$\sin(90°)= \sin(\frac{1}{2}\pi)= 0$
$\cos(90°)= \cos(\frac{1}{2}\pi)= 1$
$\sin(60°)= \sin(\frac{1}{3}\pi)=\frac{\sqrt{3}}{2}$
$\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.
unit circle, derive number for any degree, cosinus and sinus
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0This article says a few things about how Ptolemy computed these numerically, for integer multiples of half a degree: http://en.wikipedia.org/wiki/Table_of_chords – 2012-02-20
3 Answers
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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0no complex numbers, no infinite decimal places – 2012-02-20
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°).
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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0GEdgar, thanks a lot for letting me know about Casus irreducibilis. This is very intersting. – 2012-02-20