nyuuzyou/wikis huggingface dataset:
Sine 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:
:$\ds \sin n \theta = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
=== 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:
:$\ds \map \sin {n \theta} = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
from which we are to show:
:$\ds \map \sin {\paren {n + 2} \theta} = \paren {\sin^2 \frac {\paren {n + 2} \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n + 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n + 2 - k} \theta} }$
=== 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:
:$\ds \forall n \in \Z_{>0}: \sin n \theta = \paren {\sin^2 \frac {n \pi} 2} \paren {\sin \theta} + 2 \sin \theta \paren {\sum_{k \mathop = 0}^{n - 1} \paren {\sin^2 \frac {k \pi} 2} \map \cos {\paren {n - k} \theta} }$
== Examples ==
Category:Sine of Integer Multiple of Argument
Sine of Integer Multiple of Argument/Formulation 7/Examples/Sine of Quintuple Angle

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

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

== Theorem ==
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
== Proof ==
From Sine of Integer Multiple of Argument Formulation 4 we have:
Therefore $a_0 = 2 \cos \theta$
Once again, from Sine of Integer Multiple of Argument Formulation 4 we have:
In the equations above, let $n = n - k$:
Therefore $a_1 = a_2 = \cdots = a_{n-3} = 2 \cos \theta$
Finally, let $k = n - 3$, then:
Therefore $a_{n - 2} = 2 \cos \theta$
Therefore:
For $n \in \Z_{>1}$:
:$\sin n \theta = \map \sin {\paren {n - 1 } \theta} \paren { a_0 - \cfrac 1 {a_1 - \cfrac 1 {a_2 - \cfrac 1 {\ddots \cfrac {} {a_{n-3} - \cfrac 1 {a_{n-2}}} }}} }$
where $a_0 = a_1 = a_2 = \ldots = a_{n-2} = 2 \cos \theta$.
== Examples ==
Category:Sine of Integer Multiple of Argument
Category:Continued Fractions
Sine of Integer Multiple of Argument/Formulation 8/Examples/Sine of Quintuple Angle

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

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

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

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

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

== Theorem ==
Let $x \in \R$ be a real number.
Let $\sin x$ be the sine of $x$.
Then:
:$\forall n \in \Z: \sin n \pi = 0$
== Proof ==
This is established in Zeroes of Sine and Cosine.
== Also see ==
* Cosine of Integer Multiple of Pi
Category:Sine Function
Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean

== Theorem ==
Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the golden mean
:$F_n$ denotes the $n$th Fibonacci number.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $33$
Category:Fibonacci Numbers
Category:Golden Mean
Category:Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean
Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean/Proof 1

== Theorem ==
== Proof ==
Setting $n = 1$:
:$\sin z = i^0 \frac {\phi^1 - \hat \phi^1} 2 = \frac {\phi - \hat \phi} 2$
Thus:
Category:Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean
Sine of Right Angle

== Theorem ==
:$\sin 90 \degrees = \sin \dfrac \pi 2 = 1$
where $\sin$ denotes the sine function.
== Proof ==
A direct implementation of Sine of Half-Integer Multiple of Pi:
:$\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$
In this case, $n = 0$ and so:
:$\sin \dfrac 1 2 \pi = \paren {-1}^0 = 1$
== Also see ==
* Cosine 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:Sine Function
Sine of Straight Angle

== Theorem ==
:$\sin 180 \degrees = \sin \pi = 0$
where:
:$\sin$ denotes the sine function
:$180 \degrees = \pi$ is the straight angle.
== Proof ==
A direct implementation of Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
In this case, $n = 1$ and so:
:$\sin \pi = 0$
== Also see ==
* Cosine 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:Sine Function
Sine of Sum

== Theorem ==
:$\map \sin {a + b} = \sin a \cos b + \cos a \sin b$
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 ==
* Cosine of Sum
* Tangent of Sum
== Sources ==
* : $\text V$. Trigonometry: Formulae $(12)$
* : $\S 5$: Trigonometric Functions: $5.34$
* : $2$: 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:Sine of Sum
Category:Sine Function
Category:Trigonometric Addition Formulas
Sine 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 with respect to $a$, 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:Sine of Sum
Sine of Sum/Proof 7

== Theorem ==
== Proof ==
Let two triangles $\triangle ABC$ and $\triangle ADC$ be inscribed in a circle on opposite
sides of diameter $AC$.
By Thales' Theorem, they are both right triangles and $\angle ADC$ and $\angle ABC$ are right angles.
Let the diameter $AC = 1$.
Let $\angle DAC = \alpha$ and $\angle CAB = \beta$.
From the construction above, we have the following:
:$\cos \alpha = AD$
:$\cos \beta = AB$
:$\sin \alpha = DC$
:$\sin \beta = BC$
By Length of Chord of Circle:
:$DB \mathop = 2 r \map \sin {\alpha + \beta}$
Since $2r \mathop = 1$:
:$DB \mathop = \map \sin {\alpha + \beta}$
By Quadrilateral is Cyclic iff Opposite Angles sum to Two Right Angles:
:$\Box ABCD$ is a cyclic quadrilateral.
By Ptolemy's Theorem:
:$DB \times AC = AB \times CD + BC \times AD$
Substituting:
:$\map \sin {\alpha + \beta} \times 1 = \cos \beta \sin \alpha + \sin \beta \cos \alpha$
:$\map \sin {\alpha + \beta} = \sin \alpha \cos \beta + \sin \beta \cos \alpha$
By Equivalence of Definitions of Sine of Angle, the definition of sine 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:
:$ \sin {\pi} = \sin \dfrac {\pi} 2 \cos \dfrac {\pi} 2 + \sin \dfrac {\pi} 2 \cos \dfrac {\pi} 2$
:$ 0 = 0 \cdot 1 + 1 \cdot 0$
The result follows.
Category:Sine of Sum
Sine of Supplementary Angle

== Theorem ==
:$\map \sin {\pi - \theta} = \sin \theta$
where $\sin$ denotes sine.
That is, the sine of an angle equals its supplement.
== Proof ==
== Also see ==
* Cosine of Supplementary Angle
* Tangent of Supplementary Angle
* Cotangent of Supplementary Angle
* Secant of Supplementary Angle
* Cosecant of Supplementary Angle
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$: Examples
* : $\S 5$: 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:Sine Function
Category:Supplementary Angles
Sine of Three Right Angles

== Theorem ==
:$\sin 270 \degrees = \sin \dfrac {3 \pi} 2 = -1$
where $\sin$ denotes the sine function.
== Proof ==
== Also see ==
* Cosine 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:Sine Function
Sine of Three Right Angles less Angle

== Theorem ==
:$\map \sin {\dfrac {3 \pi} 2 - \theta} = -\cos \theta$
where $\sin$ and $\cos$ are sine and cosine respectively.
== Proof ==
== Also see ==
* Cosine 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:Sine Function
Sine of Twice Angle minus Sum of Sines of Twice Other Two Angles of Triangle

== Theorem ==
Let $\triangle ABC$ be a triangle.
Then:
:$\sin 2 A - \sin 2 B - \sin 2 C = -4 \sin A \cos B \cos C$
== Proof ==
First we note that:
That is, $C$ is the supplement of $A + B$.
Then:
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $19$.
Category:Triangles
Category:Sine Function
Sine of X over X as Infinite Product

== Theorem ==
Let $z \in \C$ be a non-zero complex number.
Then:
:$\ds \frac {\sin z} z = \cos \frac z 2 \cos \frac z 4 \cos \frac z 8 \cdots = \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}$
where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
== Proof ==
First we prove that:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
for $n \in \N$.
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\ds \frac {\sin z} z = \paren {\frac {2^1} z} \sin \frac z {2^1} \prod_{i \mathop = 1}^1 \cos \frac z {2^i}$
Thus $\map P 1$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \frac {\sin z} z = \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$
Then we need to show:
:$\ds \frac {\sin z} z = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$
And then:
== Sources ==
* : $\S 38$: Infinite Products: $38.8$
Category:Proofs by Induction
Category:Trigonometric Identities
Category:Examples of Infinite Products
Sine of X over X is not Continuous at 0

== Theorem ==
Let $f$ be the real function defined as:
:$\map f x := \dfrac {\sin x} x$
Then $f$ is not continuous at $x = 0$.
== Proof ==
For $f$ to be continuous at $x = 0$ it is necessary that it be defined there.
But at the point $x = 0$, we have that $\map f x = \dfrac {\sin 0} 0$.
Division by $0$ is not defined.
Hence $f$ is not continuous at $x = 0$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity
Category:Examples of Continuous Real Functions
Sine of Zero is Zero

== Theorem ==
:$\sin 0 = 0$
where $\sin$ denotes the sine.
== Proof ==
Recall the definition of the sine function:
:$\ds \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
Thus:
:$\ds \sin 0 = 0 - \frac {0^3} {3!} + \frac {0^5} {5!} - \cdots = 0$
== Also see ==
* Cosine of Zero is One
* 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
* : $\S 16.3 \ (1) \ \text{(ii)}$
* : Appendix $12$: Trigonometric formulae: Trigonometric values for some special angles
* : Appendix $14$: Trigonometric formulae: Trigonometric values for some special angles
Category:Sine Function
Sine of i

== Theorem ==
:$\sin i = \paren {\dfrac e 2 - \dfrac 1 {2 e} } i$
where $\sin$ denotes the complex sine function and $i$ is the imaginary unit.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cosine of $i$
* Tangent of $i$
* Cotangent of $i$
* Secant of $i$
* Cosecant of $i$
Category:Sine of i
Category:Sine Function
Category:Imaginary Unit
Sine plus Sine/Examples

== Examples of Use of Sine plus Sine ==
=== $\sin 40 \degrees$ plus $\sin 60 \degrees$ ===
=== $\sin 2 A$ plus $\sin 6 A$ ===
=== Solution to $\sin 2 x + \sin 5 x \sin 4 x$ ===
Category:Sine plus Sine
Sine plus Sine/Examples/Sine 2 A plus Sine 6 A

== Example of Use of Sine plus Sine ==
:$\sin 2 A + \sin 6 A = 2 \sin 4 A \cos 2 A$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $5$.
Category:Sine plus Sine
Sine plus Sine/Examples/Sine 2 x plus Sine 5 x equals Sine 4 x

== Examples of Use of Sine plus Sine ==
The equation
:$\sin 3 x + \sin 5 x = \sin 4 x$
has the general solution:
:$\set {\dfrac {n \pi} 4 : n \in \Z} \cup \set {2 n \pi \pm \dfrac \pi 3: n \in \Z}$
== Proof ==
Hence the result.
== Sources ==
* : $\text V$. Trigonometry: Solution of equations: Example $3$.
Category:Sine plus Sine
Sine plus Sine/Examples/Sine 40 plus Sine 60

== Example of Use of Sine plus Sine ==
:$\sin 40 \degrees + \sin 60 \degrees = 2 \sin 50 \degrees \cos 10 \degrees$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $1$.
Category:Sine plus Sine
Sine to Power of Even Integer

== Theorem ==
== Proof ==
First, by Euler's Sine Identity we have:
:$\sin \theta = \dfrac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }$
Therefore by Power of Product:
:$\sin^{2 n} \theta = \dfrac 1 {\paren {2 i}^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$
Now by Power of Product and the Power of Power:
:$\dfrac 1 {\paren {2 i}^{2 n} } = \dfrac 1 {2^{2 n} \paren {-1}^n} = \dfrac {\paren {-1}^n} {2^{2 n} }$
Thus:
:$\sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$
Now:
So we have:
:$\ds \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$
Now we look at each of the terms in the parentheses:
We find, for the remaining term:
Thus we have:
as required.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.72$
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $130 \ \text{(b)}$
Category:Sine Function
Single Instruction URM Programs

== Theorem ==
=== Basic Primitive Recursive Functions ===
The basic primitive recursive functions are URM computable by a single-instruction URM program:
=== Zero Function ===
=== Successor Function ===
=== Projection Function ===
=== Identity Function ===
No other functions $f: \N^k \to \N$ can be computed using a single-instruction URM program.
== Proof ==
Apart from the programs given, the only other single-instruction programs are of the form:
Line		Command
$1$ 		$\map J {m, n, q}$
The convention is that, at the start of the program, $R_1$ contains the input, and all other registers contain $0$.
Suppose $q > 1$.
Then whatever $m$ or $n$ are, the program either jumps to $q$ (a location outside the program) or steps one instruction.
In either case the program stops without changing what is in $R_1$.
Hence this is another way of computing the identity function $I_\N$.
Suppose $q = 1$.
If $m = n$, then whatever is held in the registers, $r_m = r_n$ and the program will go back to execute step 1 again.
Thus the program will loop round and do step 1 endlessly, and never terminate.
If $m \ne n$, then the program's behaviour depends on the contents of $R_n$ and $R_m$.
If $m = 1$ and $n \ne 1$, then the program will compare the input against the contents of $r_n$.
If $r_1 = 0$ the program will go into an endless loop, and never terminate.
Otherwise, i.e. if $r_1 \ne 0$, the program will never terminate with $r_1$ unchanged.
If $m \ne 1$ and $n \ne 1$ then the program will once more never terminate, as $r_m = r_n = 0$ at the start of the program.
So the program:
Line		Command
$1$ 		$\map J {m, n, 1}$
does not specify a URM computable function, as:
:when $m = n$ the program will never terminate for any input
:when $m \ne n, m = 1$ the program will never terminate on input $0$
:when $m \ne n, n = 1$ the program similarly will never terminate on input $0$
:when $m \ne n, m \ne 1, n \ne 1$ the program will never terminate for any input.
In all cases there is at least one input for which the program will never terminate.
The result follows from the definition of URM computability.
Category:URM Programs
Category:Primitive Recursive Functions
Single Instruction URM Programs/Identity Function

== Theorem ==
The identity function $I_\N: \N \to \N$ defined as:
:$\forall n \in \N: \map {I_\N} n = n$
is URM computable by a single-instruction URM program.
== Proof ==
Any of the following URM programs compute the identity function:
Line		Command
$1$ 		$\map Z m$
... where $m \ne 1$.
This sets the value $0$ into $R_m$ and then stops.
Line		Command
$1$ 		$\map S m$
... where $m \ne 1$.
The input $n$ is in $R_1$ when the program starts.
The program adds $1$ to $r_m$ and then stops.
Line		Command
$1$ 		$\map C {j, m}$
... where $m \ne 1$.
The input $n$ is in $R_1$ when the program starts.
The program copies $r_j$ to $r_m$ and then stops.
Line		Command
$1$ 		$\map C {1, 1}$
The input $n$ is in $R_1$ when the program starts.
The program copies $r_1$ to $r_1$, i.e. to itself, and then stops.
In none of these programs is $R_1$ affected, and so no change is effected to the input, which is returned as the output unchanged.
Hence they all compute the identity function.
Note that the latter program, consisting entirely of $\map C {1, 1}$, is a direct implementation of the projection function program $\pr^1_1: \N \to \N$.
This latter function is the usual way of implementing the identity function as it is well-defined and obvious, and guaranteed to have no other side-effects when embedded in a larger program.
Category:URM Programs
Category:Primitive Recursive Functions