## sin and cos of 1° [closed]

$sin(90°)\cong sin(\frac{1}{2}\pi)\cong\ 0$
$cos(90°)\cong sin(\frac{1}{2}\pi)\cong\ 1$

$sin(60°)\cong sin(\frac{1}{3}\pi)\cong\frac{\sqrt{3}}{2}$
$cos(60°)\cong sin(\frac{1}{3}\pi)\cong\frac{1}{2}$

$sin(45°)\cong sin(\frac{1}{4}\pi)\cong\frac{\sqrt{2}}{2}$
$cos(45°)\cong sin(\frac{1}{4}\pi)\cong\frac{\sqrt{2}}{2}$

$sin(30°)\cong sin(\frac{1}{6}\pi)\cong\frac{1}{2}$
$cos(30°)\cong sin(\frac{1}{6}\pi)\cong\frac{\sqrt{3}}{2}$

An heres my question (just for the purpose of curiosity): What number (not with decimals, i want numbers like for those above, square roots and fractions allowed) would $x$ be:
$sin(1°)\cong sin(\frac{1}{180}\pi)\cong\ x$
$cos(1°)\cong sin(\frac{1}{180}\pi)\cong\ x$
And how would i generally derive ANY degree, lets say $sin(3°)$ or wathever.

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I think that as it stands, the question would belong better on math.stackexchange.com – Yemon Choi Feb 19 2012 at 7:58
ok i'm going to post it there – keltik Feb 19 2012 at 8:01
To find out if there is a closed formula I would use the well-known formulas for $sin(x+y), cos(x+y)$ and express $sin(30°)=sin(2 \cdot 3 \cdot 5 \cdot 1°)$ as polynomial in $sin(1°)$. This leads to a rational polynomial in degree 15 or 30. Then compute its Galois group. If the Galois group is solvable, there is a closed formula, otherwise not. For computing the Galois group you can use, for example, GAP (also see mathoverflow.net/questions/22923/…). Unfortunately I haven't time to do it myself now. But please let us know the result. Thanks. – Ralph Feb 19 2012 at 9:02
Ralph, I think your version of the question is worthwhile, I just didn't think that an answer involving Galois theory was what the OP expected or is necessarily ready for. My reading was that the question was asked at a much more naive level, and rather than give a MO answer likely to be unsatisfactory to the OP without a crash course in solvable Galois groups, it seemed best to send this over to MSE where discussion can be had. – Yemon Choi Feb 19 2012 at 10:08
@Ralph: it’s easy to see that $\sin x$ is solvable iff $e^{ix}$ is, hence (for $x$ a rational multiple of $\pi$) the Galois group is in fact cyclic, and a fortiori solvable. However, this gives an expression (a rather trivial one) in terms of complex radicals. If one allows only real radicals, then the answer is that $\sin(\π a/b)$ (with $a,b$ coprime integers) is so expressible iff it is constructible iff $b$ is a power of $2$ multiplied by a product of distinct Fermat primes; in particular, nonquadratic radicals do not help. The reason for this is casus irreducibilis: ... – Emil Jeřábek Feb 21 2012 at 12:30