AN ATTEMPT TO SOLVE TRIGO PROBLEM.
Details including whole numbers
a, b, c
S
=2*p p =2*q
Basic premise
Long unit C c =1
C > b > a (a
+ b) > c
We have to prove that S (in a^s + b^s = c^s) isn’t bigger
than 2.
Proof: a =/= b* .1
A^2 + b^2 = 2*(b)^2 =c^2
2^.2*b =c
Continuation .2
According in “Co
sinus’s Theory”
A^2 + b^2- 2*a*b*Cos© =c^2
When© has a right angle we get
A^2 + b^2 = c^2
` and the opposite
When a^2 + b^2 =c^2
Then C is a right angle.
3.
According to Sinus Theory
A/Sin(A) = b/Sin(B) =c/Sin(C)
When C is a right angle and c=1
We get
A = Sin(A), b =Sin(B), c
= Sin(C).
.4
X^s=(X^p)^2
Y^s =(Y^p)^2
According to the trigonometry theory
When a triangle has a right angle
Sin^2(A)+Sin^2(B) = 1
A^2+b^2 =1
(a^p)^2 + (b^p)^2 =1
(a^(2*q))^2 + (b^(2*q))^2 = 1
We get:
s = 2, p =1, q = 0.5, or:
An equal on one side a^2, b^2, c^2.
And the ather side (a^p)^2, (b^p)^2, (c^p)^2.
Of the same function
Is a contradiction between the two sides.
Have been proved.
Matus
Caspi
Yagur 30065
Israel
e.m.matuscaspi@gmail.com
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