Trigonometry
Author: Thibaut BERNARD 
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Update: Monday May 5, 2003.
Alphaquark author's Note :
This page is a translation of
trigonometry
with the help of
Altavista translation.
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Part I, basic concepts
Recall of the theorem of Pythagore
Measuring units of the angles
Diagram
Basic equations
Deduction of the basic equations
Characteristic of certain angles
Summary
Part II, additions and subtractions of angles
Determination of the diagram
Basic definitions
Additions on the sides
Functions
Deductions of the functions starting from the diagram
Summary
Part I, basic concepts
Recall of the theorem of Pythagore
In a rightangled triangle, the square on the hypotenuse is equal to the sum of the squares on the
sides of the right angle.
See the demonstration.
Useless to recall celebrates it value of pi (3,1415926535) round to 3,1416.
The angles are expressed in three different ways:
Degree
Radian
Rank
In a full rotation of circle, there is
360 degrees
2 pi radian
400 ranks
Measurement is done starting from the with dimensions right and in the contrary direction of the needles of a watch.
From this definition, the point more in top is thus equal to 90° or 100 ranks or pi/2 radian.
Note: Programming languages (BASIC, Pascal, C...), some is their version, express the angles only in radian. It is thus necessary initially to convert the angle into radian before asking of it the sine, the cosine or the tangent. Idem when starting from the value of a sine or a tangent one asks of it for the angle (functions arcsinus and arctangente). In this last case the returned angle is systematically in radian, it should be made a conversion of it if one wants this angle in degree or radian.
Formulas of conversion of the angles
Angle in radian = pi * (angle in degree)/180
Angle in radian = pi * (angle in rank)/200
Angle in rank = 200 * (angle in degree)/180
Angle in rank = 200 * (angle in radian)/pi
Angle in degree = 180 * (angle in radian)/pi
Angle in degree = 180 * (angle in rank)/200
In this chapter we will express ourselves only in degree.
Rightangled triangle includes in the circle :
In a circle of equation R^{2} = X^{2} + Y^{2} :
R^{2} = X^{2} + Y^{2}
R is the ray equivalent to the hypotenuse of the rightangled triangle,
X the base of the triangle,
Y the with dimensions one opposed triangle.
With a representing the angle alpha.
Definition
sin a = Y/R (the sine equal to the division of with dimensions is
opposed by the hypotenuse).
cos a = X/R (the cosine is equal to the division of the base by the
hypotenuse).
tg a = Y/X (the tangent equal to the division of with dimensions is
opposed by the base).
cotg a = X/Y = 1/tg a (the cotangent is the
reverse of the tangent, therefore equal to the division of the base by the with dimensions opposite
one).
Deduction
tg a =
sin a / cos a
sin^{2} a + cos^{2} a = 1
(application of the theorem of Pythagore)
Deduction of the basic equations
Given that
and
then
hence
Characteristic of certain angles
In the examples which follow, one will take as ray the unit (R = 1).
Rightangled triangle of with dimensions opposite equal to the base (triangle taken in a square)
Let us take the diagonal of a square included in a circle. This diagonal represents the hypotenuse of the square (or the radius of the circle). In a square, the sides being equal and the angle a equal to 45°, one can pose R = 2n^{2}. With a ray with 1, that gives :
N =
from where
cos 45° =
sin 45° =
The tangent being the side opposed on the base, the two sides being equal, from where : tg 45° = 1. The reverse (the cotangent) is by definition so equal to 1.
Equilateral triangle
The angles all are thus equal to 60°. With a rightangled triangle applied in this equilateral triangle, one can have a base equal to half on the opposite side.
The equilateral triangle applied to a circle having like radius the unit, one b a cosine equalizes to 1/2.
For a = 60° and with x^{2} + y^{2} = 1 (Pythagore)
cos^{2} 60° + sin^{2} 60° = 1
sin^{2} 60° = 1  cos^{2} 60°
with cos 60° = 1/2, that gives
sin^{2} 60° = 1  (1/2)^{2} = 1  1/4
sin^{2} 60° = (4  1)/4
sin^{2} 60° = 3/4
hence
with tg 60° = sin 60° / cos 60°, we have
hence
With the same argument, we show that
By definition the tangent of 90° is logically 1/0. But as the divide check is impossible, it thus does not have there a tg 90°. Tg 90° would be equal ad infinitum.
Idem for cotg 0°.
Summary table :
Part II, additions and subtractions of angles
Let us take two angles (a and b) different of which :
Let us take a quadrant with an angle a: (basic diagram) 
With which we will associate there a rightangled triangle whose angle out of O is b:

What gives the resulting diagram:
In triangle HOD, we thus find ourselves with an angle equal to:
Note: For passing to the continuation, it is necessary to note in this diagram the following points: 
Notation HOD angle out of O.
Determination of the angles Angle MDK
OKD = 90°, KOD = b Thus MDK = a Angle M'K'D'
POK = a Thus M'K'D' = a 
On the same principle that on the preceding diagram, one turns over the rightangled triangle basic which one inserts in the quadrant. In the triangle H' OD', the angle out of O is thus equal to A  B.
Observations: 
In the determination of the angles, before passing to the continuation it should have been understood the fact that in rightangled triangle MDK, the angle in D is equal to A; Thus the same one as in the basic diagram (in triangle EOC). 
Angles:
AOC = a, COD = COD' = b
AOD = a+b, AOD' = ab
OKD = OKD' = 90°
OHD = OPK = OEC = OH'D' = 90°
KMD = KM'D' = 90°
MDK = M'K'D' = a
Ray :
OD = OA = OC = OD' = 1
Knowing the following points:
Righthand sides DK and KD' = sine of b
Righthand side OK = cosine of b
Righthand side EC. = sine of a
Righthand side OE = cosine of a
Righthand side HD = sine of a + b
Righthand side OH = cosine of a + b
We thus will be able to make the following deductions of them.
DH = HM + MD, MH = KP
from where DH = MD + KP
Exercise
To apply the theorem of Pythagore to the rightangled triangles as well with the sides with the goniometrical functions.
Deduction of the functions starting from the diagram