Trigonometry


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Author: Thibaut BERNARD

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Update: Monday May 5, 2003.

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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 right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the sides of the right angle.
See the demonstration.


Measuring units of the angles

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.


Diagram

Right-angled triangle includes in the circle  :

In a circle of equation R2 = X2 + Y2 :

R2 = X2 + Y2
R is the ray equivalent to the hypotenuse of the right-angled triangle,
X the base of the triangle,
Y the with dimensions one opposed triangle.

With a representing the angle alpha.


Basic equations

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
sin2 a + cos2 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).

Right-angled 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 = 2n2. 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 right-angled 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 x2 + y2 = 1 (Pythagore)

cos2 60 + sin2 60 = 1

sin2 60 = 1 - cos2 60

with cos 60 = 1/2, that gives

sin2 60 = 1 - (1/2)2 = 1 - 1/4

sin2 60 = (4 - 1)/4

sin2 60 = 3/4

hence

with tg 60 = sin 60 / cos 60, we have

hence

With the same argument, we show that


Summary

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 :


Determination of the diagram

Let us take a quadrant with an angle a:

(basic diagram)

With which we will associate there a right-angled 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:
a + b.

Note: For passing to the continuation, it is necessary to note in this diagram the following points:
right-hand side DK = sine b
right-hand side OK = cosine b
right-hand side EC = sine a
right-hand side OE = cosine a
right-hand side DH = sine (a+b)
right-hand side OH = cosine (a+b)

 

Notation

HOD angle out of O.

 

Determination of the angles

Angle MDK

OKD = 90, KOD = b
ODK = 90 - b
ODM = 90 - (a + b)
MDK = ODK - ODM
MDK = 90 - b - (90 - (a + b))
MDK = 90 - b - (90 - a - b)
MDK = 90 - b - 90 + a + b

Thus MDK = a

Angle M'K'D'

POK = a
OKM' = OKP = 90 - a
OKD' = 90
M'K'D' = OKD' - OKM'
M'K'D' = 90 - (90 - a)

Thus M'K'D' = a

On the same principle that on the preceding diagram, one turns over the right-angled 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:
right-hand side EC = sine a
right-hand side OE = cosine a
right-hand side D' H' = sine (a-b)
right-hand side OH' = cosine (a-b)

In the determination of the angles, before passing to the continuation it should have been understood the fact that in right-angled triangle MDK, the angle in D is equal to A; Thus the same one as in the basic diagram (in triangle EOC).


Basic definitions

Angles:

AOC = a, COD = COD' = b
AOD = a+b, AOD' = a-b
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:

Right-hand sides DK and KD' = sine of b
Right-hand side OK = cosine of b
Right-hand side EC. = sine of a
Right-hand side OE = cosine of a
Right-hand side HD = sine of a + b
Right-hand side OH = cosine of a + b

We thus will be able to make the following deductions of them.


Additions on the sides

DH = HM + MD, MH = KP

from where DH = MD + KP


Functions


Exercise

To apply the theorem of Pythagore to the right-angled triangles as well with the sides with the goniometrical functions.


Deduction of the functions starting from the diagram


Summary