Sine Angle Product Formula

In this post, we address the following question: does there exist a polynomial identity for $\sin(ab)$ in terms of $\sin a,\cos a,\sin b,\cos b$ like for the single angle addition formula? Formulated rigorously, does there exist a polynomial $P$ such that $\sin(ab) = P\left( \sin a,\cos a,\sin b,\cos b \right)$ for all $a,b$ ? We suspect that $P$ does not exist, which we try to show here. (Actually, a more general question that is worth discussing is whether there exists a rational identity for $\sin(ab)$ , e.g. with $P$ replaced by a rational function. But we discuss polynomial for now.)

Choose $a = \pi$ and $b = n\mathbb{\in Z}$ , so that we want $P$ such that

$\displaystyle 0 = \sin(n\pi) = P\left( \sin n,\cos n \right)\ for\ all\ n\mathbb{\in Z.}$

We want to show that this means $P$ cannot have finite degree, contradicting that $P$ exists. So let $P$ be a polynomial in two variables such that

$\displaystyle P\left( \sin n,\cos n \right) = 0\ for\ all\ n\mathbb{\in Z.}$

This looks ripe for a setup with the Factor Theorem to show that $P$ cannot have finite degree … so can we do something similar in two variables here?

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Look at https://math.stackexchange.com/questions/4155545/application-of-remainder-theorem-and-factor-theorems-to-multivariable-polynomial and algebra precalculus – Does there exist a formal statement of the Multivariable Factor Theorem? – Mathematics Stack Exchange. To apply either of the two results stated in the answer in the second link, we need a more general relation of the form

$\displaystyle P\left( x,f(x) \right) = 0,$

not just that $P$ is zero on some specific points. Is it possible to yield a result on specific points, possibly by applying the single-variable Factor Theorem?

(Aside: the OP’s question in the second link states that the second result is a generalization of the first, but that is based on their statements for complex polynomials. The first result applies to real polynomials, and it looks like more generally to commutative rings, while the second applies only to complex polynomials, or more generally to algebraically closed fields.)

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The setup we have is as follows. We have a two-variable polynomial $P$ and $a,b\mathbb{\in R}$ such that $P(a,b) = 0$ . Can we derive a factor of $P$ from this?

Let’s say we tried to apply the result that if $f$ is square-free and $f(x,y) = 0 \rightarrow P(x,y) = 0$ then $f$ is a factor of $P$ . So say we have our polynomial $P$ , where

$\displaystyle P\left( \sin n,\cos n \right) = 0\ for\ all\ n\mathbb{\in Z.}$

To apply the Factor Theorem-like result, we want to find a two-variable polynomial $f$ such that $f(x,y) = 0 \rightarrow x = \sin n,y = \cos n$ . We take inspiration from the fact that $x^{2} + y^{2} = 0 \rightarrow x = y = 0$ . But we need to be really careful here, because the theorem is stated for complex polynomials (and more generally for polynomials over algebraically closed fields), but in that case it’s not true that $x^{2} + y^{2} = 0 \rightarrow x = y = 0$ (take $x = i,y = 1$ for instance.)

We could try this way. What we’ll show is that there exists no real polynomial $P$ such that $P\left( \sin n,\cos n \right) = 0$ for all integers $n$ . But we’ll use the complex polynomial result to show this.

Let $f(x,y) = \left( x - \sin n \right)^{2} + \left( y - \cos n \right)^{2}$ . We can factorize this as

$\displaystyle f(x,y) = \left( \left( x - \sin n \right) + i\left( y - \cos n \right) \right)\left( \left( x - \sin n \right) - i\left( y - \cos n \right) \right).$

Let $f(x,y) = \left( \cos n \right)x - \left( \sin n \right)y$ . In order to have $f(x,y) = 0$ , we must have $x = \left( \sin n \right)a,y = \left( \cos n \right)a$ . Then …

It looks hard to divine a general linear relation that $P$ satisfies that would allow us to conclude what we want. Let’s try a different approach.

We have that, viewing $f$ as a real polynomial, $f(x,y) = 0 \rightarrow x = \sin n,y = \cos n$ . Can we use this somehow?

Actually, aside: deriving factors from specific points seems particularly implausible on second thought. Consider $P(x,y) = x^{2} + cy^{2}$ for any constant $c$ . Clearly, $P(0,0) = 0$ , but over the complex numbers there are only two factors of this polynomial, which are dependent on $c$ . So we probably need more trigonometry and setup beyond just $P\left( \sin n,\cos n \right) = 0$ for all integers $n$ . Maybe some deeper facts about exponentials and complex numbers and how they relate to the trigonometric functions. In fact, maybe we can turn this into a problem about exponentials and complex numbers and use some high-powered results from complex analysis to derive what we want.

This discussion is continued in a follow-up post.

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