nyuuzyou/wikis huggingface dataset:
Sine Function is Odd

== Theorem ==
For all $z \in \C$:
:$\map \sin {-z} = -\sin z$
That is, the sine function is odd.
== Proof ==
Recall the definition of the sine function:
From Sign of Odd Power, we have that:
:$\forall n \in \N: -\paren {z^{2 n + 1} } = \paren {-z}^{2 n + 1}$
The result follows directly.
== Also see ==
* Cosine Function is Even
* Tangent Function is Odd
* Cotangent Function is Odd
* Secant Function is Even
* Cosecant Function is Odd
* Hyperbolic Sine Function is Odd
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$: $(4.15)$
* : $\S 5$: Trigonometric Functions: $5.28$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.2$ Sine Function
* : $\S 16.3 \ (1) \ \text{(iv)}$
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* 
* : Appendix $12$: Trigonometric formulae: Symmetry
* : Appendix $14$: Trigonometric formulae: Symmetry
Category:Sine Function
Category:Examples of Odd Functions
Sine Inequality/Proof 1

== Theorem ==
== Proof ==
Let $\map f x = x - \sin x$.
By Derivative of Sine Function:
:$\map {f'} x = 1 - \cos x$
From Real Cosine Function is Bounded we know $\cos x \le 1$ for all $x$.
Hence $\map f x \ge 0$ for all $x$.
From Derivative of Monotone Function, $\map f x$ is increasing. 
By Sine of Zero is Zero, $\map f 0 = 0$.
It follows that $\map f x \ge 0$ for all $x \ge 0$.
Now let $\map g x = x^2 - \sin^2 x$.
From Derivative of Monotone Function, $\map g x$ is increasing for $x \ge 0$. 
By Sine of Zero is Zero, $\map g 0 = 0$.
It follows that $\map g x \ge 0$ for all $x \ge 0$.
Observe that $\map g x$ is an even function.
This implies $\map g x \ge 0$ for all $x \in \R$.
Finally note that $\sin^2 x \le x^2 \iff \size {\sin x} \le \size x$.
Category:Sine Inequality
Sine Inequality/Proof 2

== Theorem ==
== Proof ==
For $x = 0$, the inequality is trivial, as $\sin 0 = 0$.
If $\size x \ge 1$, the inequality also clear as:
Now, suppose $0 < \size x < 1$
Then on the one hand:
On the one hand:
The last expression is $\le 1$, since:
Category:Sine Inequality
Sine Integral Function is Bounded

== Theorem ==
Let $\Si: \R \to \R$ denote the sine integral function.
Then $\Si$ is bounded.
== Proof ==
By Limit at Infinity of Sine Integral Function and its corollary:
Thus there is a $M > 0$ such that for all $x \in \R$:
:$\size x > M \implies \size {\map \Si x} \le \dfrac \pi 2 + 1$
On the other hand, if $\size x \le M$, then:
Thus:
:$\forall x \in \R : \size {\map \Si x} \le \max \set {\dfrac \pi 2 + 1 , M}$
Category:Sine Integral Function
Sine Integral Function is Odd

== Theorem ==
:$\map \Si {-x} = -\map \Si x$
where:
:$\Si$ denotes the sine integral function
:$x$ is a real number.
== Proof ==
== Sources ==
* : $\S 35$: Sine Integral $\ds \map \Si x = \int_0^x \frac {\sin u} u \rd u$: $35.13$
Category:Sine Integral Function
Category:Examples of Odd Functions
Sine Integral Function of Zero

== Theorem ==
:$\map \Si 0 = 0$
where $\Si$ denotes the sine integral function.
== Proof ==
By Sine Integral Function is Odd, $\Si$ is an odd function. 
Therefore, by Odd Function of Zero is Zero:
:$\map \Si 0 = 0$
== Sources ==
* : $\S 35$: Sine Integral $\ds \map \Si x = \int_0^x \frac {\sin u} u \rd u$: $35.13$
Category:Sine Integral Function
Sine and Cosine are Periodic on Reals

== Theorem ==
The real sine function and real cosine function are periodic on the set of real numbers $\R$:
=== Real Cosine Function is Periodic ===
=== Real Sine Function is Periodic ===
:
=== Pi ===
=== Cosine of Angle plus Straight Angle ===
=== Sine of Angle plus Straight Angle ===
=== Sign of Cosine on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ ===
=== Sign of Sine on $\openint 0 \pi$ and $\openint \pi {2 \pi}$ ===
=== Zeroes of Cosine ===
=== Zeroes of Sine ===
== Note ==
Given that we have defined sine and cosine in terms of a power series, it is a plausible proposition to define $\pi$ using the same language.
$\pi$ is, of course, the famous irrational constant $3.14159 \ldots$.
== Sources ==
* : $\S 2$. Geometrical Representations
* : $\S 16.4$
Category:Sine and Cosine are Periodic on Reals
Category:Sine Function
Category:Cosine Function
Category:Periodic Functions
Sine and Cosine are Periodic on Reals/Corollary

== Corollary to Sine and Cosine are Periodic on Reals ==
Let $x \in \R$.
=== Sign of Cosine on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ ===
=== Sign of Sine on $\openint 0 \pi$ and $\openint \pi {2 \pi}$ ===
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Corollary/Cosine

== Corollary to Sine and Cosine are Periodic on Reals ==
Let $x \in \R$.
:$\cos x$ is strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
== 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 $\eta \in \R_{>0}$, where $\pi$ was defined as $\pi := 2 \eta$.
It follows that $\eta = \dfrac \pi 2$, thus:
:$\map \cos {x + \pi} = -\map \sin {x + \dfrac \pi 2} = -\cos x$
From the discussion in the proof of Real Cosine Function is Periodic:
:$\cos \eta = \map \cos {-\eta} = 0$ and $\cos x > 0$ for $-\eta < x < \eta$
It follows directly that:
:$\forall x \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}: \cos x \ge 0$
As $\map \cos {x + \pi} = -\cos x$, we have:
:$\forall x \in \closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}: \cos x \le 0$
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Corollary/Sine

== Corollary to Sine and Cosine are Periodic on Reals ==
Let $x \in \R$.
:$\sin x$ is strictly positive on the interval $\openint 0 \pi$ and strictly negative on the interval $\openint \pi {2 \pi}$
== Proof ==
From the discussion in the proof of Real Cosine Function is Periodic:
:$\map \sin {x + \eta} = \cos x$
for $\eta \in \R_{>0}$, where $\pi$ was defined as $\pi := 2 \eta$.
It follows that $\eta = \dfrac \pi 2$, thus:
:$\sin x = \map \cos {x - \dfrac \pi 2}$
From Sine and Cosine are Periodic on Reals: Corollary: Cosine, it follows that $\cos x$ is strictly positive on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ and strictly negative on $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$.
It follows directly that:
:$\forall x \in \closedint 0 \pi: \sin x \ge 0$
and:
:$\forall x \in \closedint \pi {2 \pi} : \sin x \le 0$
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Cosine

== Theorem ==
The cosine function is periodic on the set of real numbers $\R$:
:$\exists L \in \R_{\neq 0}: \forall x \in \R: \cos x = \map \cos {x + L}$
:
== Proof ==
From Real Cosine Function has Zeroes, the cosine function has at least one positive zero.
Therefore there exists a Greatest Lower Bound $\eta \in \R_{>0}$ to the set of positive zeroes.
Since Cosine Function is Continuous, $\eta$ is a zero.
Because Cosine Function is Even:
:$\cos \eta = \map \cos {-\eta} = 0$
By definition of greatest lower bound:
:$\cos x \neq 0$ for $-\eta < x < \eta$
Because Cosine of Zero is One, it follows from the Intermediate Value Theorem that:
:$\cos x > 0$ for $-\eta < x < \eta$
From Sum of Squares of Sine and Cosine:
:$\cos^2 x + \sin^2 x = 1$
Hence as $\cos \eta = 0$ it follows that:
:$\sin^2 \eta = 1$
So either $\sin \eta = 1$ or $\sin \eta = -1$.
But from Derivative of Sine Function:
:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
On the interval $\openint {-\eta} \eta$, it has been shown that $\cos x > 0$.
Thus by Derivative of Monotone Function, $\sin x$ is increasing on $\closedint {-\eta} \eta$.
Since Sine of Zero is Zero it follows that:
:$\sin \eta > 0$
So it must be the case that:
:$\sin \eta = 1$
Now we apply Sine of Sum and Cosine of Sum:
Hence it follows that:
Thus $\cos$ is periodic on $\R$ with a period $L \leq 4 \eta$.
== Sources ==
* 
* 
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Pi

== Theorem ==
Let $\sin: \R \to \R$ be the real sine function, and let $\cos: \R \to \R$ be the real cosine function.
The real number $\pi$ (called pi, pronounced pie) is uniquely defined as:
:$\pi := \dfrac p 2$
where $p \in \R$ is the period of $\sin$ and $\cos$.
== Proof 1 ==
== Proof 2 ==
== Note ==
Given that we have defined sine and cosine in terms of a power series, it is a plausible proposition to define $\pi$ using the same language.
$\pi$ is, of course, the famous irrational constant $3.14159 \ldots$.
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Pi/Proof 1

== Theorem ==
== Proof ==
From the Real Cosine Function is Periodic and Real Sine Function is Periodic, we have that $\cos x$ and $\sin x$ are periodic on $\R$ with the same period.
If we denote the period of $\cos x$ and $\sin x$ as $p$, it follows that $\pi = \dfrac p 2$ is uniquely defined.
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Pi/Proof 2

== Theorem ==
== Proof ==
By Cosine of Zero is One:
:$\cos 0 = 1$
By Cosine of 2 is Strictly Negative:
:$\cos 2 < 0$
Thus by the corollary to the Intermediate Value Theorem there exists an $h \in \openint 0 2$ such that:
:$\cos h = 0$
By Sine of Sum for all $x \in \R$:
By Cosine of Sum for all $x \in \R$:
By Sum of Squares of Sine and Cosine:
Thus for all $x \in \R$:
In particular, $\cos$ is periodic.
By Nonconstant Periodic Function with no Period is Discontinuous Everywhere, $\cos$ has a period $p \in \R_{>0}$.
In view of $(1)$ and $\sin h \ne 0$, the periodic elements of $\sin$ are exactly those of $\cos$.
Thus $p$ is also the period of $\sin$.
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Sine/Proof 1

== Theorem ==
The sine function is periodic with the same period as the cosine function.
:
== Proof ==
Since Real Cosine Function is Periodic, let $K$ be its period.
Then:
:$\cos K = \map \cos {0 + K} = \cos 0$
Because Cosine of Zero is One:
:$\cos K = 1$
Furthermore:
Then, the following holds:
Thus $\sin$ is periodic with some period $L \leq K$.
The following also hold:
So we may conclude:
Therefore the period of cosine is at most that of sine:
:$K \leq L$
But if $K \leq L \leq K$ then:
:$K = L$
Category:Sine and Cosine are Periodic on Reals
Sine and Cosine are Periodic on Reals/Sine/Proof 2

== Theorem ==
== Proof ==
Since Real Cosine Function is Periodic, let $L$ be its period.
From Primitive of Cosine Function:
for any constant $C$.
Therefore $\sin x$ is a Primitive of $\cos x$, for the special case of $C = 0$.
From Primitive of Periodic Real Function, it follows that $\sin x$ is periodic with period $L$.
Category:Sine and Cosine are Periodic on Reals
Sine in terms of Hyperbolic Sine

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$i \sin z = \map \sinh {i z}$
where:
:$\sin$ denotes the complex sine
:$\sinh$ denotes the hyperbolic sine
:$i$ is the imaginary unit: $i^2 = -1$.
== Proof ==
 
== Also see ==
* Cosine in terms of Hyperbolic Cosine
* 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: $8.80$: 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:Sine Function
Category:Hyperbolic Sine Function
Sine in terms of Secant

== Theorem ==
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
where $\sin$ denotes the sine function and $\sec$ denotes the secant function.
== Proof ==
For the first part, if there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$:
For the second part, if there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$:
When $\cos x = 0$, $\sec x$ is undefined.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Sine Function
Category:Secant Function
Sine 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 ==
For the first part, if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$:
For the second part, if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$:
When $\cos x = 0$, $\tan x$ is undefined.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Sine Function
Category:Tangent Function
Sine is Reciprocal of Cosecant

== Theorem ==
Let $\theta$ be an angle such that $\sin \theta \ne 0$.
Then:
:$\sin \theta = \dfrac 1 {\csc \theta}$
where $\sin$ denotes the sine function and $\csc$ denotes the cosecant function.
== Proof ==
== Also see ==
* Trigonometric Functions in terms of each other
Category:Sine Function
Category:Cosecant Function
Sine is of Exponential Order Zero

== Theorem ==
Let $\sin t$ be the sine of $t$, where $t \in \R$.
Then $\sin t$ is of exponential order $0$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Functions of Exponential Order
Category:Exponential Order
Category:Sine Function
Category:Sine is of Exponential Order Zero
Sine minus Sine/Examples/Sine 60 minus Sine 40

== Example of Use of Sine minus Sine ==
:$\sin 60 \degrees - \sin 40 \degrees = 2 \cos 50 \degrees \sin 10 \degrees$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercises $\text {XXXII}$: $4$.
Category:Sine minus Sine
Sine n x over Pi x Delta Sequence

== Theorem ==
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\ds \map {\delta_n} x := \frac {\map \sin {n x} }{\pi x}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof ==
From Integral to Infinity of Sine p x over x for $n \in \N_{> 0}$ we have that:
:$\ds \int_0^\infty \frac {\map \sin {n x} } x \rd x = \frac \pi 2$
Furthermore:
Hence:
:$\ds \int_{-\infty}^0 \frac {\map \sin {n y} } y \rd y + \int_0^\infty \frac {\map \sin {n x} } x \rd x = \pi$
Let $a,b \in \R$.
Then:
Suppose $0 < a < b$.
Then:
Analogously, suppose $a < b < 0$.
Then:
Let $\epsilon \in \R_{> 0}$.
Then:
$\epsilon$ is an arbitrary positive real number.
Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.
Suppose $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
:$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
Then $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.
But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.
Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.
== Sources ==
: Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
Sine of 105 Degrees

== Theorem ==
:$\sin 105^\circ = \sin \dfrac {7 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 120 Degrees

== Theorem ==
:$\sin 120 \degrees = \sin \dfrac {2 \pi} 3 = \dfrac {\sqrt 3} 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 135 Degrees

== Theorem ==
:$\sin 135 \degrees = \sin \dfrac {3 \pi} 4 = \dfrac {\sqrt 2} 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 150 Degrees

== Theorem ==
:$\sin 150 \degrees = \sin \dfrac {5 \pi} 6 = \dfrac 1 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 15 Degrees

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

== Theorem ==
:$\sin 165 \degrees = \sin \dfrac {11 \pi} {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 18 Degrees

== Theorem ==
:$\sin 18 \degrees = \sin \dfrac \pi {10} = \dfrac {\sqrt 5 - 1} 4$
where $\sin$ denotes the sine function.
== Proof ==
From Sine of $90 \degrees$:
:$\map \sin {5 \times 18 \degrees} = \sin 90 \degrees = 1$.
Consider the equation:
:$\sin 5x = 1$
where $x = 18 \degrees$ is one of the solutions.
From Quintuple Angle Formula of Sine:
:$16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$
Let $s = \sin \theta$:
:$16 s^5 - 20 s^3 + 5s - 1 = 0$
That is:
:$\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$
Therefore, either:
:$s = 1$
or by the Quadratic Formula:
:$s = \dfrac 1 4 \paren {\pm \sqrt 5 - 1}$
Since $0 < \sin 18 \degrees < 1$:
:$\sin 18 \degrees = \dfrac {\sqrt 5 - 1} 4$
Category:Sine Function
Sine of 195 Degrees

== Theorem ==
:$\sin 195 \degrees = \sin \dfrac {13 \pi} {12} = -\dfrac {\sqrt 6 - \sqrt 2} 4$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 1 Degree

== Theorem ==
where $\sin$ denotes the sine function.
== Proof ==
This is in the form:
:$a x^3 + b x^2 + c x + d = 0$
where:
:$x = \sin 1 \degrees$
:$a = 4$
:$b = 0$
:$c = -3$
:$d = \sin 3 \degrees$
From Cardano's Formula:
:$x = S + T$
where:
:$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
:$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$
where:
and:
Thus:
:$\sqrt [3] {\cis 93 \degrees}$ has $3$ unique cube roots.
$\tuple {\cis 31 \degrees, \cis 151 \degrees, \cis 271 \degrees}$
:$\sqrt [3] {\cis 87 \degrees }$ also has $3$ unique cube roots.
$\tuple {\cis 29 \degrees, \cis 149 \degrees, \cis 269 \degrees}$
We need to investigate $9$ differences and find solutions where the complex part vanishes.
In solution $2)$ above, we rotate $\sqrt [3] {\cis 87 \degrees }$ by $120 \degrees$ (multiply by $\paren {-\dfrac 1 2 + i \dfrac {\sqrt 3} 2}$)
In solution $4)$ above, we rotate $\sqrt [3] {\cis 93 \degrees }$ by $120 \degrees$ (multiply by $\paren {-\dfrac 1 2 + i \dfrac {\sqrt 3} 2}$)
In solution $9)$ above, we would rotate BOTH $\sqrt [3] {\cis 93 \degrees}$ and $\sqrt [3] {\cis 87 \degrees}$ by $240 \degrees$ (multiply by $\paren {-\dfrac 1 2 - i \dfrac {\sqrt 3} 2}$)
For the remainder of this proof, we will arbitrarily choose solution $9)$:
Squaring the numerator of the Sine of 3 Degrees, we get:
Bringing that result back into our main result, we get:
Canceling $4$ from the numerator and denominator, we get:
Moving an $8$ to the outside, we get:
Flipping the top and bottom lines, we get our main result:
We can verify that this is correct by noting that:
Plugging these into the main result, we see:
Category:Sine Function
Sine of 210 Degrees

== Theorem ==
:$\sin 210 \degrees = \sin \dfrac {7 \pi} 6 = -\dfrac 1 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 22.5 Degrees

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

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

== Theorem ==
:$\sin 240 \degrees = \sin \dfrac {4 \pi} 3 = -\dfrac {\sqrt 3} 2$
where $\sin$ denotes the sine function.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine of 240 Degrees
Category:Sine Function
Sine of 240 Degrees/Proof 2

== Theorem ==
== Proof ==
When $240 \degrees$ is embedded in a Cartesian plane, it makes an angle of $60 \degrees$ with the $x$-axis.
$240 \degrees$ can be found in the third quadrant.
Hence by definition of sine function in the third quadrant, $\sin 240 \degrees$ is negative.
Thus:
:$\sin 240 \degrees = -\sin 60 \degrees = -\dfrac {\sqrt 3} 2$
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$: Example
Category:Sine of 240 Degrees
Sine of 255 Degrees

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

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

== Theorem ==
:$\sin 300 \degrees = \sin \dfrac {5 \pi} 3 = -\dfrac {\sqrt 3} 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function
Sine of 30 Degrees

== Theorem ==
:$\sin 30 \degrees = \sin \dfrac \pi 6 = \dfrac 1 2$
where $\sin$ denotes the sine function.
== Proof ==
:
Let $\triangle ABC$ be an equilateral triangle of side $r$.
By definition, each angle of $\triangle ABC$ is equal.
From Sum of Angles of Triangle equals Two Right Angles it follows that each angle measures $60^\circ$.
Let $CD$ be a perpendicular dropped from $C$ to $AB$ at $D$.
Then $AD = \dfrac r 2$ while:
:$\angle ACD = \dfrac {60 \degrees} 2 = 30 \degrees$
So by definition of sine function:
:$\sin \paren {\angle ACD} = \dfrac {r / 2} r = \dfrac 1 2$
== 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 315 Degrees

== Theorem ==
:$\sin 315 \degrees = \sin \dfrac {7 \pi} 4 = -\dfrac {\sqrt 2} 2$
where $\sin$ denotes the sine function.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Sine Function