# Trigonometric/polynomial equations and the algebraic nature of trig functions

Prove or disprove that an equation involving one trig function (either $\sin,\cos,\tan$, etc) with an argument of the form $ax+b$ for non-zero rational $a,b$ and a polynomial with non-zero rational coefficients and a constant term not equal to $\pm1$ or zero is not solvable in closed form. For example,

$$\sin(x)=2-x-x^2$$

It has solutions near $x=0.752$ and $x=-2.242$

My reasoning for why there is no closed form solution is because

1. If $x$ is a rational multiple of $\pi$, the LHS is algebraic, while the RHS is transcendental.

2. If $x$ is of the form $x=\arcsin(u)\ne\frac ab\pi$ and $u$ is algebraic, then the LHS is algebraic but the RHS... is not algebraic? See When is ArcTan a rational multiple of pi? for some information.

3. If $x$ is none of the above, I don't think there exists a closed form solution since $\sin(x)$ cannot be calculated in closed form and neither can $2-x-x^2$.

Can someone prove this general idea?

If it is solvable, then under what conditions?

Attempting to reduce the amount of questions that ask for closed form solutions in these scenarios, like

Nonlinear algebraic equation with trigonometric function

How to solve $x+\sin(x)=b$

Solving an equation

etc.

To clarify, closed form in this context is a solution in terms of well-known constants and a finite combination of well-known functions.

• Would it be an intermediate step to think about solving $e^x = 1-x-x^2$? Change $e$ to $a$ and ask whether $a$ needs to be rational/algebraic for a closed form to exist? Just thinking out loud.... – B. Goddard Jan 6 '17 at 3:00
• @B.Goddard that is a good idea, though I'm not sure where it goes... – Simply Beautiful Art Jan 6 '17 at 12:26
• How would you look at sinx = 0? – jnyan Jan 24 '17 at 14:22
• @jnyan Do look up Niven's theorem. – Simply Beautiful Art Jan 24 '17 at 14:32
• Would thinks like $\sin x=x$, $\sin x =x-\frac{x^3}{6}$, $\sin x =x-\frac{x^3}{6}+\frac{x^5}{120}$ be allowed? These have easy to see elementary solutions. – S.C.B. Jan 29 '17 at 13:42

"Closed form" means expressions of allowed functions. If an equation is solvable in closed form depends therefore on the functions you allow.

Your idea of proof is appropriate. But let me go a more systematic way.

Let us assume an equation $trig(ax+b)=P(x)$, $a, b \in \mathbb{Q}$, $a\neq 0$, where $trig$ is one of the trigonimetric functions, and $P(x)$ is a non-constant polynomial in $x$ with rational coefficients. The trigonometric functions are the functions $\sin$, $\cos$, $\tan$, $\sec$, $\csc$ and $\cot$.

The function terms of the trigonometric functions are rational expressions of $e^{xi}$ and $e^{-xi}$, e.g.: $\sin(x)=R(e^{xi},e^{-xi})=-\frac{1}{2}i\left(e^{xi}-e^{-xi}\right)$, $R$ a rational expression of its arguments.

a) Let us allow the algebraic numbers as closed form:

If $x$ is algebraic, $P(x)$ is algebraic. But for what algebraic $x$ is $trig(ax+b)$ algebraic? $e^{(ax+b)i}$ is algebraic iff $ax+b=0$, and $e^{-(ax+b)i}$ is algebraic iff $ax+b=0$. Therefore $trig(ax+b)$ is algebraic if $ax+b=0$, that means if $x=-\frac{b}{a}$. For all other algebraic $x$, a conclusion from Lindemann-Weierstrass theorem is $trig(ax+b)$ is transcendent. Therefore the only algebraic solution of the given equation is, if the solution exists, $x=-\frac{b}{a}$.

b) Let us now allow the elementary functions as closed form:

The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).

The solutions of an equation of only one unknown are, if they exist, numbers. Each number $c$ is the value of an elementary function, of a constant function $x\mapsto c$, with $x,c\in\mathbb{C}$. Therefore each solution of each equation of only one unknown is in closed form.

Solvability of equations by closed-form expressions is related to the question of existence of inverses of the functions which are contained in the equation.

The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in [Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.

As shown above, we have $trig(ax+b)=P(x)$, that means $R(e^{(ax+b)i},e^{-(ax+b)i})=P(x)$. This equation can be transformed to an equation $P_{1}(e^{(ax+b)i},e^{-(ax+b)i},x)=0$, where $P_{1}$ is a polynomial in dependence of its arguments. The left side of this equation is an elementary function of $x$. Let us name it $f$. If the equation is solvable by applying elementary functions to the equation, the inverse of $f$ has to be an elementary function. A conclusion of the theorem of Ritt in [Ritt 1925] is: If $f$ and its inverse are both elementary, a function term for $f$ in exp-ln representation must exist that doesn't contain an algebraic function of algebraically independent elementary functions.

$P_{1}$ is an algebraic function, and with help of Lindemann-Weierstrass theorem follows its arguments $e^{(ax+b)i}$, $e^{-(ax+b)i}$, $x$ are algebraically independent. $f(x)=P_{1}(e^{(ax+b)i},e^{-(ax+b)i},x)$ cannot have an exp-ln representation that doesn't contain an algebraic function of algebraically independent elementary functions. Therefore the given equation is not solvable by applying only elementary functions.

c) You could allow algebraic expressions of known Standard functions as closed form:

If the trigonometric functions can be decomposed into compositions of algebraic functions and other known Standard functions than $\exp$ and $\ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.