# Exact Expression for Sine of 1 Degree

• I
Gold Member

## Summary:

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

fresh_42
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Infrared
Gold Member
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.

• • Abhishek11235, scottdave, bob012345 and 1 other person
Gold Member
Degree.

Gold Member
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
Mentor
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.

Gold Member
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
Mentor
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
Mentor
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.

Gold Member
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.

• jedishrfu
Gold Member
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....

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

• fresh_42
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

• Abhishek11235, Keith_McClary, bob012345 and 1 other person
Infrared
Gold Member
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.

Last edited:
• bob012345
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You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

$$\frac{\sqrt{i} - \sqrt{-i}}{2i}$$

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)

• • tworitdash, Abhishek11235, Keith_McClary and 2 others
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You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

$$\frac{\sqrt{i} - \sqrt{-i}}{2i}$$

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.

Gold Member
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.

Staff Emeritus
2019 Award
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)

• • Keith_McClary, hutchphd, jedishrfu and 1 other person
Infrared
Gold Member
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{2}## is not constructable with compass and straightedge.

Also, if someone could clear my confusion in post 15, I would be thankful.

Last edited:
Gold Member
You might be interested in https://www.physicsforums.com/threa...es-to-aops-for-prealgebra-precalculus.986266/

$$\frac{\sqrt{i} - \sqrt{-i}}{2i}$$

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?

Staff Emeritus
2019 Award
Because they aren't separable in closed form: i.e. you can't collect all the i's in one place.

• • Keith_McClary and bob012345
Infrared