Exact Expression for Sine of 1 Degree

[bob012345]

TL;DR

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 $Sin(1)$?

 

[fresh_42]

Radian or degree?

 

[Infrared]

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.

 

[bob012345]

Radiant or degree?

Degree.

 

[bob012345]

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.

 

[fresh_42]

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.

 

[bob012345]

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.

 

[jedishrfu]

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 %

 

[mfb]

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.

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.

 

[bob012345]

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.

 

[bob012345]

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...

 

[mfb]

Instead of repeatedly complaining what you think you would get, why don't you actually do the calculation and see what you really get?

 

[jedishrfu]

I found this interesting reference with a lot of cool math info:

https://www.maa.org/sites/default/files/pdf/awards/college.math.j.47.5.322.pdf

 

[willem2]

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

 

[Infrared]

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.

 

[Vanadium 50]

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

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

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)

 

[bob012345]

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

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

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.

 

[bob012345]

I found this interesting reference with a lot of cool math info:

https://www.maa.org/sites/default/files/pdf/awards/college.math.j.47.5.322.pdf

I've seen that and I suspected he was correct.

 

[bob012345]

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.

 

[Vanadium 50]

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)

 

[Infrared]

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.

 

[bob012345]

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

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

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?

 

[Vanadium 50]

Because they aren't separable in closed form: i.e. you can't collect all the i's in one place.

 

[jedishrfu]

Here's a further discussion of constructible polygons:

https://en.wikipedia.org/wiki/Constructible_polygon

 

[Infrared]

@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).

 

[Vanadium 50]

I believe the argument is in the thread I linked to.

 

[Infrared]

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.

 

[Vanadium 50]

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.

 

[Infrared]

@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.

 

[.Scott]

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
$S_{60} = sin(\pi/60) = \left(\left (-S_{20}+\sqrt{S_{20}^2-1}\right)^{1/3} + \left(-S_{20}-\sqrt{S_{20}^2-1}\right)^{1/3} \right)/2$
$S_{180} = sin(\pi/180) = \left( \left(-S_{60}+\sqrt{S_{60}^2-1}\right)^{1/3} + \left(-S_{60}-\sqrt{S_{60}^2-1}\right)^{1/3} \right)/2$