Ratio of All Angles in Exact Form (1 Viewer)

Sy123

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So I had formed a hypothesis that:



Where alpha is any integer degree.
Could I write the sine of that angle, be it any angle, in exact form
And a, b and c are all integers.

Now the above RHS is probably wrong (maybe not)

But Im just wondering if it could be done, if so is it possible to prove it? If it is wrong, can someone make an expression similar to what I have done that fits all criteria for any angle alpha, be it an integer.

And also is it possible to do this for any rational number?

Is it possible to do this even with an irrational alpha?

Ive just been wondering, and I know the maths part of BoS are really smart, so Im expecting a great solution to this :D

Thanks
 

barbernator

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there was a question a while back posted by seanieg89 i think, and you derived the parametric equations of sin@ and cos@ if i remember correctly
 

Carrotsticks

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I don't fully understand what you're saying. Is this the question you are essentially asking?

For all alpha, do there exist constants a, b and c such that:



??

EDIT: And btw a, b and c cannot be integers because the sine function oscillates between plus/minus 1 so there's no way that they can all be integers.
 
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Sy123

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Err, yes, but Im ALSO asking whether alpha needs to be an integer, or can it just be rational, or can it just be any real number of alpha.

Also, a b and c should be integers
 

Trebla

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a, b and c cannot all be integers

Example: sin 60º = (1/2)√3
 

Carrotsticks

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Sy123, it is probably also worth noting that MANY trigonometric closed forms are expressed as nested radicals.

For example: cos(x/2^n) for positive x can be expressed as a nested radical.

The thing is, some nested radicals CAN be denested but others cannot and currently (or to my knowledge) there are no definitive algorithms we can use to determine denestability except maybe that of Landau and Miller: http://en.wikipedia.org/wiki/Landau's_algorithm
 

seanieg89

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I assume you are forcing alpha to be integral and wondering if there necessarily exist rational a,b,c such that your equation holds.

The answer is in the negative:

The question can be related to one of Galois theory. Observe that the RHS is the root of a quadratic with rational coefficients. Hence such an expression is only possibly if sin(alpha) is in a quadratic extension of the rationals. However as proven in http://www.maths.manchester.ac.uk/~khudian/Etudes/Galetudes/angles2.pdf , the extension Q(sin(360/N)):Q (for an integer N) has degree a power of 2 iff N is the product of a power of 2 and a collection of distinct Fermat primes. N=360 is not such an integer and hence sin(1) (for example) cannot be written in the form claimed.
 

Fus Ro Dah

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So I had formed a hypothesis that:



Where alpha is any integer degree.
Could I write the sine of that angle, be it any angle, in exact form
And a, b and c are all integers.

Now the above RHS is probably wrong (maybe not)

But Im just wondering if it could be done, if so is it possible to prove it? If it is wrong, can someone make an expression similar to what I have done that fits all criteria for any angle alpha, be it an integer.

And also is it possible to do this for any rational number?

Is it possible to do this even with an irrational alpha?

Ive just been wondering, and I know the maths part of BoS are really smart, so Im expecting a great solution to this :D

Thanks
Sy123, to test your hypothesis you could have easily deduced a class of angles that could have been expressed in this form by having the cosine of an angle in a similar closed surd form, then using the Pythagorean Identity and equating it with 1 so we have two surdic expressions summing up to be equal to an integer.
 

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