\S 5$: 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 Supplementary Angle
Category:Cosine Function
Category:Supplementary Angles
Cosine of Supplementary Angle/Examples/Theta minus 180

== Example of Cosine of Supplementary Angle ==
:$\map \cos {\theta - 180 \degrees} = -\cos \theta$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $9.$
Category:Cosine of Supplementary Angle
Cosine of Three Right Angles

== Theorem ==
:$\cos 270 \degrees = \cos \dfrac {3 \pi} 2 = 0$
where $\cos$ denotes cosine.
== Proof ==
== Also see ==
* Sine of Three Right Angles
* Tangent of Three Right Angles
* Cotangent of Three Right Angles
* Secant of Three Right Angles
* Cosecant of Three Right Angles
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of Three Right Angles less Angle

== Theorem ==
:$\map \cos {\dfrac {3 \pi} 2 - \theta} = -\sin \theta$
where $\cos$ and $\sin$ are cosine and sine respectively.
== Proof ==
== Also see ==
* Sine of Three Right Angles less Angle
* Tangent of Three Right Angles less Angle
* Cotangent of Three Right Angles less Angle
* Secant of Three Right Angles less Angle
* Cosecant of Three Right Angles less Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cosine Function
Cosine of Zero is One

== Theorem ==
:$\cos 0 = 1$
where $\cos$ denotes the cosine.
== 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:
:$\cos 0 = 1 - \dfrac {0^2} {2!} + \dfrac {0^4} {4!} - \cdots = 1$
== Also see ==
* Sine of Zero is Zero
* Tangent of Zero
* Cotangent of Zero
* Secant of Zero
* Cosecant of Zero
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* : $\S 16.3 \ (1) \ \text{(i)}$
* : 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 i

== Theorem ==
:$\cos i = \dfrac e 2 + \dfrac 1 {2 e}$
where $\cos$ denotes the complex cosine function and $i$ is the imaginary unit.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine of $i$
* Tangent of $i$
* Cotangent of $i$
* Secant of $i$
* Cosecant of $i$
Category:Cosine of i
Category:Cosine Function
Category:Imaginary Unit
Cosine over Cosine of Complement plus Sine over Sine of Complement

== Theorem ==
:$\dfrac {\cos x} {\map \cos {90 \degrees - x} } + \dfrac {\sin x} {\map \sin {90 \degrees - x} } = \csc x \sec x$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $3.$
Category:Trigonometric Identities
Category:Sine Function
Category:Cosine Function
Cosine plus Cosine/Examples/Cosine 30 plus Cosine 70

== Example of Use of Cosine plus Cosine ==
:$\cos 30 \degrees + \cos 70 \degrees = 2 \cos 50 \degrees \cos 20 \degrees$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $3$.
Category:Cosine plus Cosine
Cosine to Power of Even Integer

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Also defined as ==
This result is also reported in a less elegant form as:
:$\ds \cos^n \theta = \frac 1 {2^{n - 1}} \sum_{k \mathop = 0}^{n / 2} \paren {\binom n k \map \cos {n - 2 k} \theta}$
for all even $n$.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.73$
Category:Cosine Function
Category:Cosine to Power of Even Integer
Cosine to Power of Even Integer/Proof 2

== Theorem ==
Let $n \in \Z$ be an even integer.
Then:
:$\ds \cos^n \theta = \frac 1 {2^{n - 1} } \sum_{k \mathop = 0}^{n / 2} \paren {\binom n k \cos \paren {n - 2 k} \theta}$
That is:
:$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } 2 \cos \paren {n - 4} \theta + \cdots + \dfrac {n!} {\paren {\paren {\frac n 2}!}^2} \cos \theta}$
== Proof ==
Matching up terms from the beginning of this expansion with those from the end:
Thus:
:$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$
Now to determine $R_n$.
The middle term of the sequence $0, 1, \ldots, n$ is $\dfrac n 2$.
So when $k = \dfrac n 2$ we have $n - 2k = 0$ and $n - k = n - \dfrac n 2 = \dfrac n 2$.
Thus:
So the middle term collapses to:
== Also defined as ==
This result is also reported in the form:
:$\ds \cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \cos \paren {2 n - 2 k + 1} \theta$
for all $n \in \Z$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(a)}$
Category:Cosine to Power of Even Integer
Cosine to Power of Odd Integer

== Theorem ==
:$\ds \cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \paren {\cos \paren {2 n + 1} \theta + \binom {2 n + 1} 1 \cos \paren {2 n - 1} \theta + \cdots + \binom {2 n + 1} n \cos \theta}$
That is:
:$\ds \cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \cos \paren {2 n - 2 k + 1} \theta$
== Proof 1 ==
== Proof 2 ==
== Also defined as ==
This result is also reported in a less elegant form as:
:$\ds \cos^n \theta = \frac 1 {2^{n - 1} } \sum_{k \mathop = 0}^{\paren {n - 1} / 2} \paren {\binom n k \cos \paren {n - 2 k} \theta}$
for all odd $n$.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.71$
Category:Cosine Function
Category:Cosine to Power of Odd Integer
Cosine to Power of Odd Integer/Mistake

== Source Work ==
: Chapter $1$: Complex Numbers
:: Supplementary Problems: $130 \ \text{(a)}$
This mistake can be seen in the 1981 printing of the second edition (1974) as published by Schaum: ISBN 0-070-84382-1
== Mistake ==
:Prove that $\cos^n \phi = \dfrac 1 {2^{n - 1} } \set {\cos n \phi + n \cos \paren {n - 2} \phi + \dfrac {n \paren {n - 1} } 2 \cos \paren {n - 4} \phi + \cdots + R_n}$
:where $R_n = \begin{cases} \cos \phi & \textit {if $n$ is odd} \\ \dfrac {n!} {\sqbrk {\paren {n / 2}!}^2} & \textit {if $n$ is even.}\end{cases}$
== Correction ==
As demonstrated in Cosine to Power of Odd Integer, the last term in the odd expansion is not $\cos \phi$, it is $\dfrac {n!} {\paren {\frac {n + 1} 2}! \paren {\frac {n - 1} 2}!} \cos \phi$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(a)}$
Category:Mistakes/Theory and Problems of Complex Variables
Cosine to Power of Odd Integer/Proof 2

== Theorem ==
== Proof ==
Matching up terms from the beginning of this expansion with those from the end:
Thus:
:$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$
Now to determine $R_n$.
The middle two terms of the sequence $0, 1, \ldots, n$ are $\dfrac {n - 1} 2$ and $\dfrac {n + 1} 2$.
Thus, when $k = \dfrac {n - 1} 2$:
Similarly, when $k = \dfrac {n + 1} 2$:
The binomial coefficient in each case is the same, because:
So:
Thus the two middle terms collapse to:
== Also defined as ==
This result is also reported in the form:
:$\ds\cos^{2 n + 1} \theta = \frac 1 {2^{2 n} } \sum_{k \mathop = 0}^n \binom {2 n + 1} k \map \cos {2 n - 2 k + 1} \theta$
for all $n \in \Z$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(a)}$
::(although see Cosine to Power of Odd Integer/Mistake for analysis of an error in that work)
Category:Cosine to Power of Odd Integer
Cotangent Additive Formula

== Theorem == 
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Let $z \in \C$ such that $\map \sin {n z} \ne 0$
Then: 
:$\ds \map \cot {n z} = \frac 1 n \sum_{k \mathop = 0}^{n - 1} \map \cot {z + \frac {k \pi} n}$
where $\cot$ denotes the cotangent function.
== Proof ==
== Also see ==
* Digamma Additive Formula
* General Harmonic Number Additive Formula
* Product Formula for Sine
Category:Cotangent Function
Cotangent Function is Odd

== Theorem ==
Let $x \in \R$ be a real number.
Let $\cot x$ be the cotangent of $x$.
Then, whenever $\cot x$ is defined:
:$\map \cot {-x} = -\cot x$
That is, the cotangent function is odd.
== Proof ==
== Examples ==
== Also see ==
* Sine Function is Odd
* Cosine Function is Even
* Tangent Function is Odd
* Secant Function is Even
* Cosecant Function is Odd
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.33$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cotangent Function is Odd
Category:Cotangent Function
Category:Examples of Odd Functions
Cotangent Function is Odd/Examples/2 x

== Examples of Cotangent Function is Odd ==
:$\map \cot {-2 x} = -\cot 2 x$
== Proof ==
From Cotangent Function is Odd:
:$\map \cot {-\theta} = -\cot \theta$
The result follows by setting $\theta = 2 x$.
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$: Examples
Category:Cotangent Function is Odd
Cotangent Function is Periodic on Reals

== Theorem ==
The real cotangent function is periodic with period $\pi$.
== Proof ==
Also, from Derivative of Cotangent Function:
:$\map {D_x} {\cot x} = -\dfrac 1 {\sin^2 x}$
provided $\sin x \ne 0$.
From Shape of Sine Function, $\sin$ is strictly positive on the interval $\openint 0 \pi$.
From Derivative of Monotone Function, $\cot x$ is strictly decreasing on that interval, and hence cannot have a period of less than $\pi$.
Hence the result.
Category:Cotangent Function
Cotangent in terms of Hyperbolic Cotangent

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$i \cot z = -\coth \paren {i z}$
where:
:$\cot$ denotes the cotangent function
:$\coth$ denotes the hyperbolic cotangent
:$i$ is the imaginary unit: $i^2 = -1$.
== Proof ==
 
== Also see ==
* Sine in terms of Hyperbolic Sine
* Cosine in terms of Hyperbolic Cosine
* Tangent in terms of Hyperbolic Tangent
* Secant in terms of Hyperbolic Secant
* Cosecant in terms of Hyperbolic Cosecant
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.85$: Relationship between Hyperbolic and Trigonometric Functions
Category:Cotangent Function
Category:Hyperbolic Cotangent Function
Cotangent is Cosecant divided by Secant

== Theorem ==
Let $\theta$ be an angle such that $\sin \theta \ne 0$.
Then:
:$\cot \theta = \dfrac {\cosec \theta} {\sec \theta}$
where $\cot$, $\cosec$ and $\sec$ mean cotangent, cosecant and secant respectively.
== Proof ==
== Sources ==
* 
* 
Category:Secant Function
Category:Cosecant Function
Category:Cotangent Function
Cotangent is Cosine divided by Sine

== Theorem ==
Let $\theta$ be an angle such that $\sin \theta \ne 0$.
Then:
:$\cot \theta = \dfrac {\cos \theta} {\sin \theta}$
where $\cot$, $\sin$ and $\cos$ mean cotangent, sine and cosine respectively.
== Proof ==
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
When $\sin \theta = 0$ the expression $\dfrac {\cos \theta} {\sin \theta}$ is not defined.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.16$
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Sine Function
Category:Cosine Function
Category:Cotangent Function
Cotangent is Reciprocal of Tangent

== Theorem ==
Let $\theta$ be an angle such that $\cos \theta \ne 0$ and $\sin \theta \ne 0$.
Then:
:$\cot \theta = \dfrac 1 {\tan \theta}$
where $\tan$ and $\cot$ mean tangent and cotangent respectively.
== Proof ==
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
$\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$.
== Sources ==
* : $\text V$. Trigonometry: Definitions of the ratios
* : $\S 5$: Trigonometric Functions: $5.16$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Cotangent Function
Category:Tangent Function
Cotangent of 105 Degrees

== Theorem ==
:$\cot 105^\circ = \cot \dfrac {7 \pi} {12} = -\left({2 - \sqrt 3}\right)$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 120 Degrees

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

== Theorem ==
:$\cot 135 \degrees = \cot \dfrac {3 \pi} 4 = -1$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 150 Degrees

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

== Theorem ==
:$\cot 15 \degrees = \cot \dfrac {\pi} {12} = 2 + \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 165 Degrees

== Theorem ==
:$\cot 165 \degrees = \cot \dfrac {11 \pi} {12} = -\paren {2 + \sqrt 3}$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 195 Degrees

== Theorem ==
:$\cot 195 \degrees = \cot \dfrac {13 \pi} {12} = 2 + \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 210 Degrees

== Theorem ==
:$\cot 210^\circ = \cot \dfrac {7 \pi} 6 = \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 225 Degrees

== Theorem ==
:$\cot 225^\circ = \cot \dfrac {5 \pi} 4 = 1$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 240 Degrees

== Theorem ==
:$\cot 240^\circ = \cot \dfrac {4 \pi} 3 = \dfrac {\sqrt 3} 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 255 Degrees

== Theorem ==
:$\cot 255 \degrees = \cot \dfrac {17 \pi} {12} = 2 - \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 285 Degrees

== Theorem ==
:$\cot 285 \degrees = \cot \dfrac {19 \pi} {12} = -\paren {2 - \sqrt 3}$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 300 Degrees

== Theorem ==
:$\cot 300 \degrees = \cot \dfrac {5 \pi} 3 = - \dfrac {\sqrt 3} 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 30 Degrees

== Theorem ==
:$\cot 30 \degrees = \cot \dfrac \pi 6 = \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 315 Degrees

== Theorem ==
:$\cot 315 \degrees = \cot \dfrac {7 \pi} 4 = -1$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 330 Degrees

== Theorem ==
:$\cot 330^\circ = \cot \dfrac {11 \pi} 6 = -\sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 345 Degrees

== Theorem ==
:$\cot 345 \degrees = \cot \dfrac {23 \pi} {12} = -\paren {2 + \sqrt 3}$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent of 45 Degrees

== Theorem ==
:$\cot 45 \degrees = \cot \dfrac \pi 4 = 1$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* 
* 
Category:Cotangent Function
Cotangent of 60 Degrees

== Theorem ==
:$\cot 60 \degrees = \cot \dfrac \pi 3 = \dfrac {\sqrt 3} 3$
where $\cot$ denotes cotangent.
== Proof ==
== Also presented as ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
* 
* 
Category:Cotangent of 60 Degrees
Category:Cotangent Function
Cotangent of 60 Degrees/Also presented as

== Cotangent of 60 Degrees: Also presented as == 
Some sources present the cotangent of $60 \degrees$ as:
:$\cot 60 \degrees = \cot \dfrac \pi 3 = \dfrac 1 {\sqrt 3}$
== Sources ==
* 
* 
Category:Cotangent of 60 Degrees
Cotangent of 75 Degrees

== Theorem ==
:$\cot 75 \degrees = \cot \dfrac {5 \pi} {12} = 2 - \sqrt 3$
where $\cot$ denotes cotangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cotangent Function
Cotangent 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:
:$\cot \theta = \dfrac x y$
where $\cot$ denotes the cotangent of $\theta$.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.10$
Category:Cotangent Function
Cotangent of Angle plus Full Angle

== Theorem ==
: $\cot \left({x + 2 \pi}\right) = \cot x$
== Proof ==
== Also see ==
* Sine of Angle plus Full Angle
* Cosine of Angle plus Full Angle
* Tangent 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:Cotangent Function
Cotangent of Angle plus Right Angle

== Theorem ==
:$\map \cot {x + \dfrac \pi 2} = -\tan x$
== Proof ==
== Also see ==
* Sine of Angle plus Right Angle
* Cosine of Angle plus Right Angle
* Tangent 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
Category:Cotangent Function
Category:Reduction Formulae (Trigonometry)
Cotangent of Angle plus Straight Angle

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

== Theorem ==
:$\map \cot {x + \dfrac {3 \pi} 2} = -\tan x$
== Proof ==
== Also see ==
* Sine of Angle plus Three Right Angles
* Cosine of Angle plus Three Right Angles
* Tangent 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:Cotangent Function
Cotangent of Complement equals Tangent

== Theorem ==
Let $\theta \ne \paren {2 n + 1} \dfrac \pi 2$
Then:
:$\map \cot {\dfrac \pi 2 - \theta} = \tan \theta$
where:
:$\cot$ and $\tan$ are cotangent and tangent respectively.
That is, the tangent of an angle is the cotangent of its complement.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
The above is valid only where $\cos \theta \ne 0$, as otherwise $\dfrac {\sin \theta} {\cos \theta}$ is undefined.
From Cosine of Half-Integer Multiple of Pi it follows that this happens when $\theta \ne \paren {2 n + 1} \dfrac \pi 2$.
== Also see ==
* Sine of Complement equals Cosine
* Cosine of Complement equals Sine
* Tangent of Complement equals Cotangent
* Secant of Complement equals Cosecant
* Cosecant of Complement equals Secant
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Cotangent of Complement equals Tangent
Category:Tangent Function
Category:Cotangent Function
Category:Complementary Angles
Cotangent of Complement equals Tangent/Corollary 1

== Corollary to Cotangent of Complement equals Tangent == 
Let $x \ne \paren {4 n + 1} \dfrac \pi 4$
Then: 
:$\ds \map \cot {\dfrac \pi 4 + x } = \map \tan {\dfrac \pi 4 - x}$
where:
:$\cot$ and $\tan$ are cotangent and tangent respectively.
== Proof ==
The above is valid only where $\cos x \ne 0$, as otherwise $\dfrac {\sin x} {\cos x}$ is undefined.
Since $\theta \ne \paren {2 n + 1} \dfrac \pi 2$. and $\theta = \dfrac \pi 4 - x$
Then $x \ne \paren {4 n + 1} \dfrac \pi 4$
Category:Cotangent of Complement equals Tangent