Exact Expression for Sine of 1 Degree

In summary, there is no closed form solution involving only real numbers for 1 degree, as a 180-sided polygon is not constructible. However, there are an infinite number of closed form solutions involving complex numbers where the imaginary part cancels out. The most compact and beautiful form is \frac{\sqrt[90]{i} - \sqrt[90]{-i}}{2i}.
  • #1
bob012345
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I believe it may be impossible but has anyone seen or come across an exact expression of the sin of 1 degree in a closed form that does not involve either complex roots, infinite series, or sines and cosines of other angles. Thanks.
For example, $$Sin(15)= \frac {(\sqrt 6 - \sqrt 2)} 4$$

and $$Sin(3)=\frac {(\sqrt 6 (\sqrt 5 -1)(3+\sqrt 3)} {48} -\frac {\sqrt 3 (3-\sqrt 3 )\sqrt{ 5+\sqrt 5 }} {24}$$

What about ##Sin(1)##?
 
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  • #3
If you know ##\sin(3^\circ)##, then the equation ##\sin(3^\circ)=3\sin(1^\circ)-4\sin^3(1^\circ)## gives you a cubic equation to solve, and cubic equations can always be solved by radicals. I used the triple angle formula ##\sin(3x)=3\sin(x)-4\sin^3(x)##, which of course follows from the angle addition formula.
 
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  • #4
fresh_42 said:
Radiant or degree?
Degree.
 
  • #5
Infrared said:
If you know ##\sin(3^\circ)##, then the equation ##\sin(3^\circ)=3\sin(1^\circ)-4\sin^3(1^\circ)## gives you a cubic equation to solve, and cubic equations can always be solved by radicals. I used the triple angle formula ##\sin(3x)=3\sin(x)-4\sin^3(x)##, which of course follows from the angle addition formula.
I know. Alas if it were that easy... I believe you end up with cubic roots and usually cubic roots of complex numbers. Then you go through a lot of work and get a simple expression involving sines and cosines of other angles which is what I'm trying to avoid. In fact going through that whole exercise allows one to derive the addition and subtraction of angles relationships.
 
  • #6
bob012345 said:
I know. Alas if it were that easy... I believe you end up with cubic roots and usually cubic roots of complex numbers. Then you go through a lot of work and get a simple expression involving sines and cosines of other angles which is what I'm trying to avoid. In fact going through that whole exercise allows one to derive the addition and subtraction of angles relationships.
There will be at least one real root. Of course the expression will be unpleasant. You cannot have both.
 
  • #7
fresh_42 said:
There will be at least one real root. Of course the expression will be unpleasant. You cannot have both.
So the issue is to find a way to get at that real root. Some kind of independent relationship. Also, there is no guarantee that it will be expressible in a straightforward and closed form with say, just radicals.

If unpleasant only involves real radicals, I would be happy. I don't think it's possible.
 
  • #8
Why not just use the Taylor approximation? of ##sin(\theta)=\theta## for very small ##\theta## in radians.

##sin(1) = 0.0174524...## vs 1degree = 0.0174533... with an accuracy of 0.005 %
 
  • #9
bob012345 said:
Also, there is no guarantee that it will be expressible in a straightforward and closed form with say, just radicals.
There is. It's the solution to a cubic equation and the coefficients are in a closed form.
jedishrfu said:
Why not just use the Taylor approximation? of ##sin(\theta)=\theta## for very small ##\theta## in radians.

##sin(1) = 0.0174524...## vs 1degree = 0.0174533... with an accuracy of 0.005 %
That's not an exact expression.
 
  • #10
jedishrfu said:
Why not just use the Taylor approximation? of ##sin(\theta)=\theta## for very small ##\theta## in radians.
If I only wanted a number sure, or just a calculator. I have been challenged by a friend to find if its possible to get a closed form exact form.
 
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  • #11
mfb said:
There is. It's the solution to a cubic equation and the coefficients are in a closed form.That's not an exact expression.
I'm fairly certain it will not give anything but another cubic root probably complex to solve before one gets the real root and that will probably involve sines and cosines of another angle. But I'll take another look...
 
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  • #12
Instead of repeatedly complaining what you think you would get, why don't you actually do the calculation and see what you really get?
 
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  • #14
You can get the answer with Wolfram alpha if you enter -4 * x ^3 + 3 * x = sin(pi/60) and click "exact expressions". There will be complex numbers in the expression. -4 * x ^3 + 3 * x = sin(pi/60) has 3 real roots, and you need to use the complex roots of a complex number in this case to get the real roots. See https://en.wikipedia.org/wiki/Casus_irreducibilis
 
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  • #15
I think there should be a formula with only real numbers, but my algebra is rusty so hopefully someone can tell me if I'm wrong. I'm going to work with ##\cos(1^\circ)## out of convenience, but this shouldn't make a difference because ##\sin^2(1^\circ)+\cos^2(1^\circ)=1## so you can solve for one in terms of the other with radicals.

Adjoining ##\cos(1^\circ)## to ##\mathbb{Q}## gives the field ##\mathbb{Q}(\zeta+\overline{\zeta})## where ##\zeta## is primitive ##360##th root of unity. This is a Galois extension of ##\mathbb{Q}## because it is a sub-extension of the Abelian extension ##\mathbb{Q}(\zeta).## It is abelian and hence solvable, so there are fields ##\mathbb{Q}\subset E_1\subset\ldots\subset \mathbb{Q}(\zeta+\overline{\zeta}),## where each extension is given by adjoining a root of a previous field, and each root is necessarily a real number because ##\mathbb{Q}(\zeta+\overline{\zeta})## is real.

Edit: I think this is wrong as it looks like it contradicts casus irreducibilis, I'll think more about it.
 
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  • #16
You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

Your answer in the most compact and beautiful form is

[tex]\frac{\sqrt[90]{i} - \sqrt[90]{-i}}{2i}[/tex]

There is no closed form solution involving only reals for 1 degree, because a 180-sided polygon is not constructable. However, there are an infinite number of closed form solutions involving complex numbers where the imaginary part cancels. (Like the above)
 
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  • #17
Vanadium 50 said:
You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

Your answer in the most compact and beautiful form is

[tex]\frac{\sqrt[90]{i} - \sqrt[90]{-i}}{2i}[/tex]

There is no closed form solution involving only reals for 1 degree, because a 180-sided polygon is not constructable. However, there are an infinite number of closed form solutions involving complex numbers where the imaginary part cancels. (Like the above)
Thanks. I suspected it was impossible. And I have seen that before but it is no answer to my quest, it just looks beautiful but evaluating it will give an answer in terms of sines and cosines of other angles which violates my original premise.
 
  • #19
mfb said:
Instead of repeatedly complaining what you think you would get, why don't you actually do the calculation and see what you really get?
I've been working on this problem on and off for weeks. I knew of those paths. Everything led to the same thing. I just wanted to see if anyone here had a unique insight. Believe me, I worked with it a long time before I posted it here.
 
  • #20
bob012345 said:
but evaluating it will give an answer in terms of sines and cosines

That's not the reason. The reason is that a necessary and sufficient condition for 1 degree to have an algebraic representation over the reals is that a regular 180-gon be constructable with compass and straightedge. It is not (Because 9 is not prime, and a theorem by Gauss)
 
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  • #21
Vanadium 50 said:
That's not the reason. The reason is that a necessary and sufficient condition for 1 degree to have an algebraic representation over the reals is that a regular 180-gon be constructable with compass and straightedge.

I don't think this is true. ##\sqrt[3]{2}## is not constructable with compass and straightedge.

Also, if someone could clear my confusion in post 15, I would be thankful.
 
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  • #22
Vanadium 50 said:
You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

Your answer in the most compact and beautiful form is

[tex]\frac{\sqrt[90]{i} - \sqrt[90]{-i}}{2i}[/tex]

There is no closed form solution involving only reals for 1 degree, because a 180-sided polygon is not constructable. However, there are an infinite number of closed form solutions involving complex numbers where the imaginary part cancels. (Like the above)
But if all the imaginary parts of a closed form solution involving complex numbers cancel, why isn't what's left a real closed form solution?
 
  • #23
Because they aren't separable in closed form: i.e. you can't collect all the i's in one place.
 
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  • #25
@Vanadium 50 Could you please explain why a regular ##180##gon is constructible if and only if ##\sin(1^\circ)## is expressible with radicals? Usually being constructible with compass and straightedge is a much stronger constraint than being radical (e.g. is ##\sqrt[3]{2}## is a radical but not constructable).
 
  • #26
I believe the argument is in the thread I linked to.
 
  • #27
I don't see that in your posted thread- could you give a post number?

In that thread, you discuss Gauss' criterion for when ##e^{2\pi i/n}## is constructible and that constructibility implies being expressible with radicals. But here it looks like you're claiming the other direction.
 
  • #28
I'm not in a good spot for long replies. Maybe it only works in one direction; that would be enough to answer the OP. I think it takes some discussion of what "it" is. 12 is perfectly algebraic yet isn't the sine of anything.
 
  • #29
@Vanadium 50 That's fine, take your time to reply. But I don't think the one direction is enough to answer the OP. Definitely ##\sin(1^\circ)## isn't constructible, but that doesn't answer the OP's question of whether it is expressible with real radicals.

I think @willem2's suggestion of citing casus irreducibilis to show impossibility is correct, but I also still don't see where my argument in post 15 fails. Perhaps @mathwonk has a resolution?

Edit: I realized the error in my previous argument: just because a Galois extension ##E/F## has solvable Galois group does not mean that ##E## is a radical extension of ##F##. It is true that every element of ##E## can be expressed from elements of ##F## with radicals, but you may also need to adjoin roots of unity. Otherwise, you can say that ##E/F## can be given by a tower of cyclic extensions, but these may not be radical.
 
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  • #30
https://www.wolframalpha.com/input/?i=sin(pi/10)
##S_{10} = sin(\pi/10) = 1/4(\sqrt{5}-1) ##

##S_{20} = sin(\pi/20) = \sqrt{ (1-\sqrt{1-S_{10}^2})/2} ##

Which gets you to sine of 9 degrees.
But then you would need to apply the one-third angle formula twice - and that involves complex roots.

https://www.quora.com/What-is-the-formula-for-sin-x-3-one-third-angle-formula
##S_{60} = sin(\pi/60) = \left(\left (-S_{20}+\sqrt{S_{20}^2-1}\right)^{1/3} + \left(-S_{20}-\sqrt{S_{20}^2-1}\right)^{1/3} \right)/2 ##
##S_{180} = sin(\pi/180) = \left( \left(-S_{60}+\sqrt{S_{60}^2-1}\right)^{1/3} + \left(-S_{60}-\sqrt{S_{60}^2-1}\right)^{1/3} \right)/2 ##
 
  • #31
.Scott said:
https://www.wolframalpha.com/input/?i=sin(pi/10)
S10=sin(π/10)=1/4(5−1)

S20=sin(π/20)=(1−1−S102)/2

Which gets you to sine of 9 degrees.
But then you would need to apply the one-third angle formula twice - and that involves complex roots.

https://www.quora.com/What-is-the-formula-for-sin-x-3-one-third-angle-formula
S60=sin(π/60)=((−S20+S202−1)1/3+(−S20−S202−1)1/3)/2
S180=sin(π/180)=((−S60+S602−1)1/3+(−S60−S602−1)1/3)/2

The Sine of ##3°## is known in closed form but when put in the third angle formula and solving the cubic, you get back to cube roots of complex numbers. I know how to solve that but it's too massive to typeset.

I found this work that shows how ugly the expressions are for those angles not multiples of ##3°## and they all involve cube roots of complex numbers.

https://www.intmath.com/blog/wp-content/images/2011/06/exact-values-sin-degrees.pdf
 
  • #32
bob012345 said:
The Sine of ##3°## is known in closed form but when put in the third angle formula and solving the cubic, you get back to cube roots of complex numbers. I know how to solve that but it's too massive to typeset.

I found this work that shows how ugly the expressions are for those angles not multiples of ##3°## and they all involve cube roots of complex numbers.

https://www.intmath.com/blog/wp-content/images/2011/06/exact-values-sin-degrees.pdf
If what you say is true, there are two closed formula ways of getting ##sin(3°)##, one using the third angle formula (described in my post above), one that avoids roots of a complex number (as described by you). That suggests (strongly suggests) that there may be a method of performing the third angle formula without taking the root of a complex number. If that is the case, then there is a solution.

Clearly it would be ugly - but not so ugly if it was expressed as a progressive series of equations - as with the series that I presented. Values are commonly expressed as a combination of operations such as add, subtract, divide, square root. Expressing a value as a series of operations just means that you are expanding your available operations before providing the final formula. It's equivalent to ##S_180 = f_a(f_b(f_c(f_d(f_e()))))##. A great way to keep the ugliness down.
 
  • #33
(edited in response to comment from @bob012345)
from this page (intmath.com), a solution that include the cube roots of complex numbers:
sine-1-degree.png

A comprehensive list is available here: Exact value from sine 1° to sine 90°
 
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  • #34
.Scott said:
Yes, but my view is that this is not really the solution I sought. There is one real root which must be the Sine of 1° but expressing it in a closed form without roots of complex numbers or Sines and Cosines of other angles which was the original intent of the post, is not possible
 
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  • #35
.Scott said:
(edited in response to comment from @bob012345)
from this page (intmath.com), a solution that include the cube roots of complex numbers:
View attachment 269775
A comprehensive list is available here: Exact value from sine 1° to sine 90°
I tried evaluating the simpler case of Sin(20°) and after a boatload of computation I got this as one of 9 possibilities;

$$Sin(20°)= \frac 1 2( (Cos(10°) - \sqrt 3 Sin(10°))$$

Alternativly, I also got;

$$Sin(20°)= -\frac 1 2 ((Cos(50°) - \sqrt 3 Sin(50°))$$

Which I could have arrived at much easier by just using the identities;

$$Sin(a ± b) = Sin(a)Cos(b) ± Cos(a)Sin(b)$$

Essentially, evaluating these cubic roots of complex numbers either gives more cubic roots of complex numbers or if you use De Moivre's theorem you end up deriving several trigonometric identities. At least that is my experience.
 
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1. What is the exact expression for sine of 1 degree?

The exact expression for sine of 1 degree is 0.01745240643728351.

2. How is the exact expression for sine of 1 degree calculated?

The exact expression for sine of 1 degree is calculated using the trigonometric identity sin(θ) = opposite/hypotenuse, where θ is the angle in degrees and the opposite side is equal to the sine of the angle.

3. Why is the exact expression for sine of 1 degree important?

The exact expression for sine of 1 degree is important because it allows us to accurately calculate the sine of 1 degree without having to rely on approximations or rounding off. This is especially useful in fields such as engineering and physics where precise calculations are necessary.

4. Can the exact expression for sine of 1 degree be simplified?

No, the exact expression for sine of 1 degree cannot be simplified any further as it is already in its simplest form.

5. How does the exact expression for sine of 1 degree compare to the value of sine in radians?

The exact expression for sine of 1 degree is equivalent to the value of sine in radians, which is approximately 0.017453292519943295. This shows that the sine function is periodic with a period of 360 degrees or 2π radians.

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