So, the question is: $\cos{n^{\circ}}$ can be expressed in real radicals iff $3 \mid n$? Is it true? The first part is easy: if $3 \mid n$ we can express it, because $\cos{36^{\circ}}=\cos{\pi\over{5}}=\frac{\sqrt{5}+1}{4}$, thus we can express $\cos{18^{\circ}}$, and now also can $\cos{48^{\circ}}=\cos{(18^{\circ}+30^{\circ})}$. From it follows that we can express $\cos{3^{\circ}}$, because $3={48\over 16}$. Thus we can express $\cos{3k^{\circ}}$ for any $k\in \mathbb N$, using formula of $\cos(nx)=...$. The main question is: why can't we express it if $n\neq3k$? Or we can?
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2Are you familiar with constructible numbers/straight-edge and compass constructions? – Alex Wertheim Dec 30 '13 at 22:34
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5See this 26 September 2005 sci.math post and the following math StackExchange question: rational angles with sines expressible with radicals. – Dave L. Renfro Dec 30 '13 at 22:36
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5I realize that you probably want $n$ to be an integer. Yet you may find this (=17-gon) at least somewhat interesting. – Jyrki Lahtonen Dec 30 '13 at 22:38
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@StefanSmith I think it was really about $3^\circ$ as halfing $48^\circ$ four times. – Berci Dec 31 '13 at 01:26
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1@DaveL.Renfro: Would you like to turn your comment into an answer? That would be helpful in dealing with related questions such as this one. – ccorn Jul 26 '16 at 17:34