MULTI-COMBINATIONAL TRIGONOMETRY

MILTI–COMBINAATIONAL TRIGONDMETRY

The goal of the research is to examine the effect of multi-combinational theory in trigonometry.

The study found that new rules, formulae, theorems and methods can be created with the influence of the multi-combinational theory (I.e. sequential comparison rule) and prove very effectively.

Looking at the selected topic discussed below, this research of the multi-combinational paradigm of trigonometry is examined:

(1)                        Can new mathematical rules, formulae and methods created in trigonometry in the name of the multi-combinational theory?

(2)                        Is the availability of the multi-combinational theory consumption affected positively in trigonometry?

DFINITION OF THE MULTI-COMBINATIONAL THEORY

Multi-Combinational theory is the study of different kinds of mathematical information which are related in multiples order of sequences and used mainly for comparison and predictive purposes.

The background of this theory is mainly depends on the “sequential comparison rule” which has main influences in all areas of trigonometry.

RATIO DEFINITION

Let ө_1* ө_2*.......* ө_i be a multiple angles of i’s-right-angles triangles in standard positions with any multiple point P[x_0*x_1*---*x_r-1,   x_1*x_2*--*x_r] on the multiple terminal side, a multiple distance x_2*x_3*---x_r+1 from the origin x_2*x_3*---*x_r+1≠ 0.

The relations of the multi-combinational functions are defined as followed.

X_r = x_0* [sin (ө_1* ө_2*.......* ө_i) / cos(ө_1* ө_2*------ ө_i)]

X_r = x_0* [sec(ө_1* ө_2*.... _* ө_i) / csc(ө_1* ө_2*......ө_i)]

For each i= 1,2,3,----, r

EXAMPLE

In the three right-angle triangles ABC, BCD, and CDE which are in linear mapping (Reference Tittle: “Maker of modern mathematics”) have sides AB = x_0, BC = x₁ and CA = x₂ for right angle triangle ABC: BC = x₁ CD = x_2, DB = x₃ for right-angle triangle BCD: CD = x_2, DE = x₃, EC = x₄ for right angle triangle CDE.

If angles ABC, BCD, and CDE are 90⁰ each and angle CAB =60⁰; angle DBC =40⁰, and angle ECD = 50⁰.

(i)  Find the length EC of right angle triangle CDE if the length AB of right angle triangle ABC is 5m.

SOLUTION

   ө_1* ө_2* ө₃ = 60⁰_* 40⁰_* 50⁰ = 120,000⁰

X₀ = 5m

Therefore

X₄ = 5_* [sin(120,000⁰) / cos(120,000⁰)]

X₄ = -8.66

But since distance don’t measure in negative sign, we neglect the negative sign and write 8.66m.