I've been playing with sine of integer-degree angles; that is, $\sin\left(\frac{k \pi}{180}\right)$, where $k$ is an integer.

I've noticed that you can divide the angle by $2$ and get sine of smaller and smaller angle by solving a quadratic equation for the correct root. One can also divide the angle by $3$ and try to solve cubic equation.

I found the expression for $\sin 3^\circ$. While it's not very short, it consists of some square roots (possibly nested, but not necessarily). The point is:

The expression for $\sin 3^\circ$ does not have to contain any mention of complex numbers. No roots of negative numbers, no $i$'s in the expression.

Of course, you can get smaller angles by dividing by $2$, so you can get a nice expression for $\sin 1.5^\circ$, etc.

Then I've searched for sine of $1^\circ$ and $2^\circ$, and those expressions are somehow odd. They involve imaginary units. However, sine of real number is a real number, so in the end, those expressions shouldn't have a imaginary part. So I thought, "Okay, I will play with these expressions, and eventually I should get something real consisting of some roots (possibly nested), but no complex numbers." And here's the thing:

No matter how hard I tried, my expression either reverted simply to "$\sin 1^\circ$" (which is nice, but does not tell anything apart from fact that the expression was correct), or else it involved complex numbers.

Here's my question:

Is there a way to express $\sin 1^\circ$ through square/cubic roots (possibly nested) of positive real numbers, without any involvement of the imaginary unit? If yes, how? If not, why not?

(In the latter case: Where is the point? What's the essence of having purely real expression vs. some kind of complex mid-steps though the answer is real? How's that connected to a $1^\circ$ vs $3^\circ$?)

Thank you! Sorry if it's a stupid question with an obvious answer.

  • $\begingroup$ There is a closed algebraic form, but it involves square roots inside cube roots, and the imaginary unit. $\endgroup$ Aug 7 '16 at 10:55
  • $\begingroup$ This depends a lot on what you consider a "real expression" to be. Apparently the square-root function qualifies for you, but the sine doesn't -- why? Or how about $f(a,b)$ defined as the largest real root of $x^3+ax+b$? $\endgroup$ Aug 7 '16 at 11:10
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    $\begingroup$ Just for reference: Here's a list of "Exact Values for the Sin of All Integers" by James Parent. (For $1^\circ$, he uses the imaginary unit.) Also, in this answer, I present a common form (involving the golden ratio) for sines of angles of multiples-of-$3^\circ$. $\endgroup$
    – Blue
    Aug 7 '16 at 11:10
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    $\begingroup$ You have rediscovered the "casus irreducibilis", en.wikipedia.org/wiki/Casus_irreducibilis When an irreducible cubic has three real roots, they can't be expressed without nonreal numbers. $\endgroup$ Aug 7 '16 at 11:14
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    $\begingroup$ Historically, this is how complex numbers came to be accepted in the first place: they showed up as intermediate steps in solving cubic equations whose coefficients were real and which were known to have three real roots. $\endgroup$
    – David K
    Aug 7 '16 at 14:24

As some comments suggest, you can't eliminate the imaginary unit from the radical expression for $\sin 1°$.

The crux of the matter is this: $x=\sin 1°$ satisfies a cubuc equation involving $\sin 3°$, to wit:

$3x-4x^3=\sin 3°$

We can express the right side with real radical expressions and plug away at Cardano's formula, only to find that the expression under the square root is negative and we are stuck with complex radicals. That's because the cubic equation above has three real roots, the other two being $\sin 121°$ and $\sin 241°$.

A rational multiple of $\pi$ can always have its trigonometric functions expressed with radicals, but the radical expressions cannot avoid the imaginary unit unless they are all square roots --meaning the angle has to be constructible!

  • $\begingroup$ So sin alpha is a real radical iff alpha is constructible angle with the compass and straightedge? $\endgroup$
    – user16320
    Aug 14 '16 at 16:59
  • $\begingroup$ That's how it works. One reason behind this is that complex square roots can be resolved algebraically into real and imaginary parts. The square root of $2+i$ is$\sqrt{1+(sqrt{5}/2)}+i\sqrt{-1+(sqrt{5}/2)}$, the cube root of $2+i$ is casus irreducibilis. $\endgroup$ Aug 14 '16 at 19:13

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