# Calculating $\sin(10^\circ)$ with a geometric method

Excuse me if this is a simple question:

What is a simple geometric method for calculating $\sin(10^\circ)$ using only the sines of $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$?

Generally, is there any geometric solution for this problem or should we use algebra?

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I doubt there is a simple method. $\sin(10^\circ)$ is a root of the irreducible cubic equation $8x^3-6x+1$. – Jonas Meyer Jan 9 '12 at 8:33
$\sin 3\alpha = 3\sin \alpha -4 \sin^3 \alpha$ – pedja Jan 9 '12 at 8:40
There is no straightedge and compass construction of the $10^\circ$ angle, so there is no "geometric method" using only the classical ools. – André Nicolas Jan 9 '12 at 9:26
@André: One should not suffer ools --classical or otherwise-- gladly. Or is it that there's no ool like a classical ool? I forget ... :) – Blue Jan 9 '12 at 9:35
If one allows for neusis (i.e. a marked straightedge), one could trisect a $30^\circ$ angle... – J. M. Jan 9 '12 at 11:18

Copy-pasting in comments that essentially answer the question, so that there's an answer here and the question might leave the unanswered list (CW so that it's editable—if you commented and want to turn your comment into an answer of your own, feel free to edit it out of this answer).

I doubt there is a simple method. $\sin(10^\circ)$ is a root of the irreducible cubic $8x^3-6x+1$. – Jonas Meyer

$\sin 3\alpha=3\sin\alpha−4\sin^3\alpha$ – pedja

There is no straightedge and compass construction of the $10^\circ$ angle, so there is no "geometric method" using only the classical [t]ools. – André Nicolas

If one allows for neusis (i.e. a marked straightedge), one could trisect a $30^\circ$ angle... – J. M.

In addition, it is a not easy theorem using Tools from Galois Theory that there is no expression for $\sin(10^\circ)$ in terms of real $n$-th roots. By Cardano's Formula, there is an expression using complex cube roots, but that is not a useful answer to your question. – André Nicolas

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Thanks, you have done a service. – André Nicolas Jan 11 '12 at 3:01