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Is there a way to solve for $x$ in $\dfrac{\cos^{-1}(ax)}{\cos^{-1}(bx)} = c$?

I guess it comes down to, are there any sine multiplication formulas I don't know about?

The motivation for this is to find $x_0$, $y_0$, $f$ and $a$ in the equation $y = a \cos(f(x - x_0)) + y_0$ given four points $(x_1, y_1)$, $(x_2, y_2)$, …
Constraining it to intersect $(x_1, y_1)$ and solving for $f$ gives $f = \dfrac{\cos^{-1}\left(\dfrac{y_1 - y_0}{a}\right)}{x_1 - x_0}$.

Eliminating $f$ for the original equation gives $y = a \cos\left(\dfrac{x - x_0}{x_1 - x_0}\cos^{-1}\left(\dfrac{y_1 - y_0}{a}\right)\right) + y_0$.

Constraining it to intersect $(x_2, y_2)$ gives a version of the simplified equation in my question:

$$\cos^{-1}\left(\dfrac{y_2 - y_0}{a}\right) = \dfrac{x_2 - x_0}{x_1 - x_0}\cos^{-1}\left(\dfrac{y_1 - y_0}{a}\right)$$

(My question replaces $\dfrac{1}{a}$ with ‘$x$’, $y_2 - y_0$ with ‘$a$’, $y_1 - y_0$ with $b$ and $\dfrac{x_2 - x_0}{x_1 - x_0}$ with $c$.)

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Looks like a good candidate for a numerical solution and a bad candidate for an algebraic solution to me. – Ross Millikan Aug 22 '11 at 4:40
@Ross: A one-dimensional numerical search is required in the end, but all but one of the variables can be eliminated algebraically; see…. – joriki Aug 22 '11 at 20:32
up vote 3 down vote accepted

If $c=m/n$ is rational, after multiplying by $n \cos^{-1}(bx)$ and taking cosine of both sides, the equation can be written as $$P_m(bx) = P_n (ax)$$ where $P_i$ is the degree $i$ Chebyshev polynomial. Thus $x$ is an algebraic function of $a$ and $b$, and if both of those parameters are algebraic numbers then so is $x$.

If $c$ is irrational or $a$ or $b$ is transcendental then there is no algebraic solution except in some easy cases such as $c = 1$ or $a/b$ rational.

In short, the only "sine multiplication formulas" that are useful here are the ones for expressing $\cos n\theta$ as a polynomial in $\cos \theta$. There is one such formula for every integer $n$.

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