nyuuzyou/wikis huggingface dataset:
Cosine of Half Angle in Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Let $s$ denote the semiperimeter of $\triangle ABC$.
Then:
:$\cos \dfrac C 2 = \sqrt {\dfrac {s \paren {s - c} } {a b} }$
== Proof ==
== Also see ==
* Sine of Half Angle in Triangle
* Tangent of Half Angle in Triangle
== Sources ==
* : $\text V$. Trigonometry: Formulae $(41)$
Category:Cosine of Half Angle in Triangle
Category:Triangles
Category:Cosine Function
Cosine of Half Side 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:
:$\cos \dfrac a 2 = \sqrt {\dfrac {\map \cos {S - B} \, \map \cos {S - C} } {\sin B \sin C} }$
where $S = \dfrac {A + B + C} 2$.
== Proof ==
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.
From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:
:not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
:but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
Let $s' = \dfrac {a' + b' + c'} 2$.
We have:
Then:
and similarly:
:$\map \sin {s' - c'} = \map \cos {S - C}$
The result follows.
== Also see ==
* The other Half Side Formulas for Spherical Triangles:
** Sine of Half Side for Spherical Triangles
** Tangent of Half Side for Spherical Triangles
* Half Angle Formulas for Spherical Triangles:
** Sine of Half Angle for Spherical Triangles
** Cosine of Half Angle for Spherical Triangles
** Tangent of Half Angle for Spherical Triangles
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.100$
* : $\S 12$: Trigonometric Functions: $12.100.$
Category:Half Side Formulas for Spherical Triangles
Cosine of Integer Multiple of Argument

== Theorem ==
For $n \in \Z_{>0}$:
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
=== Formulation 4 ===
=== Formulation 5 ===
=== Formulation 6 ===
=== Formulation 7 ===
=== Formulation 8 ===
=== Formulation 9 ===
== Also see ==
* Sine of Integer Multiple of Argument
Category:Cosine Function
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 1

== Theorem ==
For $n \in \Z_{>0}$:
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
So $\map P 1$ is seen to hold.
$\map P 2$ is the case:
So $\map P 2$ is also seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.
So this is our induction hypothesis:
:$\ds \map \cos {n \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$
from which we are to show:
:$\ds \map \cos {\paren {n + 1} \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta}^{n + 1} + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac {n + 1} k \dbinom {n + 1 - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n + 1 - 2 k} }$
=== Induction Step ===
This is our induction step:
To proceed, we will require the following Lemma:
=== Lemma ===
Dividing through by $\sin \theta$, we obtain:
We now have:
The result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta}^n + \sum_{k \mathop \ge 1} \paren {-1}^k \dfrac n k \dbinom {n - \paren {k + 1} } {k - 1} \paren {2 \cos \theta}^{n - 2 k} }$
== Examples ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.69$
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 1/Examples/Cosine of Quintuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 1 ==
:$\cos 5 \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta }^5 - 5 \paren {2 \cos \theta }^3 + 5 \paren {2 \cos \theta } }$
== Proof ==
Follows directly from the :
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 1/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 1 ==
:$\cos 6 \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta }^6 - 6 \paren {2 \cos \theta }^4 + 9 \paren {2 \cos \theta }^2 - 2 }$
== Proof ==
Follows directly from the :
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 2

== Theorem ==
For $n \in \Z_{>0}$:
== Proof ==
By De Moivre's Formula:
:$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$
As $n \in \Z_{>0}$, we use the Binomial Theorem on the , resulting in:
:$\ds \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$
When $k$ is even, the expression being summed is real.
Equating the real parts of both sides of the equation, replacing $k$ with $2 k$ to make $k$ even, gives:
== Examples ==
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 2/Examples/Cosine of Quintuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 2 ==
:$\map \cos {5 \theta } = \cos^5 \theta \paren {1 - 10 \tan^2 \theta + 5 \tan^4 \theta}$
== Proof ==
Follows directly from the :
Category:Examples of Cosine of Integer Multiple of Argument
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 2/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 2 ==
:$\map \cos {6 \theta } = \cos^6 \theta \paren {1 - 15 \tan^2 \theta + 15 \tan^4 \theta - \tan^6 \theta}$
== Proof ==
Follows directly from the :
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Category:Examples of Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 5

== Theorem ==
For $n \in \Z_{>0}$:
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \cos \theta + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
So $\map P 1$ is seen to hold.
$\map P 2$ is the case:
So $\map P 2$ is also seen to hold.
$\map P 3$ is the case:
So $\map P 3$ is also seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n > 3$ and $n$ even, then it logically follows that $\map P {n + 2}$ is true.
We also need to show that, if $\map P n$ is true, where $n > 3$ and $n$ odd, then it logically follows that $\map P {n + 2}$ is true.
So this is our induction hypothesis:
:$\ds \map \cos {n \theta} = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \paren {\cos \theta} + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
from which we are to show:
:$\ds \map \cos {\paren {n + 2} \theta} = \map \sin {\frac {\paren {n + 3} \pi} 2} + \paren {\sin \frac {\paren {n + 2} \pi} 2} \paren {\cos \theta} + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n + 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n + 2 - k} \theta} }$
=== Induction Step ===
This is our induction step:
For $n$ even or for $n$ odd: (Identical argument)
The result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \cos n \theta = \map \sin {\frac {\paren {n + 1} \pi} 2} + \paren {\sin \frac {n \pi} 2} \cos \theta + 2 \cos \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
== Examples ==
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 5/Examples/Cosine of Quintuple Angle

== Example of Use of ==
:$\cos 5 \theta = \cos \theta + 2 \cos \theta \paren {\cos 4 \theta - \cos 2 \theta}$
== Proof ==
Follows directly from :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 5/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument: Formulation $5$ ==
:$\map \cos {6 \theta } = -1 + 2 \cos \theta \paren {\cos 5 \theta - \cos 3 \theta + \cos \theta }$
== Proof ==
Follows directly from the Cosine of Integer Multiple of Argument: Formulation $5$:
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 6

== Theorem ==
For $n \in \Z$:
:$\map \cos {n \theta} = \paren {-2 \sin \theta} \map \sin {\paren {n - 1} \theta} + \map \cos {\paren {n - 2} \theta}$
== Proof ==
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 7

== Theorem ==
For $n \in \Z_{>0}$:
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds \cos n \theta = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
So $\map P 1$ is seen to hold.
$\map P 2$ is the case:
So $\map P 2$ is also seen to hold.
$\map P 3$ is the case:
So $\map P 3$ is also seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P n$ is true, where $n > 3$ and $n$ even, then it logically follows that $\map P {n + 2}$ is true.
We also need to show that, if $\map P n$ is true, where $n > 3$ and $n$ odd, then it logically follows that $\map P {n + 2}$ is true.
So this is our induction hypothesis:
:$\ds \map \cos {n \theta} = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }$
from which we are to show:
:$\ds \map \cos {\paren {n + 2} \theta} = \sin^2 \frac {\paren {n + 3} \pi} 2 + \paren {\sin^2 \frac {\paren {n + 2} \pi} 2} \paren {\cos \theta} - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n + 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n + 2 - k} \theta} }$
=== Induction Step ===
This is our induction step:
For $n$ even or for $n$ odd: (Identical argument)
The result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \cos n \theta = \sin^2 \frac {\paren {n + 1} \pi} 2 + \paren {\sin^2 \frac {n \pi} 2} \cos \theta - 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \sin {\paren {n - k} \theta} }$
== Examples ==
Category:Cosine of Integer Multiple of Argument
Cosine of Integer Multiple of Argument/Formulation 7/Examples/Cosine of Quintuple Angle

== Example of Use of ==
:$\cos 5 \theta = \cos \theta - 2 \sin \theta \paren {\sin 4 \theta + \sin 2 \theta}$
== Proof ==
Follows directly from the :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 7/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 7 ==
:$\map \cos {6 \theta } = 1 - 2 \sin \theta \paren {\sin 5 \theta + \sin 3 \theta + \sin \theta}$
== Proof ==
Follows directly from the Cosine of Integer Multiple of Argument/Formulation 7:
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 8

== Theorem ==
For $n \in \Z_{>1}$:
:$\cos n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-2} - \cfrac 1 {a_{n-1}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$ and
::A terminal $a_{n-1} = \cos \theta$ term. 
== Proof ==
From Cosine of Integer Multiple of Argument Formulation 4 we have:
Therefore $a_0 = 2 \cos \theta$
Once again, from Cosine of Integer Multiple of Argument Formulation 4 we have:
In the equations above, let $n = n - k$:
Therefore $a_1 = a_2 = \cdots = a_{n-2} = 2 \cos \theta$
Finally, let $k = n - 2$, then:
Therefore $a_{n - 1} = \cos \theta$
Therefore:
For $n \in \Z_{>1}$:
:$\cos n \theta = \map \cos {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n - 2} - \cfrac 1 {a_{n - 1} } } } } } }$
where $a_0 = a_1 = a_2 = \ldots = a_{n - 2} = 2 \cos \theta$ and
::A terminal $a_{n - 1} = \cos \theta$ term. 
== Examples ==
Category:Cosine of Integer Multiple of Argument
Category:Continued Fractions
Cosine of Integer Multiple of Argument/Formulation 8/Examples/Cosine of Quintuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 8 ==
:$\map \cos {5 \theta } = \map \cos {4 \theta} \paren { 2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2\cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {\cos \theta } } }} }$
== Proof ==
Follows directly from the :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 8/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 8 ==
:$\cos 6 \theta = \cos 5 \theta \paren {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {2 \cos \theta - \cfrac 1 {\cos \theta} } } } } }$
== Proof ==
Follows directly from the :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 9

== Theorem ==
For $n \in \Z_{>1}$:
:$\cos n \theta = \map \sin {\paren {n - 1} \theta} \paren {a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n - r} } } } } }$
where:
:$r = \begin {cases} 1 & : \text {$n$ is even} \\ 2 & : \text {$n$ is odd} \end {cases}$
:$a_k = \begin {cases} -2 \sin \theta & : \text {$k$ is even and $k < n - 1$} \\ 2 \sin \theta & : \text {$k$ is odd} \\ \sin \theta & : k = n - 1 \end {cases}$
== Proof ==
By comparing Line 2 to Line 8, we see that:
Therefore, the terminal denominator will be:
:$-2 \sin \theta + \dfrac {\map \cos {\paren {n - 2 k} \theta} } {\map \sin {\paren {n - \paren {2 k - 1} } \theta} }$
Assume $n$ even:
:$n = 2 k$
Assume $n$ odd:
:$n = 2 k + 1$
Therefore:
:$\cos n \theta = \map \cos {\paren {n - 1} \theta} \paren {a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots \cfrac {} {a_{n - r} } } } } }$
where:
:$r = \begin {cases} 1 & : \text {$n$ is even} \\ 2 & : \text {$n$ is odd} \end {cases}$
:$a_k = \begin {cases} -2 \sin \theta & : \text {$k$ is even and $k < n - 1$} \\ 2 \sin \theta & : \text {$k$ is odd} \\ \sin \theta & : k = n - 1 \end {cases}$
== Examples ==
Category:Cosine of Integer Multiple of Argument
Category:Continued Fractions
Cosine of Integer Multiple of Argument/Formulation 9/Examples/Cosine of Quintuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 9 ==
:$\map \cos {5 \theta } = \map \sin {4 \theta} \paren { -2 \sin \theta + \cfrac 1 {2 \sin \theta + \cfrac 1 {-2\sin \theta + \cfrac 1 {2 \sin \theta } }} }$
== Proof ==
Follows directly from the :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Quintuple Angle Formula for Cosine
Category:Quintuple Angle Formulas
Cosine of Integer Multiple of Argument/Formulation 9/Examples/Cosine of Sextuple Angle

== Example of Use of Cosine of Integer Multiple of Argument/Formulation 9 ==
:$\map \cos {6 \theta } = \map \sin {5 \theta} \paren { -2 \sin \theta + \cfrac 1 {2 \sin \theta + \cfrac 1 {-2\sin \theta + \cfrac 1 {2 \sin \theta + \cfrac 1 {-2\sin \theta + \cfrac 1 {\sin \theta } } } }} }$
== Proof ==
Follows directly from the :
Explicit derivation illustrated below:
Category:Examples of Cosine of Integer Multiple of Argument
Category:Sextuple Angle Formula for Cosine
Category:Sextuple Angle Formulas
Cosine of Integer Multiple of Pi

== Theorem ==
Let $x \in \R$ be a real number.
Let $\cos x$ be the cosine of $x$.
Then:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
or:
== Proof ==
Recall the definition of the cosine function:
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
From Cosine of Zero is One, we have that:
:$\cos 0 = 1$
This takes care of the case $n = 0$.
From Sine and Cosine are Periodic on Reals:
: $\cos \paren {x + 2 \pi}= \cos x$
and thus:
:$\forall m \in \Z: \cos \paren {x + 2 m \pi} = \cos x$.
The above observations combine to establish that:
:$\cos 2 m \pi = 1$
Again from Sine and Cosine are Periodic on Reals:
:$\cos \paren {x + \pi} = -\cos x$
Combining this with the above reasoning, it follows that:
:$\forall m \in \Z: \cos \paren {2 m + 1} \pi = -1$
Note that $\forall n \in \Z$:
:If $n$ is even, it is of the form $n = 2m$ for some $m \in \Z$, and so $\cos n \pi = 1$
:If $n$ is odd, it is of the form $n = 2 m + 1$ for some $m \in \Z$, and so $\cos n \pi = -1$
This leads to the desired conclusion:
:$\cos n \pi = \paren {-1}^n$
== Also see ==
* Sine of Integer Multiple of Pi
Category:Cosine Function
Cosine of Right Angle

== Theorem ==
:$\cos 90 \degrees = \cos \dfrac \pi 2 = 0$
where $\cos$ denotes cosine.
== Proof ==
A direct implementation of Cosine of Half-Integer Multiple of Pi:
:$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$
In this case, $n = 0$ and so:
:$\cos \dfrac 1 2 \pi = 0$
== Also see ==
* Sine of Right Angle
* Tangent of Right Angle
* Cotangent of Right Angle
* Secant of Right Angle
* Cosecant of Right Angle
== 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 Straight Angle

== Theorem ==
:$\cos 180 \degrees = \cos \pi = -1$
where $\cos$ denotes cosine.
== Proof ==
A direct implementation of Cosine of Multiple of Pi:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
In this case, $n = 1$ and so:
:$\cos \pi = -1^1 = -1$
== Also see ==
* Sine of Straight Angle
* Tangent of Straight Angle
* Cotangent of Straight Angle
* Secant of Straight Angle
* Cosecant of Straight Angle
== Sources ==
* : $\text V$. Trigonometry: Special angles
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Cosine Function
Cosine of Sum

== Theorem ==
:$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$
where $\sin$ denotes the sine and $\cos$ denotes the cosine.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Proof 6 ==
== Proof 7 ==
== Historical Note ==
== Also see ==
* Sine of Sum
* Tangent of Sum
== Sources ==
* : $\text V$. Trigonometry: Formulae $(11)$
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
* : $\S 5$: Trigonometric Functions: $5.35$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* 
* : Chapter $5$: Eternal Triangles: Ptolemy
* 
* : Appendix $12$: Trigonometric formulae: Addition formulae
* : Appendix $14$: Trigonometric formulae: Addition formulae
Category:Cosine of Sum
Category:Cosine Function
Category:Trigonometric Addition Formulas
Cosine of Sum/Mistake

== Source Work ==
:
:compound angle formulae (in trigonometry)
== Mistake ==
:$\map \cos {a \pm b} = \cos a \cos b \pm \sin a \sin b$
== Correction ==
This should read:
:$\map \cos {a \pm b} = \cos a \cos b \mp \sin a \sin b$
Otherwise it would be interpreted as:
:$\map \cos {a + b} = \cos a \cos b + \sin a \sin b$
:$\map \cos {a - b} = \cos a \cos b - \sin a \sin b$
which is incorrect.
The correct identities are:
:$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$
:$\map \cos {a - b} = \cos a \cos b + \sin a \sin b$
In the $6$th edition, the structure of the dictionary has changed to put these formulae into an appendix where they are specified in full, error-free.
== Sources ==
* 
Category:Mistakes/The Concise Oxford Dictionary of Mathematics
Cosine of Sum/Proof 2

== Theorem ==
== Proof ==
Recall the analytic definitions of sine and cosine:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$
:$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
Let:
Let us differentiate these $a$, keeping $b$ constant.
Then from Derivative of Sine Function and Derivative of Cosine Function, we have:
Hence:
Thus from Derivative of Constant:
:$\forall a \in \R: \map g a^2 + \map h a^2 = c$
In particular, it is true for $a = 0$, and so:
:$\map g 0^2 + \map h 0^2 = 0$
So:
:$\map g a^2 + \map h a^2 = 0$
But from Square of Real Number is Non-Negative:
:$\map g a^2 \ge 0$
and:
:$\map h a^2 \ge 0$
So it follows that:
:$\map g a = 0$
and:
:$\map h a = 0$
Hence the result.
== Sources ==
* : $\S 16.3 \ (2)$
Category:Cosine of Sum
Cosine of Sum/Proof 4

== Theorem ==
== Proof ==
:File:Tri1.PNG
$AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$.
Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.
By Euclid's First Postulate, we can construct line segments $BD$ and $CE$.
By Euclid's second common notion, $\angle DAB = \angle CAE$.
Thus by Triangle Side-Angle-Side Congruence, $\triangle DAB \cong \triangle CAE$.
Therefore, $DB = CE$.
 
We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$:
We use the definition of the distance function on the Euclidean space $\struct {\R^2, d}$ as defined by the Euclidean metric:
:$\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$.
Thus:
:$DB \cong CE \iff \map d {D, B} = \map d {C, E}$
So, plugging in the coordinates of $B, C, D, E$, we get:
Category:Cosine of Sum