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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!

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(Incidentally, I would just use '$\sin(x)$' rather than '$\sin(n^\circ)$'; the latter is a bit of a red herring since there are constructable values $\sin(x)$ where $x$ isn't a whole number of degrees. There's nothing in the proof that uses the fact that $n$ is an integer.) –  Steven Stadnicki Dec 7 '12 at 23:25

4 Answers 4

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

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Not quite - sin(2n) is not polynomial in sin(n). –  Steven Stadnicki Dec 7 '12 at 23:15
    
Yes you are right –  Amr Dec 7 '12 at 23:15
    
Just edited my post, right on time:) –  Haruki Dec 7 '12 at 23:16

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.

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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.

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I noticed that with the spell-checker too - I'd swear I've always seen 'constructible' but Firefox doesn't like that and does like 'constructable'. –  Steven Stadnicki Dec 7 '12 at 23:26
    
@StevenStadnicki, I suspect I've always used constructible. Maybe it is a holdover from constructive. Interesting that you think it is Firefox doing the checking. I note there is a preference for British spelling in my stuff, I'm not sure how that happened. –  Will Jagy Dec 7 '12 at 23:29
    
Well, I can't say with any surety that FF isn't 'farming out' its spell-checking to some system library, but I presume it's FF (or at least webkit) and not math.SE because it's far from unique to this site; the same results show up in any input form and not just the ones here. –  Steven Stadnicki Dec 7 '12 at 23:31
    
@StevenStadnicki, both Hartshorne and Greenberg sent me their geometry books as i was involved. They both use constructible. So does the O.E.D. –  Will Jagy Dec 7 '12 at 23:39

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,

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