To specify $a,b,k$ in $a\cos(kx)+b$

Question

If we have known that $f(x)=a\cos(kx)+b$ for some $ f:(0,\infty)\to \mathbb C$, is there a method of choosing finite $(x,f(x))$ to know what $a,b,k$ are?

Answer 1

Even if you require $k>0$ to avoid the sign ambiguity, there will be trouble. After eliminating $a$ and $b$ from the equations $f(x_1) = y_1$, $f(x_2) = y_2$, $f(x_3) = y_3$ you get

$$ (y_2 - y_3) \cos(k x_1) + (y_3 - y_1) \cos(k x_2) + (y_1 - y_2) \cos(k x_3) = 0 $$

from which you want to determine $k$. However, the left side will usually be $0$ for infinitely many $k$. The exceptions are when one of the $x_i$, let's say $x_1$, is $0$, and $y_3-y_1$ and $y_1-y_2$ have the same sign, in which case the only way to satisfy the equation is for $\cos(k x_1)$ and $\cos(k x_2)$ to both be $1$, and therefore (if $x_1/x_2$ is irrational) $k = 0$.

Thus if the answer happens to be $k = 0$ you can choose the $x_i$ and $y_i$ to prove that, but if the correct $k$ is nonzero you can't determine it.