$\sin n$ having a closed form expression with no complex terms?

Asked Jan 11 '14 at 14:15

Active Jan 11 '14 at 14:15

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[All angles are in degrees] I have heard that we cannot express $\sin 1$ in closed form with no complex terms. However, I know that we can derive $\sin 18$ by solving $\cos 3x = \sin 2x$. Thus we can get trig values for $72$ degrees and then get to $\sin 3$ as we know every term of the expansion of $\sin 75-72$. And so we can have sine of every integral multiple of $3$ in closed form with no complex terms.

But what about others such as $\sin 5$? Well if we know $\sin 1$, our problems would be solved, which I have heard is not closed form with no complex terms. Can anyone prove that? Maybe we can use some other method or some sort of geometric solution to find out, $\sin 1$ or $\sin n$ for any natural number but it turns out not. Why?

asked Jan 11 '14 at 14:15

Sawarnik

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I think you might find your answer here. – David Mitra Jan 11 '14 at 14:27
@DavidMitra Is not there something more elementary? – Sawarnik Jan 11 '14 at 14:33
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Perhaps for specific angles; I'm not sure... – David Mitra Jan 11 '14 at 14:36
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What is a "closed form with no complex terms"? – Hagen von Eitzen Jan 11 '14 at 15:01
@HagenvonEitzen Well something like a real number with an expression having finite number of terms and does not contain the symbol $i$. – Sawarnik Jan 11 '14 at 15:03
A list of exact values of trigonometric functions, for all multiples of $3^0$. – Lucian Jan 12 '14 at 00:22
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@Lucian: You might want to use "3^\circ". (I misread it as 3^0 = 1 at first.) – user21820 Jan 23 '14 at 01:50

$\sin n$ having a closed form expression with no complex terms?

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