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I've just encountered a problem that seems to me interesting enough so that some result exists on the subject. I was working on a problem in complex analysis, in which I needed the fifth root of a complex number in rectangular coordinates.

Long story short, I eventually reasoned that, being in rectangular form I could easily know $\cos\theta$, $\theta$ being the argument. So I now just needed to find $$\cos{\frac{\theta}5}$$

So I said: Oh, I'll just look up one of those handy trig identities and that's that.

Well, the identity for this one wasn't to be found anywhere, or any identities of fractions of angles other than the half angle. So, taking inspiration from the half angle formula's deduction, I expanded $\cos5x$ using sum and double angle identities, which gave: $$\cos5x=16\cos ^5x-20\cos^3x+5\cos x$$ Unsurprisingly I later found that there were tables of such polynomials, and even recursion relations for the general form, and similarly for other trig functions. To my dismay though, I now had to solve a $5$-th degree polynomial in order to find $\cos x/5$. Every time this happens to me I recall the original exasperation I felt when I learned that no closed form existed for degree $5$ and up. It really pains me.

Back to the main question though, is the problem of finding expressions for trigonometric functions of fractions of angles really so non-trivial? Something tells me yes, because then we could easily calculate cosines and sines of any rational multiple of $\pi$, which doesn't seem so trivial.

If someone knows about this, or even just the solution for the specific polynomial I have written, their help will be appreciated.

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  • $\begingroup$ The solutions are precisely when $\cos 5x =0$. That says enough about the polynomial... If you want to get $\cos\pi/10$, just set $3x=\pi/2-2x$, get sines of both sides, and you will get a factorable cubic(in $\sin x$), with only one admissible solution. There are half-angle formulas, but since no 5-degree or higher polynomial is always solvable, I think that they can't exist in the general case. $\endgroup$ – chubakueno Feb 16 '14 at 23:36
  • $\begingroup$ Is your problem related to solving, for $x$, $cos(nx)=k$ ? Are you looking for numerical methods for that ? $\endgroup$ – Claude Leibovici Feb 17 '14 at 9:19
  • $\begingroup$ @ClaudeLeibovici Alas, no it's not related to that. I'm studying mathematics and the question arose in an exercise in complex analysis. The question was unrelated really, but in an intermediate step this came up. $\endgroup$ – GPerez Feb 17 '14 at 14:04
  • $\begingroup$ When something elemental does not appear in the books is because it is not available because impracticable. There are many examples in mathematics, one of them being that no explicit formula for the sinus of the n-th of the angle for $n≥3$ appears but yes for $\frac x2$. In the case of the n-th angle giving an explicit formula is equivalent to give a formula for a polynomial of the degree n, something it is known does not exist. $\endgroup$ – Piquito Aug 11 '16 at 16:17
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As you seem to have already got it, the problem translates into the problem of finding the roots of the 5-th degree Chebyshev polynomial of the first kind T_5 with the Cosine of the fraction-of-an-angle as the argument.

Infact applying the Cosine to any angle x/n with integer n>0, requires us to find the roots of the corresponding n-th Chebyshev polynomial and attempt to solve over these roots.

For the Sine if you don't want to convert it into Cosine you can use instead the Spread polynomials in a similar way.

For Hyperbolic functions I cannot locate any source for an answer but I guess that a similar strategy applies as long as we work with complex numbers (given the cross-identities between functions).

ouch... quite old question...

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This is how I would act: $$|16x^5-2x^3+5x|\le 1$$ Besides, to know what the value of $\cos \frac x5$ is one must first know what is the value of $x$. So I can plot the polynomial, calculate numerical approximations or view directly in the graph an approximate value of $\frac x5$

enter image description here

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  • $\begingroup$ Thank you for the response, this is quite the old question! However I can't quite consider this an answer to it. I am well aware of numerical methods of solving polynomial equations. $\endgroup$ – GPerez Aug 13 '16 at 16:57
  • $\begingroup$ I wrotte "This is how I would act". However, it could suggest how to do in similar problems. For example from the graph you get an approximate value that you can refine with well known iterations of successive aproximations. Regards. $\endgroup$ – Piquito Aug 14 '16 at 15:49

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