avewright/proofwiki-math huggingface dataset:
Theorem: Sine of 15 Degrees

:$\sin 15 \degrees = \sin \dfrac \pi {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 15 \degrees
      | r = \map \sin {45 \degrees - 30 \degrees}
      | c = 
}}
{{eqn | r = \sin 45 \degrees \cos 30 \degrees - \cos 45 \degrees \sin 30 \degrees
      | c = Sine of Difference
}}
{{eqn | r = \paren {\frac {\sqrt 2} 2} \paren {\frac {\sqrt 3} 2} - \paren {\frac {\sqrt 2} 2} \paren {\dfrac 1 2}
      | c = Sine of $45 \degrees$, Cosine of $30 \degrees$, Cosine of $45 \degrees$, Sine of $30 \degrees$
}}
{{eqn | r = \frac {\sqrt 6} 4 - \frac {\sqrt 2} 4
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} 4
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 30 Degrees

:$\sin 30 \degrees = \sin \dfrac \pi 6 = \dfrac 1 2$

Proof:
:200px
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:
:$\sin \paren {\angle ACD} = \dfrac {r / 2} r = \dfrac 1 2$
{{qed}}
Theorem: Sine of 45 Degrees

:$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
:300px
Let $ABCD$ be a square of side $r$.
By definition, each angle of $\triangle ABCD$ is equal to $90 \degrees$.
Let $AC$ be a diagonal of $ABCD$.
As $\triangle ABC$ is a right angled triangle, it follows from Pythagoras's Theorem that $AC = \sqrt 2 A B$.
As $AC$ is a bisector of $\angle DAB$ it follows that $\angle CAB = 45 \degrees$.
So by definition of sine function:
:$\map \sin {\angle CAB} = \dfrac r {r \sqrt 2} = \dfrac {\sqrt 2} 2$
{{qed}}
Theorem: Sine of 45 Degrees

:$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 45 \degrees
      | r = \map \sin {30 \degrees + 15 \degrees}
      | c = 
}}
{{eqn | r = \sin 30 \degrees \cos 15 \degrees + \cos 30 \degrees \sin 15 \degrees
      | c = Sine of Sum
}}
{{eqn | r = \paren {\frac 1 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} + \paren {\frac {\sqrt 3} 2} \paren {\dfrac {\sqrt 6 - \sqrt 2} 4}
      | c = {{sin|30}}, {{cos|15}}, {{cos|30}}, {{sin|15}}
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 6 + \sqrt 2 + \sqrt 3 \paren {\sqrt 6 - \sqrt 2} }
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 3 \sqrt 2 + \sqrt 2 + \sqrt 3 \sqrt 3 \sqrt 2 - \sqrt 3 \sqrt 2}
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 2 + 3 \sqrt 2}
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {4 \sqrt 2}
      | c = 
}}
{{eqn | r = \frac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 45 Degrees

:$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 45 \degrees
      | r = \map \sin {3 \times 15 \degrees}
      | c = 
}}
{{eqn | r = 3 \sin 15 \degrees - 4 \sin^3 15 \degrees
      | c = Triple Angle Formula for Sine
}}
{{eqn | r = 3 \paren {\frac {\sqrt 6 - \sqrt 2} 4} - 4 \paren {\frac {\sqrt 6 - \sqrt 2} 4}^3
      | c = {{sin|15}}
}}
{{eqn | r = \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {6 \sqrt 6 - 18 \sqrt 2 + 6 \sqrt 6 - 2 \sqrt 2} {16}
      | c = 
}}
{{eqn | r = \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {12 \sqrt 6 - 20\sqrt 2} {16}
      | c = 
}}
{{eqn | r = \frac {3 \sqrt 6 - 3 \sqrt 2} 4 - \frac {3 \sqrt 6 - 5 \sqrt 2} 4
      | c = 
}}
{{eqn | r = \frac {2 \sqrt 2} 4
      | c = 
}}
{{eqn | r = \frac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 45 Degrees

:$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 45 \degrees
      | r = \map \sin {60 \degrees - 15 \degrees}
}}
{{eqn | r = \sin 60 \degrees \cos 15 \degrees - \cos 60 \degrees \sin 15 \degrees
      | c = Sine of Difference
}}
{{eqn | r = \paren {\frac {\sqrt 3} 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} - \paren {\frac 1 2} \paren {\frac {\sqrt 6 - \sqrt 2} 4}
      | c = Sine of $60 \degrees$, Cosine of $15 \degrees$, Cosine of $60 \degrees$, Sine of $15 \degrees$
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 3 \paren {\sqrt 6 + \sqrt 2} - \paren {\sqrt 6 - \sqrt 2} }
      | c = simplifying
}}
{{eqn | r = \frac 1 8 \paren {3 \sqrt 2 + \sqrt 6 - \sqrt 6 + \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac 1 8 \paren {3 \sqrt 2 + \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac 1 8 \paren {4 \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 45 Degrees

:$\sin 45 \degrees = \sin \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 90 \degrees
      | r = 1
      | c = Sine of Right Angle
}}
{{eqn | ll= \leadsto
      | l = \map \sin {2 \times 45 \degrees}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 \sin 45 \degrees \cos 45 \degrees
      | r = 1
      | c = Double Angle Formula for Sine
}}
{{eqn | ll= \leadsto
      | l = 2 \sin 45 \degrees \map \sin {90 \degrees - 45 \degrees}
      | r = 1
      | c = Sine of Complement equals Cosine
}}
{{eqn | ll= \leadsto
      | l = 2 \sin 45 \degrees \sin 45 \degrees
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 \sin^2 45 \degrees
      | r = 1
      | c = by $(1)$
}}
{{eqn | ll= \leadsto
      | l = \sin 45 \degrees
      | r = \pm \frac 1 {\sqrt 2}
      | c = 
}}
{{end-eqn}}
The negative solution is rejected because $45 \degrees$ is an acute angle and Sine of Acute Angle is Positive.
Therefore:
:$\sin 45 \degrees = \dfrac 1 {\sqrt 2} = \dfrac {\sqrt 2} 2$
{{qed}}
Theorem: Sine of 60 Degrees

:$\sin 60 \degrees = \sin \dfrac \pi 3 = \dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 60 \degrees
      | r = \map \cos {90 \degrees - 60 \degrees}
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \cos 30 \degrees
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 3} 2
      | c = Cosine of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 75 Degrees

:$\sin 75 \degrees = \sin \dfrac {5 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 75 \degrees
      | r = \map \sin {60 \degrees + 15 \degrees}
      | c = 
}}
{{eqn | r = \sin 60 \degrees \cos 15 \degrees + \cos 60 \degrees \sin 15 \degrees
      | c = Sine of Sum
}}
{{eqn | r = \paren {\frac {\sqrt 3} 2} \paren {\frac {\sqrt 6 + \sqrt 2} 4} + \paren {\frac 1 2} \paren {\dfrac {\sqrt 6 - \sqrt 2} 4}
      | c = Sine of $60 \degrees$, Cosine of $15 \degrees$, Cosine of $60 \degrees$, Sine of $15 \degrees$
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 3 \paren {\sqrt 6 + \sqrt 2} + \sqrt 6 - \sqrt 2}
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {\sqrt 3 \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 + \sqrt 3 \sqrt 2 - \sqrt 2}
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {3 \sqrt 2 + 2 \sqrt 3 \sqrt 2 - \sqrt 2}
      | c = 
}}
{{eqn | r = \frac 1 8 \paren {2 \sqrt 2 + 2 \sqrt 6}
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 75 Degrees

:$\sin 75 \degrees = \sin \dfrac {5 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 75 \degrees
      | r = \map \cos {90 \degrees - 75 \degrees}
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \cos 15^\circ
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = Cosine of $15 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Right Angle

:$\sin 90 \degrees = \sin \dfrac \pi 2 = 1$

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:
:$\sin \dfrac 1 2 \pi = \paren {-1}^0 = 1$
{{qed}}
Theorem: Sine of Angle plus Right Angle

: $\sin \paren {x + \dfrac \pi 2} = \cos x$

Proof:
{{begin-eqn}}
{{eqn | l = \sin \paren {x + \frac \pi 2}
      | r = \sin x \cos \frac \pi 2 + \cos x \sin \frac \pi 2
      | c = Sine of Sum
}}
{{eqn | r = \sin x \cdot 0 + \cos x \cdot 1
      | c = Cosine of Right Angle and Sine of Right Angle
}}
{{eqn | r = \cos x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 105 Degrees

:$\sin 105^\circ = \sin \dfrac {7 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 105^\circ
      | r = \sin \left({90^\circ + 15^\circ}\right)
      | c = 
}}
{{eqn | r = \cos 15^\circ
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = Cosine of 15 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 120 Degrees

:$\sin 120 \degrees = \sin \dfrac {2 \pi} 3 = \dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 120 \degrees
      | r = \map \sin {90 \degrees + 30 \degrees}
      | c = 
}}
{{eqn | r = \cos 30 \degrees
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \frac {\sqrt 3} 2
      | c = Cosine of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 135 Degrees

:$\sin 135 \degrees = \sin \dfrac {3 \pi} 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 135 \degrees
      | r = \map \sin {90 \degrees + 45 \degrees}
      | c = 
}}
{{eqn | r = \cos 45 \degrees
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \frac {\sqrt 2} 2
      | c = Cosine of $45 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 150 Degrees

:$\sin 150 \degrees = \sin \dfrac {5 \pi} 6 = \dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 150 \degrees
      | r = \map \sin {90 \degrees + 60 \degrees}
      | c = 
}}
{{eqn | r = \cos 60 \degrees
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \frac 1 2
      | c = Cosine of $60 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 165 Degrees

:$\sin 165 \degrees = \sin \dfrac {11 \pi} {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 165 \degrees
      | r = \map \sin {90 \degrees + 75 \degrees}
      | c = 
}}
{{eqn | r = \cos 75 \degrees
      | c = Sine of Angle plus Right Angle
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} 4
      | c = Cosine of $75 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Straight Angle

:$\sin 180 \degrees = \sin \pi = 0$

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:
:$\sin \pi = 0$
{{qed}}
Theorem: Sine of Supplementary Angle

:$\map \sin {\pi - \theta} = \sin \theta$
where $\sin$ denotes sine.
That is, the sine of an angle equals its supplement.

Proof:
{{begin-eqn}}
{{eqn | l = \map \sin {\pi - \theta}
      | r = \sin \pi \cos \theta - \cos \pi \sin \theta
      | c = Sine of Difference
}}
{{eqn | r = 0 \times \cos \theta - \paren {-1} \times \sin \theta
      | c = Sine of Straight Angle and Cosine of Straight Angle
}}
{{eqn | r = \sin \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Supplementary Angle

:$\map \cos {\pi - \theta} = -\cos \theta$
where $\cos$ denotes cosine.
That is, the cosine of an angle is the negative of its supplement.

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {\pi - \theta}
      | r = \cos \pi \cos \theta + \sin \pi \sin \theta
      | c = Cosine of Difference
}}
{{eqn | r = \paren {-1} \times \cos \theta + 0 \times \sin \theta
      | c = Cosine of Straight Angle and Sine of Straight Angle
}}
{{eqn | r = -\cos \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Conjugate Angle

:$\map \sin {2 \pi - \theta} = -\sin \theta$
where $\sin$ denotes sine.
That is, the sine of an angle is the negative of its conjugate.

Proof:
{{begin-eqn}}
{{eqn | l = \map \sin {2 \pi - \theta}
      | r = \map \sin {2 \pi} \cos \theta - \map \cos {2 \pi} \sin \theta
      | c = Sine of Difference
}}
{{eqn | r = 0 \times \cos \theta - 1 \times \sin \theta
      | c = Sine of Full Angle and Cosine of Full Angle
}}
{{eqn | r = -\sin \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Conjugate Angle

:$\map \cos {2 \pi - \theta} = \cos \theta$
where $\cos$ denotes cosine.
That is, the cosine of an angle equals its conjugate.

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {2 \pi - \theta}
      | r = \map \cos {2 \pi} \cos \theta + \map \sin {2 \pi} \sin \theta
      | c = Cosine of Difference
}}
{{eqn | r = 1 \times \cos \theta + 0 \times \sin \theta
      | c = Cosine of Full Angle and Sine of Full Angle
}}
{{eqn | r = \cos \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 195 Degrees

:$\sin 195 \degrees = \sin \dfrac {13 \pi} {12} = -\dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 195 \degrees
      | r = \map \sin {360 \degrees - 165 \degrees}
      | c = 
}}
{{eqn | r = -\sin 165 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 6 - \sqrt 2} 4
      | c = Sine of $165 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 210 Degrees

:$\sin 210 \degrees = \sin \dfrac {7 \pi} 6 = -\dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 210 \degrees
      | r = \map \sin {360 \degrees - 150 \degrees}
      | c = 
}}
{{eqn | r = -\sin 150 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\frac 1 2
      | c = Sine of $150 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 225 Degrees

:$\sin 225 \degrees = \sin \dfrac {5 \pi} 4 = -\dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 225 \degrees
      | r = \map \sin {360 \degrees - 135 \degrees}
      | c = 
}}
{{eqn | r = -\sin 135 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 2} 2
      | c = Sine of $135 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 240 Degrees

:$\sin 240 \degrees = \sin \dfrac {4 \pi} 3 = -\dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 240 \degrees
      | r = \map \sin {360 \degrees - 120 \degrees}
      | c = 
}}
{{eqn | r = -\sin 120 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 3} 2
      | c = Sine of $120 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 240 Degrees

:$\sin 240 \degrees = \sin \dfrac {4 \pi} 3 = -\dfrac {\sqrt 3} 2$

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:
:$\sin 240 \degrees = -\sin 60 \degrees = -\dfrac {\sqrt 3} 2$
{{qed}}
Theorem: Sine of 255 Degrees

:$\sin 255^\circ = \sin \dfrac {17 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 255^\circ
      | r = \sin \left({360^\circ - 105^\circ}\right)
      | c = 
}}
{{eqn | r = - \sin 105^\circ
      | c = Sine of Conjugate Angle
}}
{{eqn | r = - \frac {\sqrt 6 + \sqrt 2} 4
      | c = Sine of 105 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Three Right Angles

:$\sin 270 \degrees = \sin \dfrac {3 \pi} 2 = -1$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 270 \degrees
      | r = \map \sin {360 \degrees - 90 \degrees}
      | c = 
}}
{{eqn | r = -\sin 90 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -1
      | c = Sine of Right Angle
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 285 Degrees

:$\sin 285^\circ = \sin \dfrac {19 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 285^\circ
      | r = \sin \left({360^\circ - 75^\circ}\right)
      | c = 
}}
{{eqn | r = - \sin 75^\circ
      | c = Sine of Conjugate Angle
}}
{{eqn | r = - \dfrac {\sqrt 6 + \sqrt 2} 4
      | c = Sine of $75^\circ$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 300 Degrees

:$\sin 300 \degrees = \sin \dfrac {5 \pi} 3 = -\dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 300 \degrees
      | r = \map \sin {360 \degrees - 60 \degrees}
      | c = 
}}
{{eqn | r = -\sin 60 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\dfrac {\sqrt 3} 2
      | c = Sine of $60 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 315 Degrees

:$\sin 315 \degrees = \sin \dfrac {7 \pi} 4 = -\dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 315 \degrees
      | r = \map \sin {360 \degrees - 45 \degrees}
      | c = 
}}
{{eqn | r = -\sin 45 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\dfrac {\sqrt 2} 2
      | c = Sine of $45 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 330 Degrees

:$\sin 330 \degrees = \sin \dfrac {11 \pi} 6 = -\dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 330 \degrees
      | r = \map \sin {360 \degrees - 30 \degrees}
      | c = 
}}
{{eqn | r = -\sin 30 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\dfrac 1 2
      | c = Sine of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of 345 Degrees

:$\sin 345 \degrees = \sin \dfrac {23 \pi} {12} = - \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \sin 345 \degrees
      | r = \map \sin {360 \degrees - 15 \degrees}
      | c = 
}}
{{eqn | r = -\sin 15 \degrees
      | c = Sine of Conjugate Angle
}}
{{eqn | r = -\dfrac {\sqrt 6 - \sqrt 2} 4
      | c = {{sin|15}}
}}
{{end-eqn}}
{{qed}}
Theorem: Sine of Full Angle

:$\sin 360^\circ = \sin 2 \pi = 0$

Proof:
A direct implementation of Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
In this case, $n = 2$ and so:
:$\sin 2 \pi = 0$
{{qed}}
Theorem: Cosine of 15 Degrees

:$\cos 15 \degrees = \cos \dfrac \pi {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 15 \degrees
      | r = \cos \frac {30 \degrees} 2
      | c = 
}}
{{eqn | r = \sqrt {\frac {1 + \cos 30 \degrees} 2}
      | c = Half Angle Formula for Cosine: $\theta$ is in Quadrant I
}}
{{eqn | r = \sqrt {\frac {1 + \frac {\sqrt 3} 2} 2}
      | c = Cosine of $30 \degrees$
}}
{{eqn | r = \sqrt {\frac {2 + \sqrt 3} 4}
      | c = 
}}
{{eqn | r = \sqrt {\frac {8 + 4 \sqrt 3} {16} }
      | c = 
}}
{{eqn | r = \sqrt {\frac {6 + 2 + 2 \sqrt 2 \sqrt 6} {16} }
      | c = 
}}
{{eqn | r = \sqrt {\frac {\paren {\sqrt 6 + \sqrt 2}^2} {16} }
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = positive because $\theta$ is in the first quadrant
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 15 Degrees

:$\cos 15 \degrees = \cos \dfrac \pi {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 15 \degrees
      | r = \map \cos {45 \degrees - 30 \degrees}
      | c = 
}}
{{eqn | r = \cos 45 \degrees \cos 30 \degrees + \sin 45 \degrees \sin 30 \degrees
      | c = Cosine of Difference
}}
{{eqn | r = \paren {\frac {\sqrt 2} 2} \paren {\frac {\sqrt 3} 2} + \paren {\frac {\sqrt 2} 2} \paren {\dfrac 1 2}
      | c = Cosine of $45 \degrees$, Cosine of $30 \degrees$, Sine of $45 \degrees$, Sine of $30 \degrees$
}}
{{eqn | r = \frac {\sqrt 6} 4 + \frac {\sqrt 2} 4
      | c = 
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 30 Degrees

:$\cos 30 \degrees = \cos \dfrac \pi 6 = \dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\cos 30 \degrees}^2
      | r = 1 - \paren {\sin 30 \degrees}^2
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 1 - \paren {\frac 1 2}^2
      | c = {{sin|30}}
}}
{{eqn | r = \frac 3 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cos 30 \degrees
      | r = \sqrt {\frac 3 4}
      | c = 
}}
{{eqn | r = \dfrac {\sqrt 3} 2
      | c = positive because $\cos 30 \degrees$ is in Quadrant I
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 45 Degrees

:$\cos 45 \degrees = \cos \dfrac \pi 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\cos 45 \degrees}^2
      | r = 1 - \paren {\sin 45 \degrees}^2
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = 1 - \paren {\frac {\sqrt 2} 2}^2
      | c = Sine of $45 \degrees$
}}
{{eqn | r = \frac 1 2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cos 45 \degrees
      | r = \sqrt {\frac 1 2}
      | c = positive because $\cos 45 \degrees$ is in Quadrant I
}}
{{eqn | r = \dfrac {\sqrt 2} 2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 60 Degrees

:$\cos 60 \degrees = \cos \dfrac \pi 3 = \dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 60 \degrees
      | r = \map \cos {90 \degrees - 30 \degrees}
      | c = 
}}
{{eqn | r = \sin 30 \degrees
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \dfrac 1 2
      | c = Sine of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 75 Degrees

:$\cos 75^\circ = \cos \dfrac {5 \pi}{12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 75^\circ
      | r = \cos \left({90^\circ - 15^\circ}\right)
      | c = 
}}
{{eqn | r = \sin 15^\circ
      | c = Cosine of Complement equals Sine
}}
{{eqn | r = \dfrac {\sqrt 6 - \sqrt 2} 4
      | c = Sine of $15^\circ$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Right Angle

:$\cos 90 \degrees = \cos \dfrac \pi 2 = 0$

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:
:$\cos \dfrac 1 2 \pi = 0$
{{qed}}
Theorem: Cosine of 105 Degrees

:$\cos 105 \degrees = \cos \dfrac {7 \pi} {12} = - \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 105 \degrees
      | r = \map \cos {90 \degrees + 15 \degrees}
      | c = 
}}
{{eqn | r = -\sin 15 \degrees
      | c = Cosine of Angle plus Right Angle
}}
{{eqn | r = -\frac {\sqrt 6 - \sqrt 2} 4
      | c = Sine of $15 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Angle plus Right Angle

:$\map \cos {x + \dfrac \pi 2} = -\sin x$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cos {x + \frac \pi 2}
      | r = \cos x \cos \frac \pi 2 - \sin x \sin \frac \pi 2
      | c = Cosine of Sum
}}
{{eqn | r = \cos x \cdot 0 - \sin x \cdot 1
      | c = Cosine of Right Angle and Sine of Right Angle
}}
{{eqn | r = -\sin x
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 120 Degrees

:$\cos 120 \degrees = \cos \dfrac {2 \pi} 3 = -\dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 120 \degrees
      | r = \map \cos {90 \degrees + 30 \degrees}
      | c = 
}}
{{eqn | r = -\sin 30 \degrees
      | c = Cosine of Angle plus Right Angle
}}
{{eqn | r = -\frac 1 2
      | c = {{sin|30}}
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 135 Degrees

:$\cos 135 \degrees = \cos \dfrac {3 \pi} 4 = -\dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 135 \degrees
      | r = \map \cos {90 \degrees + 45 \degrees}
      | c = 
}}
{{eqn | r = -\sin 45 \degrees
      | c = Cosine of Angle plus Right Angle
}}
{{eqn | r = -\frac {\sqrt 2} 2
      | c = Sine of $45 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 150 Degrees

:$\cos 150 \degrees = \cos \dfrac {5 \pi} 6 = -\dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 150 \degrees
      | r = \map \cos {90 \degrees + 60 \degrees}
      | c = 
}}
{{eqn | r = -\sin 60 \degrees
      | c = Cosine of Angle plus Right Angle
}}
{{eqn | r = -\frac {\sqrt 3} 2
      | c = Sine of $60 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 165 Degrees

:$\cos 165 \degrees = \cos \dfrac {11 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 165 \degrees
      | r = \map \cos {90 \degrees + 75 \degrees}
      | c = 
}}
{{eqn | r = -\sin 75 \degrees
      | c = Cosine of Angle plus Right Angle
}}
{{eqn | r = -\dfrac {\sqrt 6 + \sqrt 2} 4
      | c = Sine of $75 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Straight Angle

:$\cos 180 \degrees = \cos \pi = -1$

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:
:$\cos \pi = -1^1 = -1$
{{qed}}
Theorem: Cosine of 195 Degrees

:$\cos 195 \degrees = \cos \dfrac {13 \pi} {12} = - \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 195 \degrees
      | r = \cos \paren {360 \degrees - 165 \degrees}
      | c = 
}}
{{eqn | r = \cos 165 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = - \frac {\sqrt 6 + \sqrt 2} 4
      | c = Cosine of $165 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 210 Degrees

:$\cos 210 \degrees = \cos \dfrac {7 \pi} 6 = -\dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 210 \degrees
      | r = \map \cos {360 \degrees - 150 \degrees}
      | c = 
}}
{{eqn | r = \cos 150 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 3} 2
      | c = Cosine of $150 \degrees$
}}
{{end-eqn}}
{{qed}}