The question is linked with the previous one:

I have $3 \times 3$ rotation matrix $ R=\begin{bmatrix} \dfrac{1}{\sqrt{2}} & -\dfrac{ 1}{2} & \dfrac{1}{2} \\ \dfrac {1}{\sqrt{2}} & \dfrac {1}{2} & -\dfrac{ 1}{2} \\ 0 & \dfrac{1}{\sqrt{2}} & \dfrac{1}{ \sqrt{2} } \end{bmatrix} $

and I would like to be sure that this matrix can't be a n-th root of Identity for any natural $n$
i.e. $R^n \neq I$ for any $n$.
The most straightforward method, it seems, is to calculate from the trace of $R$ cosine for rotation angle $ \theta $ of this matrix and to prove that $\cos (n\theta) \neq 1$ for any $n$ (axis remains fixed for $R^n$ operation).

In this case we have $\text{trace}(R)=\sqrt{2}+\dfrac{1}{2}=1+2\cos(\theta)$
and hence $\cos(\theta)=\dfrac{1}{2}\sqrt{2}-\dfrac{1}{4}$.

However trigonometric identities from Wikipedia seem to be quite difficult to apply and I don't know how to move from this point.

  • Is it possible to prove that if $\cos(\theta)=\dfrac{1}{2}\sqrt{2}-\dfrac{1}{4}$ then $\cos(n\theta) \neq 1$ ?

  • Could the proposition be generalized for any expression of the form $\cos(\theta)=a\sqrt{b}-c$ where $a,c$ are non-zero rational numbers, $b$- natural $>1$ ? if not what are conditions when $\cos(n\theta)=1$ ?


If $\theta$ is a rational multiple of $\pi$ then $2\cos\theta=e^{i\theta} +e^{-i\theta}$ is an algebraic integer, as $e^{\pm i\theta}$ will be roots of unity.

Here, $2\cos\theta=\sqrt2-1/2$ is not an algebraic integer (as $\sqrt2$ is an algebraic integer but $1/2$ is not).

  • $\begingroup$ You mean probably 'cos' instead of 'sin'.. what means this construct 'algebraic integer' ? The leaps are too big for me in your answer... $\endgroup$ – Widawensen Dec 20 '17 at 12:02
  • $\begingroup$ Now I'm studying en.wikipedia.org/wiki/Algebraic_integer $\endgroup$ – Widawensen Dec 20 '17 at 12:04
  • $\begingroup$ Probably this is a naive question: why we can't multiply additionally sides of equation by two receiving $4\cos\theta=2\sqrt2-1 $ . $2\sqrt2-1$ is "algebraic integer"? $\endgroup$ – Widawensen Dec 20 '17 at 12:10
  • $\begingroup$ $2\sqrt2-1$ is an algebraic integer: just as $\sqrt2-1/2$ isn't. $\endgroup$ – Lord Shark the Unknown Dec 20 '17 at 12:11
  • 1
    $\begingroup$ Indeed it's an error. They might have meant $(1+\sqrt{\color{Red}{\bf -}7})/2$ is an algebraic integer. $\endgroup$ – anon Dec 20 '17 at 15:03

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