Definition of a Natural Number - Needed to solve a Trignometric Equation

[KayJay]

I've been working on this off and on for a very very long time now. I'm stuck Can the below be solved at all

Sqr(A*Tan(t))  = N ; Where N is natural number
for a given natural constant "A", with "t" ranging from 45 degrees to 90 degrees

Is it solvable?

[DiGiTaIErRoR]

Hmmm

well

a=-n^2/tan(t) and n =< 0

[KayJay]

Thanxs. Will that result in "N" being "Natural"? My question again; What is a Natural Number?

Anyways its going to take a while to digest that.

Now. my question was wrong Sorry. It should read

Sqr(A*Tan(t))  = N ; Where N is natural number
for a given natural constant "A", with "t" ranging from 0 degrees to 45 degrees

You would have realised by now this is to with Prime Numbers

[bugzpodder]

tan0=0
tan45=1

A*tan(t) - goes from 0 - A
sqrt(A*tan(t)) - goes from 0 - sqrt(A)

so your answer would be all natural numbers between 0 - sqrt(A)

[KayJay]

Thanx for the reply. I do understand UR solution. But in that range, lies all solutions for the equation, not only Postive Integers. So how does one define a Positive Integer, such that its curve may be intersected with the above equation, to find points of intersection.

Am I out of my mind? or do I make any sense at all

Thanx

[jim mcnamara]

Here is the definition - and it's only integers, not all of the reals.
It's denoted usually as N

http://mathworld.wolfram.com/NaturalNumber.html

[KayJay]

Thanx. Not much help, but a darned good site. Bookmark worthy

Thanx again

[RAEsquivelC]

KayJay posted:

Sqr(A*Tan(t)) = N ; Where N is natural number for a given natural constant "A", with "t" ranging from 45 degrees to 90 degrees

First, note that the tangent of 45 degrees =1, and the tangent of 90 degrees = infinite, So, for “t”
Ranging from 45 degrees to 90 degrees, we can re-write the equation as:

Sqr(A*B) = N, where B is any real number from 1 to infinity.

The next thing to know is, does “Sqr” stand for “the square of” or for “the square root of”?

1) For “the square root of”, we have:
(A*B)^1/2 = N
A*B = N^2
B = N^2/A
There are an infinite solutions to this equation, it seems obvious to me. For example:

For N = 1, then B = N^2/A = 1/A
If A = 1, then B = 1/1 = 1
If A = 2, then B = 1/2
Etc.

For N = 2, then B = N^2/A = 4/A
If A = 1, = 4/1
If A = 2, then B = 4/2 = 2
Etc.

2) 2) For “the square of”, we have:
(A*B)^2 = N^2
A*B = N, or A*B = -N
B = N/A, or B = -N/A
This also has infinite solutions, and I think that obtaining a vlue of B for each combination of N and A is obvious, so, I will stop at this point.

Oh, yes! If you MUST have the angle, than you would proceed to obtain the angle by stating that
arctangent (t) = B, and proceeding to solve for t, in radians or in degrees, as desired.