Constructible numbers If we know that $\sin(n^\circ)$ is constructible where $n$ is some integer,then is $\sin((an)^\circ)$ also constructible for any integer $a$ ?
I am thinking it should be but not sure how to show it? Maybe using some recurrence relation for sin, to express it entirely in terms of powers of $\sin(n^\circ)$? 
Got it!
 A: yes Using the trigonemetric addition fromulae $sin(an)$ is a polynomial in $sin (n),cos(n)$(both of which areconstructible).  Since the set of constructible numbers is a field, therefore $sin(an)$ is constructible
A: First of all, note that $\sin(x)$ is constructable iff $\cos(x)$ is constructable: since $\sin(x)$ is, $\sin^2(x)$ is, so $\cos^2(x) = 1-\sin^2(x)$ is, and you can use one of the classic ruler-and-compass methods for square roots (e.g. http://www.cs.cas.cz/portal/AlgoMath/Geometry/PlaneGeometry/GeometricConstructions/SquareSquareRootConstruction.htm) to construct $\cos(x)$ from $\cos^2(x)$.  Now, just use the addition formula for $\sin$ to show the claim by induction: $\sin((n+1)x) = \sin(nx)\cos(x)+\cos(nx)\sin(x)$.
Also, note that the converse is not true; with $x=20^\circ$, $\sin(3x) = \sin(60^\circ) = \frac{1}{2} $ is constructable but $\sin(x)$ isn't.
A: You can construct the sine of an angle $\theta$ if and only if you can construct $e^{i\theta}$, because if you can construct a length $\sin(\theta)$, you can build the line $y = \sin(\theta)$, and by building the circle centered at the origin of radius $1$, you get $e^{i\theta}$. Conversely, given that $e^{i\theta}$ is constructible, since the constructible numbers form a field, you can also produce 
$$
\sin(\theta) = \frac{e^{i\theta} + \frac 1{e^{i\theta}}}2.
$$
Constructing $\sin(a \theta)$ is therefore equivalent to constructing $e^{i (a\theta)} = (e^{i\theta})^a$, which is just a product of constructible numbers if $a$ is a positive integers. Therefore $\sin(a \theta)$ is constructible.
If $a$ is negative, use the formula $\sin(-\theta) = - \sin(\theta)$.
Hope that helps,
A: Bit of a tautology, I suppose. The pretty statement comes from the observation that an angle $\theta$ is constructible (as an angle in a right triangle, for example) if and only if $\sin \theta$ or $\cos \theta$ or $\tan \theta$ is constructible, these three conditions being equivalent for, say, acute angles. It comes more or less for free that the sum or difference of constructible angles is also constructible. 
The attractive part is this: the constructible angles on the surface of the unit sphere, and the constructible angles in the hyperbolic plane of curvature $-1,$ are exactly the same as the constructible angles in the traditional Euclidean plane.
The spell checker prefers constructable. i thought it was i. 
Just for flavour, or flavor, it is generally impossible in the hyperbolic plane to trisect a line segment, which seems mysterious. However, if you put a line segment on the equator of the unit sphere and consider the triangle made with the North Pole, you see that you are asked to trisect the angle at the North pole. And angle trisection is often impossible; we know about that from the Euclidean plane.  
