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How can we express $\cos(\frac{x}{k})$ ($k \in \mathbb{N}$) in terms of $\cos(x)$?

And $\sin(\frac{x}{k})$ in terms of $\sin(x)$?

Edit

Maybe this another question helps. Is there a $T_n(x)$ inverse? (Link don't say anything about this)

If $\cos(x)=T_k(\cos(\frac{x}{k}))$, is there $T^{-1}_n(x)$ that $\cos(\frac{x}{k})=T^{-1}_k(\cos(x))$ is at least one solution for the question? What is the $T^{-1}_n(x)$ formula?

Later
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GarouDan
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    Let $x=ky$. You want to express $\cos y$ in terms of $\cos ky$. – Pedro May 09 '13 at 21:43
  • I agree, but I will use with fractions, so I wrote in this way. I had seen some time before the inverse of this question, but I don't know how to solve this one. – GarouDan May 09 '13 at 21:50
  • Your problem reduces to find the inverse of the Chebyshev polynomials, but it is well known that for degree $5$ or higher, there is no formula to solve such equations. – jjagmath Sep 07 '23 at 14:30

1 Answers1

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For any positive integer $k$, the trigonometric function $\cos ky$ is a polynomial in $\cos y$. For example, with $k = 2$, $$ \cos 2y = 2\cos^2 y - 1. $$ The polynomial here is $T_2(u) = 2u^2 - 1$. In order to express $\cos y$ in terms of $\cos 2y$ (this is equivalent to expressing $\cos \frac{x}{2}$ in terms of $\cos x$ via the substitution $y = 2x$), we have to invert the polynomial function $T_2$.

In other words, you have to solve $v = 2u^2 - 1$ for $u$. This doesn't have a unique solution, but we can describe both possible solutions: $$ u = \pm \sqrt{\frac{1 + v}{2}}. $$ This is the well-known "half-angle formula for cosine" $$ \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}. $$

For larger $k \in \mathbb{N}$, you have to solve the higher degree polynomial equations $$ v = T_k(u) $$ for $u$ in order to express $\cos \frac{x}{k}$ in terms of $\cos x$.

Sammy Black
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    You added a question about the Chebyshev polynomial having inverses. Polynomials don't generally have inverses because they don't tend to be one-to-one functions. The sketches of the graphs of $T_k$ that I linked to in my answer should convince you that these functions are $k$-to-one on the interval $(0, 1)$. The upshot is that for all angles, there are $k$ possible values for $\cos \frac{x}{k}$. – Sammy Black May 09 '13 at 22:12
  • SammyBlack, this I already know, but the problem is I would like an abstract way to solve the question, if it is possible. Something like $\cos(\frac{x}{k}) = \sum_{i=0}^{k}f(k)\cos(x)^k$ – GarouDan May 09 '13 at 22:16
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    Already for $k=2$, such an expression is not possible. The half-angle formula involves a radical! – Sammy Black May 09 '13 at 22:17
  • I think find at least one solution (avoiding the other k-1 solutions) on abstract way will hold to me. I agree that we will not have a defined inverse (like you wrote $cos{\frac{x}{2}} = ^+_-$ above), but $cos{\frac{x}{2}} = ^+$ is a valid solution. – GarouDan May 09 '13 at 22:18
  • Well, if we have something like $cos{\frac{x}{k}} = T^{-1}{k}(\cos(x))$ is at least one solution of the question is I'm looking for. But what is $T^{-1}{k}$? Is this a well known function? What is its form? – GarouDan May 09 '13 at 22:21
  • Sometimes, it is correct. An example: with $x = \frac{2\pi}{3}$, $\cos \frac{x}{2} = \cos \frac{\pi}{3} = \frac{1}{2}$. Since, $\cos{x} = -\frac{1}{2}$, using the half-angle formula, $\cos \frac{\pi}{3} = \pm \sqrt{\frac{1 + (-\frac{1}{2})}{2}} = \pm \frac{1}{2}$. However, with $x = \frac{4\pi}{3}$, where $\cos x = -\frac{1}{2}$ as well, the actual values of $\cos \frac{x}{2} = - \frac{1}{2}$. The half-angle formula is ambiguous! – Sammy Black May 09 '13 at 22:22
  • The point is that there are $k$ different points on the circle $x_1, \dots, x_k$ such that $k x_i$ is a given point on the circle for each one. (These points actually form the vertices of a regular $k$-gon.) – Sammy Black May 09 '13 at 22:27