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First published on Saturday, Jul 6, 2024 and last modified on Thursday, Apr 10, 2025
Mathedu SAS
In that note, we discover the oriented angles between two vectors, their measure in Radian modulo \( 2\pi\) , and their cosine and sine.
We learn to manipulate the angles and we study many useful trigonometric formulae.
Even if that note is about geometry and contains many drawings, it is also full of calculations, to prove the trigonometric formulae in the whole set \( \mathbb{R}\) .
The unit circle of the euclidean plane is made of the extremities of the “unit vectors”, that are the vectors of norm \( 1\) .
Its center is the origin \( O\) , that is assimilated with the null vector, its ray is \( R=1\) , and its circumference is \( 2\pi\) .
Consider the unit vectors as rays of the unit circle:
The first vector \( \overrightarrow{i}\) of the canonical base, corresponding to the ray \( (OI)\) .
Another unit vector \( \overrightarrow{u}\) , such as \( \left\| \overrightarrow{u} \right\|=1\) , corresponding to the ray \( (OM)\).
Consider the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) .
Its measure in radians is the length of the \( \overset{\frown}{IM}\) , with the following sign:
‘\( +\) ’ if the rotation from \( \overrightarrow{i}\) to \( \overrightarrow{u}\) is counter clockwise: it is the direct direction of rotation,
‘\( -\) ’ if the rotation from \( \overrightarrow{i}\) to \( \overrightarrow{u}\) is clockwise: it is the reverse direction of rotation,
The table 1 gives the angles shown in the figure 3 in Radians, from \( -\pi\) excluded to \( \pi\) included, and in degrees, from \( 0^{{\circ}}\) included to \( 360^{{\circ}}\) excluded.
| Radians | Degrees | Radians | Degrees |
| \( 0\) | \( 0^{{\circ}}\) | \( \pi\) | \( 180^{{\circ}}\) |
| \( \frac{\pi}{2}\) | \( 90^{{\circ}}\) | \( -\frac{\pi}{2}\) | \( 270^{{\circ}}\) |
| \( \frac{\pi}{4}\) | \( 45^{{\circ}}\) | \( -\frac{\pi}{4}\) | \( 315^{{\circ}}\) |
| \( \frac{3\pi}{4}\) | \( 135^{{\circ}}\) | \( -\frac{3\pi}{4}\) | \( 225^{{\circ}}\) |
| \( \frac{\pi}{6}\) | \( 30^{{\circ}}\) | \( -\frac{\pi}{6}\) | \( 330^{{\circ}}\) |
| \( \frac{\pi}{3}\) | \( 60^{{\circ}}\) | \( -\frac{\pi}{3}\) | \( 300^{{\circ}}\) |
| \( \frac{2\pi}{3}\) | \( 120^{{\circ}}\) | \( -\frac{2\pi}{3}\) | \( 240^{{\circ}}\) |
| \( \frac{5\pi}{6}\) | \( 150^{{\circ}}\) | \( -\frac{5\pi}{6}\) | \( 210^{{\circ}}\) |
From now on, the angles will be assimilated with their measures in Radians, without mention of the unit.
Now, we shall define the cosine and sine of the angles we defined in the unit circle.
Consider the three unit vectors:
The vectors \( \overrightarrow{i}\) and \( \overrightarrow{j}\) of the canonical base.
Another unit vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) , where \( (x,y)\in\mathbb{R}^2\) are real numbers such as \( \sqrt{x^2+y^2}=1\) , or simply \( x^2+y^2=1\) .
Consider the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) .
Its cosine and sine are related to the coordinates of \( \overrightarrow{u}\) the following way:
its cosine \( \cos(\theta)\) is equal to the abscissa \( x\) of \( \overrightarrow{u}\) ,
and its sine \( \sin(\theta)\) is equal to the ordinate \( y\) of \( \overrightarrow{u}\)
Because \( x^2+y^2=1\) , we have \( \cos^2(\theta)+\sin^2(\theta)=1\) .
We give in the table 1 the cosine and sine of the angles shown on the figure 3.
| Radians | Cosine | Sine | Radians | Sine | Cosine |
| \( 0\) | \( 1\) | \( 0\) | \( \pi\) | \( -1\) | \( 0\) |
| \( \frac{\pi}{2}\) | \( 0\) | \( 1\) | \( -\frac{\pi}{2}\) | \( 0\) | \( -1\) |
| \( \frac{\pi}{4}\) | \( \frac{\sqrt{2}}{2}\) | \( \frac{\sqrt{2}}{2}\) | \( -\frac{\pi}{4}\) | \( \frac{\sqrt{2}}{2}\) | \( -\frac{\sqrt{2}}{2}\) |
| \( \frac{3\pi}{4}\) | \( -\frac{\sqrt{2}}{2}\) | \( \frac{\sqrt{2}}{2}\) | \( -\frac{3\pi}{4}\) | \( -\frac{\sqrt{2}}{2}\) | \( -\frac{\sqrt{2}}{2}\) |
| \( \frac{\pi}{3}\) | \( \frac{1}{2}\) | \( \frac{\sqrt{3}}{2}\) | \( -\frac{\pi}{3}\) | \( \frac{1}{2}\) | \( -\frac{\sqrt{3}}{2}\) |
| \( \frac{2\pi}{3}\) | \( -\frac{1}{2}\) | \( \frac{\sqrt{3}}{2}\) | \( -\frac{2\pi}{3}\) | \( -\frac{1}{2}\) | \( -\frac{\sqrt{3}}{2}\) |
| \( \frac{\pi}{6}\) | \( \frac{\sqrt{3}}{2}\) | \( \frac{1}{2}\) | \( -\frac{\pi}{6}\) | \( \frac{\sqrt{3}}{2}\) | \( -\frac{1}{2}\) |
| \( \frac{5\pi}{6}\) | \( -\frac{\sqrt{3}}{2}\) | \( \frac{1}{2}\) | \( -\frac{5\pi}{6}\) | \( -\frac{\sqrt{3}}{2}\) | \( -\frac{1}{2}\) |
Consider the ray \( (OM)\) in the first quarter of the unit circle. Its length is equal to \( 1\) .
Consider the oriented angle \( \theta\) of the ray \( (OM)\) with the \( Ox\) axis.
Because the ray \( (OM)\) is in the first quadrant of the unit circle, \( 0< \theta< \frac{\pi}{2}\) .
If \( M'\) is the symmetrical of \( M\) across the \( Ox\) axis then, as may be observed on the figure 6, the following assertions hold:
\( M'\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM'\) is \( -\theta\) ,
the abscissa of \( M'\) is \( \cos(\theta)\) ,
and the ordinate of \( M'\) is \( -\sin(\theta)\) .
We have so proved that:
\( \cos(-\theta)=\cos(\theta)\)
and \( \sin(-\theta)=-\sin(\theta)\)
If \( M''\) is the symmetrical of \( M\) across the \( Oy\) axis then, as may be observed on the figure 7, the following assertions hold:
\( M''\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM^{''}\) is \( \pi-\theta\) ,
the abscissa of \( M''\) is \( -\cos(\theta)\) ,
and the ordinate of \( M''\) is \( \sin(\theta)\) .
We have so proved that:
\( \cos(\pi-\theta)=-\cos(\theta)\)
and \( \sin(\pi-\theta)=\sin(\theta)\)
If \( M^{(3)}\) is the symmetrical of \( M\) around the center \( O\) then, as may be observed on the figure 8, the following assertions hold:
\( M^{(3)}\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM^{(3)}\) is \( \theta+\pi\) ,
the abscissa of \( M^{(3)}\) is \( -\cos(\theta)\) ,
and the ordinate of \( M^{(3)}\) is \( -\sin(\theta)\) .
We have so proved that:
\( \cos(\theta+\pi)=-\cos(\theta)\)
and \( \sin(\theta+\pi)=-\sin(\theta)\)
If \( M^{(4)}\) is the symmetrical of \( M\) across the bisector of the axes of equation \( y=x\) then, as may be observed on the figure 9, the following assertions hold:
\( M^{(4)}\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM^{(4)}\) is \( \frac{\pi}{2}-\theta\) ,
the abscissa of \( M^{(4)}\) is \( \sin(\theta)\) ,
and the ordinate of \( M^{(4)}\) is \( \cos(\theta)\) .
We have so proved that:
\( \cos(\frac{\pi}{2}-\theta)=\sin(\theta)\)
and \( \sin(\frac{\pi}{2}-\theta)=\cos(\theta)\)
If \( M^{(5)}\) is the rotated of \( M\) of angle \( \frac{\pi}{2}\) then, as may be observed on the figure 10, the following assertions hold:
\( M^{(5)}\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM^{(5)}\) is \( \theta+\frac{\pi}{2}\) ,
the abscissa of \( M^{(5)}\) is \( -\sin(\theta)\) ,
and the ordinate of \( M^{(5)}\) is \( \cos(\theta)\) .
We have so proved that:
\( \cos(\theta+\frac{\pi}{2})=-\sin(\theta)\)
and \( \sin(\theta+\frac{\pi}{2})=\cos(\theta)\)
If \( M^{(6)}\) is the rotated of \( M\) of angle \( -\frac{\pi}{2}\) then, as may be observed on the figure 11, the following assertions hold:
\( M^{(6)}\) is on the unit circle,
the oriented angle from the \( Ox\) axis to the ray \( OM^{(6)}\) is \( \theta-\frac{\pi}{2}\) ,
the abscissa of \( M^{(6)}\) is \( \sin(\theta)\) ,
and the ordinate of \( M^{(6)}\) is \( -\cos(\theta)\) .
We have so proved that:
\( \cos(\theta+\frac{\pi}{2})=\sin(\theta)\)
and \( \sin(\theta+\frac{\pi}{2})=-\cos(\theta)\)
We have defined \( \cos(\theta)\) and \( \sin(\theta)\) for \( \theta\in(-\pi,\pi]\) . Let’s define them for any \( \theta\in\mathbb{R}\) .
Assume \( \theta\in(-\pi,\pi]\) is a real number such as \( -\pi<\theta\le\pi\) . It is the measure in Radians of an oriented angle in the unit circle.
Then we may note that:
If we add \( 2\pi\) to \( \theta\) , we make one complete turnaround in the counterclockwise direction, and the physical angle is unchanged.
If we subtract \( 2\pi\) to \( \theta\) , we make one complete turnaround in the clockwise direction, and the physical angle is unchanged.
And generally, for any integer \( k\in\mathbb{Z}\) , if we add \( 2k\pi\) to \( \theta\) , the physical angle is unchanged.
Reversely, assume \( \theta\in\mathbb{R}\) is any real number.
Then there exists some integer \( k\in\mathbb{Z}\) such as, if we denote \( \theta_0=\theta-2k\pi\) , then \( -\pi<\theta_{0}\le\pi\) .
Proof (of the previous assertion)
Assume \( \theta\in\mathbb{R}\) is any real number.
Then we are in one of the following cases=
\( -\pi<\theta\le\pi\) , and the assertion is true for \( k=0\) and \( \theta_{0}=\theta\) ,
or \( \theta>\pi\) , and the assertion is true for \( k\) being the entire part of \( \frac{\theta}{2\pi}\) and \( \theta_{0}=\theta-2k\pi\) ,
or else \( \theta\le -\pi\) , so that \( -\theta\ge \pi\) and the assertion is true for \( k\) being the opposite of the entire part of \( \frac{-\theta}{2\pi}\) and \( \theta_{0}=\theta-2k\pi\) .
As a consequence of the previous assertion, as \( \theta=\theta_0+2k\pi\) , \( \theta\) measures the same angle as , and we may define:
\( \cos(\theta)\) as equal to \( \cos(\theta_{0})\) ,
and \( \sin(\theta)\) as equal to \( \sin(\theta_{0})\)
The angles modulo \( 2\pi\) :
Assume \( (\theta,\mu)\in\mathbb{R}^{2}\) are real numbers.
Then, the following assertions hold:
If \( \theta\equiv\mu \mod 2\pi\) , then \( \cos(\theta)=\cos(\mu)\) and \( \sin(\theta)=\sin(\mu)\) .
And reversely, if \( \cos(\theta)=\cos(\mu)\) and \( \sin(\theta)=\sin(\mu)\) , then there exists some integer \( k\in\mathbb{Z}\) such as \( \theta=\mu+2k\pi\) .
That may be said as:
The cosine and the sine of an angle determine that angle modulo \( 2\pi\) .
Proof (of the last assertion)
Assume \( (\theta,\mu)\in\mathbb{R}^{2}\) are real numbers.
Assume \( \theta\equiv\mu \mod 2\pi\) . Then, by definition of the congruence, there exists some integer \( k\in\mathbb{Z}\) such as \( \theta=\mu+2k\pi\) .
Consider the integer \( k_{1}\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k_{1}\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
Let’s denote \( k_{2}=k_{1}+k\) , so that \( \mu-2k_{2}\pi=\theta_{0}\) , that is in the interval \( (-\pi,\pi]\) .
Consequently, \( \cos(\mu)=\cos(\theta_{0})=\cos(\theta)\) and \( \sin(\mu)=\sin(\theta_{0})=\sin(\theta)\) .
Assume now \( \cos(\theta)=\cos(\mu)\) and \( \sin(\theta)=\sin(\mu)\) .
Consider the integer \( k_{1}\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k_{1}\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
Consider also the integer \( k_{2}\in\mathbb{Z}\) such as \( \mu_{0}=\theta-2k_{2}\pi\) is such as \( -\pi<\mu_{0}\le\pi\) , so that \( \cos(\mu)=\cos(\mu_{0})\) and \( \sin(\mu)=\sin(\mu_{0})\) .
Then, because \( \cos(\theta)=\cos(\mu)\) and \( \sin(\theta)=\sin(\mu)\) , \( \cos(\theta_{0})=\cos(\mu_{0})\) and \( \sin(\theta_{0})=\sin(\mu_{0})\) .
Thus, as the angle, between \( -\pi\) excluded and \( \pi\) included, with the \( Ox\) axis of a unit vector with some abscissa and ordinate is unique, \( \theta_{0}=\mu_{0}\) .
Consequently, if we consider the integer \( k=k_{1}-k_{2}\) , \( \theta=\mu+2k\pi\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then \( \cos^2(\theta)+\sin^2(\theta)=1\) .
Proof (of the fundamental trigonometric identity)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
Then, because of the formula at the end of the paragraph 3.1, \( \cos^2(\theta_{0})+\sin^2(\theta_{0})=1\) , so that \( \cos^2(\theta)+\sin^2(\theta)=1\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos(-\theta)=\cos(\theta)\) : the cosine function is even.
\( \sin(-\theta)=-\sin(\theta)\) : the sine function is odd.
Proof (of the parity of the cosine and sine fuctions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos(-\theta_{0})=\cos(\theta_{0})\) and \( \sin(-\theta_{0})=-\sin(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos(-\theta_{0})=\cos(0)=1=\cos(\theta_{0})\) ,
and \( \sin(-\theta_{0})=\sin(0)=0=-0=-\sin(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.2 applies for \( \theta_{0}\) .
Consequently, \( \cos(-\theta_{0})=\cos(\theta_{0})\) and \( \sin(-\theta_{0})=-\sin(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos(-\theta_{1})=\cos(\theta_{1})\) and \( \sin(-\theta_{1})=-\sin(\theta_{1})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of pragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( -\theta_{0}=-\theta_{1}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7,
\( \cos(-\theta_{0})=\sin(-\theta_{1})=-\sin(\theta_{1})\) and \( \sin(-\theta_{0})=-\cos(-\theta_{1})=-\cos(\theta_{1})\)
Consequently, \( \cos(-\theta_{0})=\cos(\theta_{0})\) and \( \sin(-\theta_{0})=-\sin(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos(-\theta_{2})=\cos(\theta_{2})\) and \( \sin(-\theta_{2})=-\sin(\theta_{2})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( -\theta_{0}=-\theta_{2}+\frac{\pi}{2}\) so that, because of pragraph 3.3.6,
\( \cos(-\theta_{0})=-\sin(-\theta_{2})=\sin(\theta_{2})\) and \( \sin(-\theta_{0})=\cos(-\theta_{2})=\cos(\theta_{2})\)
Consequently, \( \cos(-\theta_{0})=\cos(\theta_{0})\) and \( \sin(-\theta_{0})=-\sin(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos(\pi-\theta_{0})=\cos(\theta_{00})\) and \( \sin(\pi-\theta_{0})=-\sin(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi=\theta_{0}+\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos(-\theta_{3})=\cos(\theta_{3})\) and \( \sin(-\theta_{3})=-\sin(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( -\theta_{0}=-\theta_{3}+\pi\) so that, because of paragraph 3.3.4,
\( \cos(-\theta_{0})=-\cos(-\theta_{3})=-\cos(\theta_{3})\) and \( \sin(-\theta_{0})=-\sin(-\theta_{3})=\sin(\theta_{3})\)
Consequently, \( \cos(-\theta_{0})=\cos(\theta_{00})\) and \( \sin(-\theta_{0})=-\sin(\theta_{00})\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos(\pi-\theta)=-\cos(\theta)\) .
\( \sin(\pi-\theta)=\sin(\theta)\) .
Proof (of the previous assertions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos(\pi-\theta_{0})=-\cos(\theta_{0})\) and \( \sin(\pi-\theta_{0})=\sin(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos(\pi-\theta_{0})=\cos(\pi)=-1=-\cos(\theta_{0})\) , and \( \sin(\pi-\theta_{0})=\sin(\pi)=0=\sin(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.3 applies for \( \theta_{0}\) .
Consequently, \( \cos(\pi-\theta_{0})=-\cos(\theta_{0})\) and \( \sin(\pi-\theta_{0})=\sin(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\pi-\theta_{1})=-\cos(\theta_{1})\) and \( \sin(\pi-\theta_{1})=\sin(\theta_{1})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( \pi-\theta_{0}=\pi-\theta_{1}-\frac{\pi}{2}=-\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6,
\( \cos(\pi-\theta_{0})=-\sin(-\theta_{1})=\sin(\theta_{1})\) and \( \sin(\pi-\theta_{0})=\cos(-\theta_{1})=\cos(\theta_{1})\)
Consequently, \( \cos(\pi-\theta_{0})=-\cos(\theta_{0})\) and \( \sin(\pi-\theta_{0})=\sin(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\pi-\theta_{2})=-\cos(\theta_{2})\) and \( \sin(\pi-\theta_{2})=\sin(\theta_{2})\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos(\pi-\theta_{00})=-\cos(\theta_{0})\) and \( \sin(\pi-\theta_{00})=\sin(\theta_{0})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( \pi-\theta_{00}=\pi-\theta_{0}-2\pi=-\theta_{0}-\pi=-\theta_{2}-\frac{\pi}{2}\)
so that, because of paragraph 3.3.7,
\( \cos(\pi-\theta_{00})=\sin(-\theta_{2})=-\sin(\theta_{2})\) and \( \sin(\pi-\theta_{00})=-\cos(-\theta_{2})=-\cos(\theta_{2})\) .
Consequently, \( \cos(\pi-\theta_{00})=-\cos(\theta_{0})\) and \( \sin(\pi-\theta_{00})=\sin(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos(\pi-\theta_{00})=-\cos(\theta_{00})\) and \( \sin(\pi-\theta_{00})=\sin(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\pi-\theta_{3})=-\cos(\theta_{3})\) and \( \sin(\pi-\theta_{3})=\sin(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( \pi-\theta_{00}=-\theta_{3}\) so that, because of paragraph 3.3.2, \( \cos(\pi-\theta_{00})=\cos(\theta_{3})\) and \( \sin(\pi-\theta_{00})=-\sin(\theta_{3})\)
Consequently, \( \cos(\pi-\theta_{00})=-\cos(\theta_{00})\) and \( \sin(\pi-\theta_{00})=\sin(\theta_{00})\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos(\theta+\pi)=-\cos(\theta)\) .
\( \sin(\theta+\pi)=-\sin(\theta)\) .
Proof (of the previous assertions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos(\theta_{0}+\pi)=-\cos(\theta_{0})\) and \( \sin(\theta_{0}+\pi)=-\sin(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos(\theta_{0}+\pi)=\cos(\pi)=-1=-\cos(\theta_{0})\) , and \( \sin(\theta_{0}+\pi)=\sin(\pi)=0=-0=-\sin(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.4 applies for \( \theta_{0}\) .
Consequently, \( \cos(\theta_{0}+\pi)=-\cos(\theta_{0})\) and \( \sin(\theta_{0}+\pi)=-\sin(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\theta_{1}+\pi)=-\cos(\theta_{1}))\) and \( \sin(\theta_{1}+\pi)=-\sin(\theta_{1})\) .
Consider \( \theta_{00}=\theta_{0}-2\pi\) , and lets prove that
\( \cos(\theta_{00}+\pi)=-\cos(\theta_{0})\) and \( \sin(\theta_{00}+\pi)=-\sin(\theta_{0})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( \theta_{00}+\pi=\theta_{0}-2\pi+\pi=\theta_{0}-\pi=\theta_{1}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7,
\( \cos(\theta_{00}+\pi)=\sin(\theta_{1})\) and \( \sin(\theta_{00}+\pi)=-\cos(\theta_{1})\)
Consequently, \( \cos(\theta_{00}+\pi)=-\cos(\theta_{0})\) and \( \sin(\theta_{00}+\pi)=-\sin(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\theta_{2}+\pi)=-\cos(\theta_{2})\) and \( \sin(\theta_{2}+\pi)=\sin(\theta_{2})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( \theta_{0}+\pi=\theta_{2}-\frac{\pi}{2}+\pi=\theta_{2}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0}+\pi)=-\sin(\theta_{2})\) and \( \sin(\theta_{0}+\pi)=\cos(\theta_{2})\)
Consequently, \( \cos(\theta_{0}+\pi)=-\cos(\theta_{0})\) and \( \sin(\theta_{0}+\pi)=-\sin(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos(\theta_{0}+\pi)=-\cos(\theta_{00})\) and \( \sin(\theta_{0}+\pi)=-\sin(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi=\theta_{0}+\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos(\theta_{3}+\pi)=-\cos(\theta_{3})\) and \( \sin(\theta_{3}+\pi)=-\sin(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( \theta_{0}+\pi=\theta_{3}\) so that \( \cos(\theta_{0}+\pi)=\cos(\theta_{3})\) and \( \sin(\theta_{0}+\pi)=\sin(\theta_{3})\).
Consequently, \( \cos(\theta_{0}+\pi)=-\cos(\theta_{00})\) and \( \sin(\theta_{0}+\pi)=-\sin(\theta_{00})\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)\) .
\( \sin\left(\frac{\pi}{2}-\theta\right)=\cos(\theta)\) .
Proof (of the previous assertions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=\sin(\theta_{0})\) and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\cos(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=\cos\left(\frac{\pi}{2}\right)=0=\sin(\theta_{0})\) ,
and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\sin\left(\frac{\pi}{2}\right)=1=\cos(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.5 applies for \( \theta_{0}\) .
Consequently, \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=\sin(\theta_{0})\) and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\sin\left(\frac{\pi}{2}\right)=1=\cos(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\frac{\pi}{2}-\theta_{1}\right)=\sin(\theta_{1})\)
and \( \sin\left(\frac{\pi}{2}-\theta_{1}\right)=\cos(\theta_{1})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( \frac{\pi}{2}-\theta_{0}=\frac{\pi}{2}-\theta_{1}-\frac{\pi}{2}=-\theta_{1}\) ,
so that \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=-\cos(-\theta_{1})=\cos(\theta_{1})\)
and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\sin(-\theta_{1})=-\sin(\theta_{1})\)
Consequently, \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=\sin(\theta_{0})\) and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\cos(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\frac{\pi}{2}-\theta_{2}\right)=\sin(\theta_{2})\)
and \( \sin\left(\frac{\pi}{2}-\theta_{2}\right)=\cos(\theta_{2})\) .
and \( \sin\left(\frac{\pi}{2}-\theta_{2}\right)=\cos(\theta_{2})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( \frac{\pi}{2}-\theta_{0}=\frac{\pi}{2}-\theta_{2}+\frac{\pi}{2}=-\theta_{2}+\pi\), so that, because of paragraph 3.3.4,
\( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=-\cos(-\theta_{2})=-\cos(\theta_{2})\) and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=-\sin(-\theta_{2})=\sin(\theta_{2})\)
Consequently, \( \cos\left(\frac{\pi}{2}-\theta_{0}\right)=\sin(\theta_{0})\) and \( \sin\left(\frac{\pi}{2}-\theta_{0}\right)=\cos(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos\left(\frac{\pi}{2}-\theta_{00}\right)=\sin(\theta_{00})\) and \( \sin\left(\frac{\pi}{2}-\theta_{00}\right)=\cos(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\frac{\pi}{2}-\theta_{3}\right)=\sin(\theta_{3})\) and \( \sin\left(\frac{\pi}{2}-\theta_{3}\right)=\cos(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( \frac{\pi}{2}-\theta_{00}=\frac{\pi}{2}-\theta_{3}-\pi=-\theta_{3}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7,
\( \cos\left(\frac{\pi}{2}-\theta_{00}\right)=\sin(-\theta_{3})=-\sin(\theta_{3})\) and \( \sin\left(\frac{\pi}{2}-\theta_{00}\right)=-\cos(-\theta_{3})=-\cos(\theta_{3})\)
Consequently, \( \cos\left(\frac{\pi}{2}-\theta_{00}\right)=\sin(\theta_{00})\) and \( \sin\left(\frac{\pi}{2}-\theta_{00}\right)=\cos(\theta_{00})\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos\left(\theta+\frac{\pi}{2}\right)=-\sin(\theta)\) .
\( \sin\left(\theta+\frac{\pi}{2}\right)=\cos(\theta)\) .
Proof (of the previous assertions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{0})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=\cos\left(\frac{\pi}{2}\right)=0=-0=-\sin(\theta_{0})\) ,
and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1=\cos(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.6 applies for \( \theta_{0}\) .
Consequently, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{0})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{1}+\frac{\pi}{2}\right)=-\sin(\theta_{1})\) and \( \sin\left(\theta_{1}+\frac{\pi}{2}\right)=\cos(\theta_{1})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( \theta_{0}+\frac{\pi}{2}=\theta_{1}+\frac{\pi}{2}+\frac{\pi}{2}=\theta_{1}+\pi\) so that, because of paragraph 3.3.4,
\( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\cos(\theta_{1})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{1})\)
Consequently, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{0})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{2}+\frac{\pi}{2}\right)=-\sin(\theta_{2})\) and \( \sin\left(\theta_{2}+\frac{\pi}{2}\right)=\cos(\theta_{2})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( \theta_{0}+\frac{\pi}{2}=\theta_{2}\) so that, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{2})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\sin(\theta_{2})\)
Consequently, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{0})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{00})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi=\theta_{0}+\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{3}+\frac{\pi}{2}\right)=-\sin(\theta_{3})\) and \( \sin\left(\theta_{3}+\frac{\pi}{2}\right)=\cos(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( \theta_{0}+\frac{\pi}{2}=\theta_{3}-\pi+\frac{\pi}{2}=\theta_{3}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=\sin(\theta_{3})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=-\cos(\theta_{3})\)
Consequently, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{00})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{00})\) .
Assume \( \theta\in\mathbb{R}\) is a real number.
Then the following assertions hold:
\( \cos\left(\theta-\frac{\pi}{2}\right)=\sin(\theta)\) .
\( \sin\left(\theta-\frac{\pi}{2}\right)=-\cos(\theta)\) .
Proof (of the previous assertions)
Assume \( \theta\in\mathbb{R}\) is a real number.
Consider the integer \( k\in\mathbb{Z}\) such as \( \theta_{0}=\theta-2k\pi\) is such as \( -\pi<\theta_{0}\le\pi\) , so that \( \cos(\theta)=\cos(\theta_{0})\) and \( \sin(\theta)=\sin(\theta_{0})\) .
We have to prove that \( \cos\left(\theta_{0}-\frac{\pi}{2}\right)=\sin(\theta_{0})\) and \( \sin\left(\theta_{0}-\frac{\pi}{2}\right)=-\cos(\theta_{0})\) .
Assume \( 0\le\theta_{0}<\frac{\pi}{2}\) .
If \( \theta_{0}=0\) , then \( \cos\left(\theta_{0}-\frac{\pi}{2}\right)=\cos\left(-\frac{\pi}{2}\right)=0=\sin(\theta_{0})\) ,
and \( \sin\left(\theta_{0}-\frac{\pi}{2}\right)=\sin\left(-\frac{\pi}{2}\right)=-1=-\cos(\theta_{0})\) .
And if \( 0<\theta_{0}<\frac{\pi}{2}\) , we are in the first quadrant and the paragraph 3.3.7 applies for \( \theta_{0}\) .
Consequently, \( \cos\left(\theta_{0}+\frac{\pi}{2}\right)=-\sin(\theta_{0})\) and \( \sin\left(\theta_{0}+\frac{\pi}{2}\right)=\cos(\theta_{0})\) .
Assume \( \frac{\pi}{2}\le\theta_{0}<\pi\) .
Consider \( \theta_{1}=\theta_{0}-\frac{\pi}{2}\) .
Then \( 0\le\theta_{1}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{1}-\frac{\pi}{2}\right)=\sin(\theta_{1})\) and \( \sin\left(\theta_{1}-\frac{\pi}{2}\right)=-\cos(\theta_{1})\) .
But \( \theta_{0}=\theta_{1}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos(\theta_{0})=-\sin(\theta_{1})\) and \( \sin(\theta_{0})=\cos(\theta_{1})\) .
And \( \theta_{0}-\frac{\pi}{2}=\theta_{1}\) so that \( \cos\left(\theta_{0}-\frac{\pi}{2}\right)=\cos(\theta_{1})\) and \( \sin\left(\theta_{0}-\frac{\pi}{2}\right)=\sin(\theta_{1})\)
Consequently, \( \cos\left(\theta_{0}-\frac{\pi}{2}\right)=\sin(\theta_{0})\) and \( \sin\left(\theta_{0}-\frac{\pi}{2}\right)=-\cos(\theta_{0})\) .
Assume \( -\frac{\pi}{2}\le\theta_{0}<0\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=\sin(\theta_{0})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=-\cos(\theta_{0})\) .
Consider \( \theta_{2}=\theta_{0}+\frac{\pi}{2}\) .
Then \( 0\le\theta_{2}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{2}-\frac{\pi}{2}\right)=\sin(\theta_{2})\) and \( \sin\left(\theta_{2}-\frac{\pi}{2}\right)=-\cos(\theta_{2})\) .
But \( \theta_{0}=\theta_{2}-\frac{\pi}{2}\) so that, because of paragraph 3.3.7, \( \cos(\theta_{0})=\sin(\theta_{2})\) and \( \sin(\theta_{0})=-\cos(\theta_{2})\) .
And \( \theta_{00}-\frac{\pi}{2}=\theta_{2}-\frac{\pi}{2}+2\pi-\frac{\pi}{2}=\theta_{2}+\pi\) so that, because of paragraph 3.3.4,
\( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=-\cos(\theta_{2})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=-\sin(\theta_{2})\)
Consequently, \( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=\sin(\theta_{0})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=-\cos(\theta_{0})\) .
Assume \( -\pi\le\theta_{0}<\frac{\pi}{2}\) .
Consider \( \theta_{00}=\theta_{0}+2\pi\) , and lets prove that
\( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=-\sin(\theta_{00})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=\cos(\theta_{00})\) .
Consider \( \theta_{3}=\theta_{00}-\pi=\theta_{0}+\pi\) .
Then \( 0\le\theta_{3}<\frac{\pi}{2}\) , and because of item (I), \( \cos\left(\theta_{3}-\frac{\pi}{2}\right)=\sin(\theta_{3})\) and \( \sin\left(\theta_{3}-\frac{\pi}{2}\right)=-\cos(\theta_{3})\) .
But \( \theta_{00}=\theta_{3}+\pi\) so that, because of paragraph 3.3.4, \( \cos(\theta_{00})=-\cos(\theta_{3})\) and \( \sin(\theta_{00})=-\sin(\theta_{3})\) .
And \( \theta_{00}-\frac{\pi}{2}=\theta_{3}+\pi-\frac{\pi}{2}=\theta_{3}+\frac{\pi}{2}\) so that, because of paragraph 3.3.6, \( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=-\sin(\theta_{3})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=\cos(\theta_{3})\)
Consequently, \( \cos\left(\theta_{00}-\frac{\pi}{2}\right)=\sin(\theta_{00})\) and \( \sin\left(\theta_{00}-\frac{\pi}{2}\right)=-\cos(\theta_{00})\) .
We sum up the trigonometric formulae given in the table 2 below.
| \( \forall\; \theta\in\mathbb{R}\) | \( \cos^2(\theta)+\sin^2(\theta)=1\) | ||
| \( \cos(-\theta)=\cos(\theta)\) | \( \sin(-\theta)=-\sin(\theta)\) | \( \cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)\) | \( \sin\left(\frac{\pi}{2}-\theta\right)=\cos(\theta)\) |
| \( \cos(\pi-\theta)=-\cos(\theta)\) | \( \sin(\pi-\theta)=\sin(\theta)\) | \( \cos\left(\theta+\frac{\pi}{2}\right)=-\sin(\theta)\) | \( \sin\left(\theta+\frac{\pi}{2}\right)=\cos(\theta)\) |
| \( \cos(\theta+\pi)=-\cos(\theta)\) | \( \sin(\theta+\pi)=-\sin(\theta)\) | \( \cos\left(\theta-\frac{\pi}{2}\right)=\sin(\theta)\) | \( \sin\left(\theta-\frac{\pi}{2}\right)=-\cos(\theta)\) |
Assume \( (x,y)\in(\mathbb{R}_+^*)^2\) are real numbers such as \( x>0\) and \( y>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the first quadrant of the plane.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) . is a unit vector positively aligned with the vector \( \overrightarrow{u}\) , and it is in the first quadrant of the plane as well.
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( 0\) and \( \frac{\pi}{2}\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) .
Proof (of the previous assertions)
Assume \( (x,y)\in(\mathbb{R}_+^*)^2\) are real numbers such as \( x>0\) and \( y>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the first quadrant of the plane.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|>0\) because \( \overrightarrow{u}\ne\overrightarrow{0}\) , \( \overrightarrow{u_0}\) is positively aligned with the vector \( \overrightarrow{u}\) .
And \( \left\| \overrightarrow{u_{0}} \right\|=\frac{\left\| \overrightarrow{u} \right\|} {\left\| \overrightarrow{u} \right\|}=1\) , so that the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector.
Moreover, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+y^{2}}\) , the coordinates of the vector \( \overrightarrow{u_{0}}\) are the following:
its abscissa \( x_{0}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) , that is positive becasue \( x>0\) ,
and its ordinate \( y_{0}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) , that is positive becasue \( y>0\) .
Consequently, the vector \( \overrightarrow{u_{0}}\) is in the first quadrant of the plane.
And because of that, the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( 0\) and \( \frac{\pi}{2}\) (modulo \( 2\pi\) ).
Assume \( (x,y)\in\mathbb{R}_-^*\times\mathbb{R}_+^*\) are real numbers such as \( x<0\) and \( y>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the second quadrant of the plane.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector positively aligned with the vector \( \overrightarrow{u}\) , and it is in the second quadrant of the plane as well.
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( \frac{\pi}{2}\) and \( \pi\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) .
Proof (of the previous assertions)
Assume \( (x,y)\in\mathbb{R}_-^*\times\mathbb{R}_+^*\) are real numbers such as \( x<0\) and \( y>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the second quadrant of the plane.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|>0\) because \( \overrightarrow{u}\ne\overrightarrow{0}\) , \( \overrightarrow{u_0}\) is positively aligned with the vector \( \overrightarrow{u}\) .
And \( \left\| \overrightarrow{u_{0}} \right\|=\frac{\left\| \overrightarrow{u} \right\|} {\left\| \overrightarrow{u} \right\|}=1\) , so that the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector.
Moreover, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+y^{2}}\) , the coordinates of the vector \( \overrightarrow{u_{0}}\) are the following:
its abscissa \( x_{0}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) , that is negative becasue \( x<0\) ,
and its ordinate \( y_{0}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) , that is positive becasue \( y>0\) .
Consequently, the vector \( \overrightarrow{u_{0}}\) is in the second quadrant of the plane.
And because of that, the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_{0}})}\) is strictly comprised between \( \frac{\pi}{2}\) and \( \pi\) (modulo \( 2\pi\) ).
Assume \( (x,y)\in(\mathbb{R}_-^*)^2\) are real numbers such as \( x<0\) and \( y<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the third quadrant of the plane.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector positively aligned with the vector \( \overrightarrow{u}\) , and it is in the third quadrant of the plane as well.
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( -\pi\) and \( -\frac{\pi}{2}\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) .
Proof (of the previous assertions)
Assume \( (x,y)\in(\mathbb{R}_-^*)^2\) are real numbers such as \( x<0\) and \( y<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the third quadrant of the plane.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|>0\) because \( \overrightarrow{u}\ne\overrightarrow{0}\) , \( \overrightarrow{u_0}\) is positively aligned with the vector \( \overrightarrow{u}\) .
And \( \left\| \overrightarrow{u_{0}} \right\|=\frac{\left\| \overrightarrow{u} \right\|} {\left\| \overrightarrow{u} \right\|}=1\) , so that the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector.
Moreover, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+y^{2}}\) , the coordinates of the vector \( \overrightarrow{u_{0}}\) are the following:
its abscissa \( x_{0}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) , that is negative becasue \( x<0\) ,
and its ordinate \( y_{0}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) , that is negative becasue \( y<0\) .
Consequently, the vector \( \overrightarrow{u_{0}}\) is in the third quadrant of the plane.
And because of that, the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( -\pi\) and \( -\frac{\pi}{2}\) (modulo \( 2\pi\) ).
Assume \( (x,y)\in\mathbb{R}_+^*\times\mathbb{R}_-^*\) are real numbers such as \( x>0\) and \( y<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the fourth quadrant of the plane.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector positively aligned with the vector \( \overrightarrow{u}\) , and it is in the fourth quadrant of the plane as well.
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( -\frac{\pi}{2}\) and \( 0\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) .
Proof (of the previous assertions)
Assume \( (x,y)\in\mathbb{R}_+^*\times\mathbb{R}_-^*\) are real numbers such as \( x>0\) and \( y<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) in the third quadrant of the plane.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|>0\) because \( \overrightarrow{u}\ne\overrightarrow{0}\) , \( \overrightarrow{u_0}\) is positively aligned with the vector \( \overrightarrow{u}\) .
And \( \left\| \overrightarrow{u_{0}} \right\|=\frac{\left\| \overrightarrow{u} \right\|} {\left\| \overrightarrow{u} \right\|}=1\) , so that the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is a unit vector.
Moreover, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+y^{2}}\) , the coordinates of the vector \( \overrightarrow{u_{0}}\) are the following:
its abscissa \( x_{0}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) , that is positive becasue \( x>0\) ,
and its ordinate \( y_{0}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) , that is negative becasue \( y<0\) .
Consequently, the vector \( \overrightarrow{u_{0}}\) is in the fourth quadrant of the plane.
And because of that, the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is strictly comprised between \( -\frac{\pi}{2}\) and \( 0\) (modulo \( 2\pi\) ).
Assume \( x\in\mathbb{R}_+^*\) is a real number such as \( x>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\0\end{bmatrix}\) positively aligned with the \( x\) axis.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is equal to the vector \( \overrightarrow{i}\) , that is itself positively aligned with the vector \( \overrightarrow{u}\) .
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is equal to \( 0\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=0\) .
Proof (of the previous assertions)
Assume \( x\in\mathbb{R}_+^*\) is a real number such as \( x>0\) and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\0\end{bmatrix}\) positively aligned with the \( x\) axis.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+0^{2}}=x\) because \( x>0\) , \( \overrightarrow{u_0}=\begin{bmatrix}\frac{x}{x}\\0\end{bmatrix} =\begin{bmatrix}1\\0\end{bmatrix}=\overrightarrow{i}\) , that is positively aligned with the \( x\) axis, as is the vector \( \overrightarrow{u}\) .
Moreover, \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\widehat{(\overrightarrow{i},\overrightarrow{i})}\) is equal to \( 0\) (modulo \( 2\pi\) ).
Assume \( y\in\mathbb{R}_+^*\) is a real number such as \( y>0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}0\\y\end{bmatrix}\) positively aligned with the \( y\) axis.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is equal to the vector \( \overrightarrow{j}\) , that is itself positively aligned with the vector \( \overrightarrow{u}\) .
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is equal to \( \frac{\pi}{2}\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\frac{\pi}{2}\) .
Proof (of the previous assertions)
Assume \( y\in\mathbb{R}_+^*\) is a real number such as \( y>0\) and consider the vector \( \overrightarrow{u}=\begin{bmatrix}0\\y\end{bmatrix}\) positively aligned with the \( y\) axis.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|=\sqrt{0^{2}+y^{2}}=y\) because \( y>0\) , \( \overrightarrow{u_0}=\begin{bmatrix}0\\\frac{y}{y}\end{bmatrix} =\begin{bmatrix}0\\1\end{bmatrix}=\overrightarrow{j}\) , that is positively aligned with the \( y\) axis, as is the vector \( \overrightarrow{u}\) .
Moreover, \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\widehat{(\overrightarrow{i},\overrightarrow{j})}\) is equal to\( \frac{\pi}{2}\) (modulo \( 2\pi\) ).
Assume \( x\in\mathbb{R}_-^*\) is a real number such as \( x<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\0\end{bmatrix}\) negatively aligned with the \( x\) axis.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is equal to the vector \( -\overrightarrow{i}\) , that is itself positively aligned with the vector \( \overrightarrow{u}\) .
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is equal to \( \pi\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\pi\) .
Proof (of the previous assertions)
Assume \( x\in\mathbb{R}_-^*\) is a real number such as \( x<0\) and consider the vector \( \overrightarrow{u}=\begin{bmatrix}x\\0\end{bmatrix}\) negatively aligned with the \( x\) axis.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+0^{2}}=-x\) because \( x<0\) , \( \overrightarrow{u_0}=\begin{bmatrix}\frac{x}{-x}\\0\end{bmatrix} =\begin{bmatrix}-1\\0\end{bmatrix}=-\overrightarrow{i}\) , that is negatively aligned with the \( x\) axis, as is the vector \( \overrightarrow{u}\) .
Moreover, \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\widehat{(\overrightarrow{i},-\overrightarrow{i})}\) is equal to \( \pi\) (modulo \( 2\pi\) ).
Assume \( y\in\mathbb{R}_-^*\) is a real number such as \( y<0\) , and consider the vector \( \overrightarrow{u}=\begin{bmatrix}0\\y\end{bmatrix}\) negatively aligned with the \( y\) axis.
Then the following assertions hold:
The vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) is equal to the vector \( -\overrightarrow{j}\) , that is itself positively aligned with the vector \( \overrightarrow{u}\) .
The oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}\) is equal to \( -\frac{\pi}{2}\) (modulo \( 2\pi\) ).
Then we define the oriented angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) of the vector \( \overrightarrow{u}\) with the \( x\) axis as \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=-\frac{\pi}{2}\) .
Proof (of the previous assertions)
Assume \( y\in\mathbb{R}_-^*\) is a real number such as \( y<0\) and consider the vector \( \overrightarrow{u}=\begin{bmatrix}0\\y\end{bmatrix}\) negatively aligned with the \( y\) axis.
Consider the vector \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Then, as \( \left\| \overrightarrow{u} \right\|=\sqrt{0^{2}+y^{2}}=-y\) because \( y<0\) , \( \overrightarrow{u_0}=\begin{bmatrix}0\\\frac{y}{-y}\end{bmatrix} =\begin{bmatrix}0\\{-1}\end{bmatrix}=-\overrightarrow{j}\) , that is negatively aligned with the \( y\) axis, as is the vector \( \overrightarrow{u}\) .
Moreover, \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u_0})}=\widehat{(\overrightarrow{i},-\overrightarrow{j})}\) is equal to \( -\frac{\pi}{2}\) (modulo \( 2\pi\) ).
Assume \( \overrightarrow{u}\in\mathbb{P}^*\) is a non zero vector, and consider the unit vector \( \overrightarrow{u}_0=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) .
Denote \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) , were \( (x,y)\in\mathbb{R}^2\) are the cartesian coordinates of the vector \( \overrightarrow{u}\) .
Then, as \( \left\| \overrightarrow{u} \right\|=\sqrt{x^{2}+y^{2}}\) the cartesian coordinates of the vector \( \overrightarrow{u}_0\) are respectively:
its abscissa \( x_0=\frac{x}{\sqrt{x^2+y^2}}\) ,
and its ordinate \( y_0=\frac{y}{\sqrt{x^2+y^2}}\) .
Consider the angle \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})} =\widehat{(\overrightarrow{i},\overrightarrow{u}_0)}\) .
Then the cartesian coordinates of the vector \( \overrightarrow{u}_0\) are also:
its abscissa \( x_0=\cos(\theta)\) ,
and its ordinate \( y_0=\sin(\theta)\) .
Consequently, if \( \overrightarrow{u}=\begin{bmatrix}x\\y\end{bmatrix}\) and \( \theta=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) , then \( \cos(\theta)=\frac{x}{\sqrt{x^2+y^2}}\) and \( \sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}\) .
Assume \( (\overrightarrow{u},\overrightarrow{v}) \in(\mathbb{P}^*)^2\) are non zero vectors, and consider their angles with the \( x\) axis:
\( \theta_u=\widehat{(\overrightarrow{i},\overrightarrow{u})}\)
and \( \theta_v=\widehat{(\overrightarrow{i},\overrightarrow{v})}\)
Then we may define the angle between \( \overrightarrow{u}\) and \( \overrightarrow{v}\) as \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}=\theta_v-\theta_u\) .
Theorem 1
Assume \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are non zero vectors, and consider their angle \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) .
Then the following assertions hold:
\( \overrightarrow{u}\) and \( \overrightarrow{v}\) are aligned if and only if \( \theta\) is equal to \( 0\) or \( \pi\) modulo \( 2\pi\) , i.e. \( \theta\equiv 0\) modulo \( \pi\) .
\( \overrightarrow{u}\) and \( \overrightarrow{v}\) are orthogonal if and only if \( \theta\) is equal to \( \pm\frac{\pi}{2}\) modulo \( 2\pi\) , i.e. \( \theta\equiv \frac{\pi}{2}\) modulo \( \pi\) .
Proof
Assume \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are non zero vectors, and consider their angle \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) .
Consider their angles with the \( x\) axis \( \theta_u=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) and \( \theta_v=\widehat{(\overrightarrow{i},\overrightarrow{v})}\) .
And consider the unit vectors \( \overrightarrow{u_0}=\frac{\overrightarrow{u}}{\left\| \overrightarrow{u} \right\|}\) and \( \overrightarrow{v_0}=\frac{\overrightarrow{v}}{\left\| \overrightarrow{v} \right\|}\) ,
Then \( \theta_u=\widehat{(\overrightarrow{i},\overrightarrow{u_{0}})}\) and \( \theta_v=\widehat{(\overrightarrow{i},\overrightarrow{v_{0}})}\) .
Assume \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are aligned, and consider their factor of proportionality (that is non zero because they are non zero vectors): \( \lambda\in\mathbb{R^{*}}\) is such as \( \overrightarrow{v}=\lambda\overrightarrow{u}\) .
If \( \lambda>0\) , then \( \left\| \overrightarrow{u} \right\|=\lambda\left\| \overrightarrow{v} \right\|\) , so that \( \overrightarrow{u_0}=\frac{\lambda\overrightarrow{v}}{\lambda\left\| \overrightarrow{v} \right\|} =\frac{\overrightarrow{v}}{\left\| \overrightarrow{v} \right\|} = \overrightarrow{v_0}\) , so that \( \theta_{v}=\theta_{u}\) modulo \( 2\pi\), and \( \theta=0\) modulo \( 2\pi\) .
And if \( \lambda<0\) , then \( \left\| \overrightarrow{u} \right\|=-\lambda\left\| \overrightarrow{v} \right\|\) , so that \( \overrightarrow{u_0}=\frac{\lambda\overrightarrow{v}}{-\lambda\left\| \overrightarrow{v} \right\|} =-\frac{\overrightarrow{v}}{\left\| \overrightarrow{v} \right\|} = -\overrightarrow{v_0}\) , the reflected of \( \overrightarrow{v_0}\) around the center \( 0\) , so that \( \theta_{v}=\theta_{u}+\pi\) modulo \( 2\pi\) , and \( \theta=\pi\) modulo \( 2\pi\) .
Reversely, assume \( \theta\) is equal to \( 0\) or \( \pi\) modulo \( 2\pi\) .
If \( \theta=0\) modulo \( 2\pi\) , then \( \theta_{v}=\theta_{u}\) , so that \( \overrightarrow{v_0}=\overrightarrow{u_0}\) and \( \overrightarrow{v}=\frac{\left\| \overrightarrow{v} \right\|}{\left\| \overrightarrow{u} \right\|} \overrightarrow{u}\) . Consequently, \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are aligned.
And if \( \theta=\pi\) modulo \( 2\pi\) , then \( \theta_{v}=\theta_{u}+\pi\) , so that \( \overrightarrow{v_0}=-\overrightarrow{u_0}\) , the reflected of \( \overrightarrow{v_0}\) around the center \( 0\) , and \( \overrightarrow{v}=-\frac{\left\| \overrightarrow{v} \right\|}{\left\| \overrightarrow{u} \right\|} \overrightarrow{u}\) . Consequently, \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are aligned.
Assume \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are orthogonal, i.e. \( \overrightarrow{u}\cdot\overrightarrow{v}=0\) .
Then, as the dot product is a bilinear form: \( \overrightarrow{u_{0}}\cdot\overrightarrow{v_{0}} =\frac{\overrightarrow{u}\cdot\overrightarrow{v}} {\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|} =0\) , so that \( \overrightarrow{u_0}\) and \( \overrightarrow{v_0}\) are orthogonal as well.
If they are oriented in the direct direction, then \( \theta_{v}=\theta_{u}+\frac{\pi}{2}\) modulo \( 2\pi\) , and \( \theta=\frac{\pi}{2}\) modulo \( 2\pi\) .
And if they are oriented in the reverse direction, then \( \theta_{v}=\theta_{u}-\frac{\pi}{2}\) modulo \( 2\pi\) , and \( \theta=-\frac{\pi}{2}\) modulo \( 2\pi\) .
Reversely, assume \( \theta\) is equal to \( \pm\frac{\pi}{2}\) modulo \( 2\pi\) .
If \( \theta\) is equal to \( \frac{\pi}{2}\) modulo \( 2\pi\), then \( \theta_{v}=\theta_{u}+\frac{\pi}{2}\) .
But \( \overrightarrow{u_{0}}=\begin{bmatrix}\cos(\theta_{u})\\\sin(\theta_{u})\end{bmatrix}\) and \( \overrightarrow{v_{0}}=\begin{bmatrix}\cos(\theta_{v})\\\sin(\theta_{v})\end{bmatrix} =\begin{bmatrix}\cos(\theta_{u}+\frac{\pi}{2})\\\sin(\theta_{u}+\frac{\pi}{2})\end{bmatrix} =\begin{bmatrix}-\sin(\theta_{u})\\\cos(\theta_{u})\end{bmatrix}\) .
Consequently, if we denote \( R_{u}=\left\| \overrightarrow{u} \right\|\) and \( R_{v}=\left\| \overrightarrow{v} \right\|\) ,
\( \overrightarrow{u}=\begin{bmatrix}R_{u}\cos(\theta_{u})\\R_{u}\sin(\theta_{u})\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}-R_{v}\sin(\theta_{u})\\R_{v}\cos(\theta_{u})\end{bmatrix}\) .
So that \( \overrightarrow{u}\cdot\overrightarrow{v} =-R_{u}R_{v}\cos(\theta_{u})\sin(\theta_{v})+R_{u}R_{v}\sin(\theta_{u})\cos(\theta_{v})=0\) , and \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are orthogonal.
And if \( \theta\) is equal to \( -\frac{\pi}{2}\) modulo \( 2\pi\), then \( \theta_{v}=\theta_{u}-\frac{\pi}{2}\) .
But \( \overrightarrow{u_{0}}=\begin{bmatrix}\cos(\theta_{u})\\\sin(\theta_{u})\end{bmatrix}\) and \( \overrightarrow{v_{0}}=\begin{bmatrix}\cos(\theta_{v})\\\sin(\theta_{v})\end{bmatrix} =\begin{bmatrix}\cos(\theta_{u}-\frac{\pi}{2})\\\sin(\theta_{u}-\frac{\pi}{2})\end{bmatrix} =\begin{bmatrix}\sin(\theta_{u})\\{-\cos(\theta_{u})}\end{bmatrix}\) .
Consequently, if we denote \( R_{u}=\left\| \overrightarrow{u} \right\|\) and \( R_{v}=\left\| \overrightarrow{v} \right\|\) ,
\( \overrightarrow{u}=\begin{bmatrix}R_{u}\cos(\theta_{u})\\R_{u}\sin(\theta_{u})\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}-R_{v}\sin(\theta_{u})\\R_{v}\cos(\theta_{u})\end{bmatrix}\) .
So that \( \overrightarrow{u}\cdot\overrightarrow{v} =R_{u}R_{v}\cos(\theta_{u})\sin(\theta_{v})-R_{u}R_{v}\sin(\theta_{u})\cos(\theta_{v})=0\) , and \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are orthogonal.
Theorem 2
Assume \( (\overrightarrow{u},\overrightarrow{v},\overrightarrow{w})\in(\mathbb{P}^*)^3\) are non zero vectors.
Then the following assertions hold:
\( \widehat{(\overrightarrow{u},\overrightarrow{u})}=0\) modulo \( 2\pi\) .
\( \widehat{(\overrightarrow{u},-\overrightarrow{u})}=\pi\) modulo \( 2\pi\) .
\( \widehat{(\overrightarrow{v},\overrightarrow{u})}=-\widehat{(\overrightarrow{u},\overrightarrow{v})}\) modulo \( 2\pi\) .
\( \widehat{(\overrightarrow{u},\overrightarrow{w})} =\widehat{(\overrightarrow{u},\overrightarrow{v})}+\widehat{(\overrightarrow{v},\overrightarrow{w})}\) modulo \( 2\pi\) .
\( \widehat{(\overrightarrow{u},\overrightarrow{v})} =\widehat{(\overrightarrow{w},\overrightarrow{v})}-\widehat{(\overrightarrow{w},\overrightarrow{u})}\) modulo \( 2\pi\) .
Proof
Assume \( (\overrightarrow{u},\overrightarrow{v},\overrightarrow{w})\in(\mathbb{P}^*)^3\) are non zero vectors.
Consider their angles with the \( x\) axis \( \theta_u=\widehat{(\overrightarrow{i},\overrightarrow{u})}\) , \( \theta_v=\widehat{(\overrightarrow{i},\overrightarrow{v})}\) and
\( \theta_w=\widehat{(\overrightarrow{i},\overrightarrow{w})}\) .
Denote \( \theta_{uu}=\widehat{(\overrightarrow{u},\overrightarrow{u})}\) . Then \( \theta_{uu}=\theta_{u}-\theta_{u}=0\) modulo \( 2\pi\) .
Denote \( \theta_{-u}=\widehat{(\overrightarrow{i},-\overrightarrow{u})}\) . Then \( \theta_{-u}=\theta_{u}+\pi\) modulo \( 2\pi\) .
And denote \( \theta_{u-u}=\widehat{(\overrightarrow{u},-\overrightarrow{u})}\) . Then \( \theta_{u-u}=\theta_{-u}-\theta_{u}=\pi\) modulo \( 2\pi\) .
Denote \( \theta_{uv}=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) and \( \theta_{vu}=\widehat{(\overrightarrow{v},\overrightarrow{u})}\) . Then \( \theta_{uv}=\theta_{v}-\theta_{u}\) and
\( \theta_{vu}=\theta_{u}-\theta_{v}\) , so that \( \theta_{vu}=-\theta_{uv}\) .
Denote \( \theta_{uv}=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) , \( \theta_{vw}=\widehat{(\overrightarrow{v},\overrightarrow{w})}\) and \( \theta_{uw}=\widehat{(\overrightarrow{u},\overrightarrow{w})}\) .
Then \( \theta_{uv}=\theta_{v}-\theta_{u}\) ,
\( \theta_{vw}=\theta_{w}-\theta_{v}\) , and \( \theta_{uw}=\theta_{w}-\theta_{v}\) , so that \( \theta_{uw}=\theta_{uv}+\theta_{vw}\) .
Denote \( \theta_{uv}=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) , \( \theta_{wv}=\widehat{(\overrightarrow{w},\overrightarrow{v})}\) and \( \theta_{wu}=\widehat{(\overrightarrow{w},\overrightarrow{u})}\) .
Then, because of item (II), \( \theta_{wv}=\theta_{wu}+\theta_{uv}\) , so that \( \theta_{uv}=\theta_{wv}-\theta_{wu}\)
Theorem 3
Assume \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are non zero vectors, and consider their angle \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) . Then \( \overrightarrow{u}\cdot\overrightarrow{v} =\left\| \overrightarrow{u} \right\|\left\| \overrightarrow{v} \right\|\cos(\theta)\)
Corollary 1
Assume \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are unit vectors, and consider their angle \( \theta=\widehat{(\overrightarrow{u},\overrightarrow{v})}\) . Then \( \overrightarrow{u}\cdot\overrightarrow{v}=\cos(\theta)\)
The corollary 1 is a direct consequence of the theorem 3 because, if \( \overrightarrow{u}\) and \( \overrightarrow{v}\) are unit vectors, then \( \left\| \overrightarrow{u} \right\|=\left\| \overrightarrow{v} \right\|=1\) .
The theorem 3 will be proved in section 11 about the orthonormal bases.
Assume \( (\overrightarrow{u},\overrightarrow{v})\in(\mathbb{P}^*)^2\) are unit vectors, and consider their angles with the axis, \( \widehat{(\overrightarrow{i},\overrightarrow{u})}=b\) and \( \widehat{(\overrightarrow{i},\overrightarrow{v})}=a\) .
Then we may calculate the dot product of \( \overrightarrow{u}\) and \( \overrightarrow{v}\) the two following ways:
As \( \overrightarrow{u}=\begin{bmatrix}\cos(b)\\\sin(b)\end{bmatrix}\) and \( \overrightarrow{v}=\begin{bmatrix}\cos(a)\\\sin(a)\end{bmatrix}\) , then \( \overrightarrow{u}\cdot\overrightarrow{v}=\cos(a)\cos(b)+\sin(a)\sin(b)\) .
And as \( \widehat{(\overrightarrow{u},\overrightarrow{v})}=a-b\) , then, because of theorem 3, \( \overrightarrow{u}\cdot\overrightarrow{v}=\cos(a-b)\)
Consequently, \( \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\) .
Theorem 4
Assume \( (a,b)\in\mathbb{R}^2\) are real numbers.
Then the following trigonometric formulae hold:
\( \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\)
\( \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\)
\( \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\)
\( \sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)\)
Corollary 2
Assume \( a\in\mathbb{R}\) is real number.
Then the following trigonometric formulae hold:
\( \cos(2a)=\cos^2(a)-\sin^2(a)\)
\( \cos(2a)=2\cos^2(a)-1\)
\( \cos(2a)=1-2\sin^2(a)\)
\( \sin(2a)=2\cos(a)\sin(a)\)
Proof (of the theorem 4)
Assume \( (a,b)\in\mathbb{R}^2\) are real numbers.
Because of item (III): \( \cos(a+b)=\cos(a-(-b)) =\cos(a)\cos(-b)+\sin(a)\sin(-b)\)
\( =\cos(a)\cos(b)-\sin(a)\sin(b)\)
The formula \( \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\) has been proved previously geometrically.
Because of item (III): \( \sin(a+b)=\cos\left(\frac{\pi}{2}-(a+b)\right) =\cos\left(\left(\frac{\pi}{2}-a\right)-b\right)\)
\( =\cos\left(\frac{\pi}{2}-a\right)\cos(b)+\sin\left(\frac{\pi}{2}-a\right)\sin(b) =\sin(a)\cos(b)+\cos(a)\sin(b)\)
Because of item (IV): \( \sin(a-b)=\sin(a+(-b)) =\sin(a)\cos(-b)+\cos(a)\sin(-b) =\sin(a)\cos(b)-\cos(a)\sin(b)\)
Proof (of the corollary 2)
Assume \( a\in\mathbb{R}\) is real number.
Setting \( b=a\) in the item (II) of the theorem 4, we obtain directly:
\( \cos(2a)=\cos^2(a)-\sin^2(a)\)
Let’s remember that \( \cos^{2}(a)+\sin^{2}(a)=1\) , so that \( \sin^{2}(a)=1-\cos^{2}(a)\) . Then we obtain, using item (V),:
\( \cos(2a)=\cos^{2}(a)-(1-\cos^{2}(a))=2\cos^2(a)-1\)
As \( \cos^{2}(a)+\sin^{2}(a)=1\) , then \( \cos^{2}(a)=1-\sin^{2}(a)\) . Then we obtain, using item (V),:
\( \cos(2a)=1-\sin^{2}(a)-\sin^{2}(a)=1-2\sin^2(a)\)
Setting \( b=a\) in the item (IV) of the theorem 4, we obtain directly:
\( \sin(2a)=2\cos(a)\sin(a)\)
We sum up the trigonometric formulae given in the theorem 4 and the corollary 2 in the table 3 below.
| \( \forall\; (a,b)\in\mathbb{R}^2\) | \( \forall\; a\in\mathbb{R}\) |
| \( \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\) | \( \cos(2a)=\cos^2(a)-\sin^2(a)\) |
| \( \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\) | \( \cos(2a)=2\cos^2(a)-1\) |
| \( \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\) | \( \cos(2a)=1-2\sin^2(a)\) |
| \( \sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)\) | \( \sin(2a)=2\cos(a)\sin(a)\) |
Now that we know everything needed about the angles and the trigonometry, we are ready for a new geometrical tool, the polar coordinates of non zero vectors.
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