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

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

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

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

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 $x$ and $y$ be:
$\sin(1°)= \sin(\frac{1}{180}\pi)=\ x$
$\cos(1°)= \cos(\frac{1}{180}\pi)=\ y$
And how would I generally derive ANY degree, lets say $\sin(3°)$ or wathever.

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Why "$\cong$" instead of plain old "$=$"? – Hans Lundmark Feb 19 '12 at 8:35
just for kicks, having the geometrical pictures in mind – kiltek Feb 19 '12 at 9:12
Well, $\sin(\pi/6)$ and $1/2$ are both numbers (not triangles), so the proper relation is equality (not congruence). – Hans Lundmark Feb 19 '12 at 9:22
This page doesn't answer the question but may be of interest: In particular, it gives us this: $\sin\frac{\pi}{60}=\sin 3^\circ=\tfrac{1}{16} \left[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)\right]$ – Michael Hardy Feb 20 '12 at 20:27
This article says a few things about how Ptolemy computed these numerically, for integer multiples of half a degree: – Michael Hardy Feb 20 '12 at 20:31
up vote 4 down vote accepted

The Chebyshev polynomials of the first kind $T_n(x)$ satisfy the relation $$\cos(nx) = T_n(\cos x).$$ This shows that $\cos x$ is algebraic if and only if $\cos (nx)$ is. In your particular case, taking $n = 180$ and $x = \pi/180$ shows that $\cos (\pi/180)$ is algebraic (because $\cos \pi = -1$ is), and gives you a polynomial with integer coefficients satisfied by $\cos (\pi/180)$.

The roots of these polynomials are solvable by radicals, though not necessarily with just square roots. Without writing out a lot of details, this is because of the relation $$e^{2 \pi i/n} = \cos(2 \pi/n) + i \sin(2 \pi /n).$$ The left side is a root of unity and therefore has abelian Galois group over $\mathbb{Q}$, hence it's solvable by radicals.

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seems like no fine numbers possible, thanks – kiltek Feb 19 '12 at 16:38
what do you mean by "fine number" ? – Ted Feb 19 '12 at 17:31
no complex numbers, no infinite decimal places – kiltek Feb 20 '12 at 6:00

Using the usual formulas for $\sin(x+y),\cos(x+y)$, one obtains $\cos(3x) = 4\cos^3x-3\cos x$. Thus $\cos(10°)$ is a root of the polynomial $4x^3-3x-\sqrt{3}/2$. Cardano's formula now yields $$\cos(10°) = \frac{1}{2}\big (\;\sqrt[3]{\frac{\sqrt{3} + i}{2}} + \sqrt[3]{\frac{\sqrt{3} - i}{2}}\;\big)$$ where the 3rd root is defined by the function $z \mapsto e^{z/3}$.

Applying the formula $\sin(x/2) = \sqrt{\frac{1-\cos x}{2}}$ then yields

$$\sin(5°) = \frac{1}{2} \sqrt{2 - \sqrt[3]{\frac{\sqrt{3} + i}{2}} - \sqrt[3]{\frac{\sqrt{3} - i}{2}}}\;\;.$$

Note: This also shows casus irreducibilis in action (cf. GEdgar's answer).

A closed formula for $\sin(1°)$ can be derived as follows: Express $\sin(5x)$ as polynomial in $\sin x$. Then $x:=6°$ yields a polynomial of degree 5 with linear factor $x-1/2$. Thus $\sin(6°)$ is a root of a polynomial of degree 4. That can be computed by Ferrari's formula. By using the $\sin(3x)$-formula from above and solving a degree 3 equation similar as above, one obtains an expression for $\sin(2°)$. Applying the $\sin(x/2)$-formula finally yields the searched formula for $\sin(1°)$.

But as the expression for $\sin(5°)$ already shows, the result will be a rather ugly formula. Therefore it's - in my opinion - waste of time to figure it out explicitely.

Added: The wikipedia article in Michael Hardy's comment above gives $\sin(3°), \cos(3°)$. In the same manner as $\sin(5°)$ before, one can compute $\cos(5°)$. Then one can use compute $\sin(2°) = \sin(5°-3°) = \sin(5°)\cos(3°)-\sin(3°)\cos(5°)$. Applying the $\sin(x/2)$-formula finally yields the searched formula for $\sin(1°)$.

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The angle $3^\circ$ can be constructed with Euclidean tools, so there is a closed form formula for its sine and cosine using only square-roots. A complicated formula, to be sure, but a formula. To go on to $1^\circ$ you have to solve a cubic equation, so the formula will involve cube roots and complex numbers...

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Can you give a reason why you think the formula will involve complex numbers ? I would expect that cubic roots, square roots and possibly fourth roots of (positive) rational numbers should do. – Ralph Feb 19 '12 at 19:02
@Ralph: Casus irreducibilis ( ) – GEdgar Feb 20 '12 at 0:10
GEdgar, thanks a lot for letting me know about Casus irreducibilis. This is very intersting. – Ralph Feb 20 '12 at 19:46

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