Algebraic Number?
Phillip Cameron Sisson
In a class this summer (it was over two days ago, so I'm not asking
anyone to do my homework) I was asked to prove that sin(1 degree) is an
algebraic number. The proof involves producing a polynomial of positive
integer degree and integer (or rational) coefficients that has sin(1 degree)
as a root. That is, for some P(x), P(sin(1)) = 0. I've been working on it
for a week and haven't been able to find a polynomial yet. Just wanted to
see what you guys might come up with, if think the problem is interesting.
Thanks for all response,
cameron
Dave L. Renfro
= imaginary part of exp[(pi/180)*i] {i = sqrt(-1)}.
Now if we put z = exp[(pi/180)*i],
then
z^180 = exp[pi*i] = -1,
which gives you z^180 + 1 = 0.
Hence, z is algebraic (of degree at most 180).
The imaginary part of any algebraic number is
imaginary [Im(z) = (1/2)*(z - conjugate of z),
and the algebraic numbers form a field], and
so it follows that sin(1 deg) = Im(z) is algebraic.
If you want a more explicitly given polynomial,
start with the multiple angle expansions
for cosine:
cos(2x) = 2*cos^2(x) - 1
cos(3x) = 4*cos^3(x) - 3*cos(x)
cos(4x) = 8*cos^4(x) - 8*cos^2(x) + 1
etc.
So, how do we come up with these? [Important
question, since I'm going to use the
expansion for cos(90x)!]
Recall [cos(x) + i*sin(x)]^n = cos(nx) + i*sin(nx).
Thus, for example,
[cos(x) + i*sin(x)]^3 = cos(3x) + i*sin(3x).
Now two complex expressions are equal iff both
their real and imaginary parts are equal, so
we have (in particular)
Re{ [cos(x) + i*sin(x)]^3 } = cos(3x).
If you work out the left hand side (cube the
binomial, put i^2 = -1 and i^3 = -i) and gather together
all the terms without an i factor (i.e. take the
real part after expanding the cube), you'll get
cos^3(x) - 3*cos(x)*sin^2(x).
Replacing sin^2(x) with 1 - cos^2(x) gives the
expansion for cos(3x) that I gave earlier.
Now, instead of 3, do this with 90 as the exponent.
Letting C(n,r) be the r'th binomial coefficient in
the (usual ordered) expansion of (a+b)^n, you'll get
cos(90x) = cos^90(x) - C(90,2)*cos^88(x)*sin^2(x)
+ C(90,4)*cos^86(x)*sin^4(x)
- C(90,6)*cos^84(x)*sin^6(x)
+ ...
Now we're in luck, since all the occurrences of
sin(x) show up in even powers. Thus, when you
make the replacement sin^2(x) = 1 - cos^2(x),
you'll obtain cos(90x) as a POLYNOMIAL (i.e. no
fractional powers show up) in cos(x).
Call this polynomial P[cos(x)]. [It's actually the
Chebyshev polynomial of degree 90, if you must
know.] Then
cos(90x) = P[cos(x)]. Now, when you let x = 1 deg,
you'll get
cos(90x) = cos(90 deg) = 0 = P[cos(1 deg)].
This gives you a polynomial that cos(1 deg)
is a root of, namely P(z). [Ooops! I almost
forgot. Note that P is not just any polynomial,
but P is actually a polynomial with integer
coefficients.]
What about sin(1 deg)? Consider the function
Q(z) = P[(1-z^2)^(1/2)], that is, P evaluated at
sqrt(1 - z^2).
Since P has only even powers in its expansion
(go back and look at the details of getting
P again), Q(z) is a POLYNOMIAL. [Note that in
carrying this out you'll have a certain 45'th
degree polynomial evaluated at 1 - z^2, but this
still results in a polynomial.] Moreover, Q has
integer coefficients. Now observe that
Q[sin(1 deg)] = P[cos(1 deg)] = 0,
which shows that sin(1 deg) is a root of Q(z).
Dave L. Renfro
Dave L. Renfro
yesterday, the proof that sin(1 deg) is an algebraic number,
to include some additional remarks. (I'm thinking that someone
may later stumble upon that post while searching the internet
for information on this issue.)
First, it true more generally that any trigonometric
function evaluated at a rational degree measure angle
is, when defined for that angle, an algebraic number.
The proof I gave for cos(1 deg) and sin(1 deg) in
my earlier post clearly generalizes to show that
cos[(180/n) deg] and sin[(180/n) deg] are algebraic
numbers for any nonzero integer n. Regarding other
rational degree measures, the following example
illustrates the general procedure to follow for
cosine and sine.
To prove that sin(5/7 deg) is algebraic, first expand sin(5x):
sin(5x) = 16*cos^4(x)*sin(x) - 12*cos^2(x)*sin(x) + sin(x).
Putting x = (1/7 deg) gives you sin(5/7 deg) expressed as
sums and products of algebraic numbers [e.g. -12,
cos(1/7 deg), sin(1/7 deg), etc. are algebraic]. Since the
algebraic numbers form a subfield of the real numbers, it
therefore follows that sin(5/7 deg) is algebraic.
As for the other trigonometric functions (tangent, cotangent,
secant, and cosecant), just observe that each is a rational
combination of cosine and/or sine, and so their rational
degree evaluations being algebraic is a consequence of the
fact that the algebraic numbers are a subfield of the reals.
I stumbled upon this about a year ago while trying to find
some nontrivial applications of Euler's formula
exp(i*x) = cos(x) + i*sin(x) {Note that raising both sides
to the n'th power and using an identity for exponentiation
gives you De Moivre's formula [cos(x) + i*sin(x)]^n =
cos(nx) + i*sin(nx).} for my high school calculus class.
(One encounters Euler's formula in the sequences/series
chapter of most calculus texts.) I couldn't find this in
any books I had (Later I learned otherwise. It's in Nivin's
book (see below), but I had overlooked the passage initially.),
and so I became curious as to whether this result was generally
known. I posted the question of whether this result was
generally known on the Amer. Math. Soc. Real analysis
discussion group (to which I belonged) and I also sent
it to someone in charge of a number theory discussion
group. I got many responses, perhaps 8 or 10, which indicated
to me that the result was generally known, or at least,
easily proved. (I was thinking this might make an interesting
short note for the Mathematics Magazine or the College
Math. Journal.) One person mentioned that it was in a number
theory text by Niven, Montgomery, and Zuckerman, but I
have not yet been able to verify this. However, when I got a
chance to visit a research library, I did find the following
previously published proofs in the literature.
R. W. Hamming, "The transcendental character of cos x",
Amer. Math. Monthly 52 (1945), 336-337.
D. H. Lehmer, "A note on trigonometric algebraic numbers",
Amer. Math. Monthly 40 (1933), 165-166.
Lei Lin, "An algebraic property of trigonometric functions"
(Chinese), J. East China Norm. Univ. Natur. Sci. Ed. 1994,
no. 4, 10-12. [From Math. Reviews 96k:11093 … "In this
paper, the author proves that the values of rational degrees
of triangle functions are all algebraic numbers; the method
used is elementary."]
Ivan Niven, IRRATIONAL NUMBERS, Carus Math.
Monographs 11, MAA, 1956. [See the second and
third pages of chapter 3.]
Elijah Swift, "Note on trigonometric functions", Amer.
Math. Monthly 29 (1922), 404-405.
Hamming's paper refers to Lehmer's paper, but otherwise
none of these papers seem to refer to any of the others.
[This is speculation on my part regarding Lin's paper,
by the way.] Niven mentions in his bibliographic notes
the papers by Hamming, Lehmer, and Swift.
Lehmer proves (1) If k/n is reduced with n > 2, then
2*cos(360*k/n deg) is algebraic of degree (1/2)*phi(n),
where phi(n) is the number of positive integers less than
n that are relatively prime to n; AND (2) If k/n is reduced
with n > 2 (Lehmer omits "n > 2", but his proof seems to
require this as he makes use of his first result), then
2*sin(360*k/n) is algebraic of degree (1/2)*phi(n) or
of degree phi(n), according to whether n is or n isn't a
multiple of 4.
Phillip Cameron Sisson
Dave L. Renfro wrote in message ...
>I thought it might be useful regarding my post
>yesterday, the proof that sin(1 deg) is an algebraic number,
>to include some additional remarks. (I'm thinking that someone
>may later stumble upon that post while searching the internet
>for information on this issue.)
>
Dave --
Thanks for the extensive response to my question. I am aware of many of
the general consequences of the result (i.e. trig functions evaluated at
rational degree measures are algebraic), but I was having quite a bit of
trouble getting the initial result, that sine(1 degree) is algebraic. Your
method (or the method that you used, if it is a standard technique) for
constructing the polynomial is a new type of thing for me -- thanks for the
detail. I believe that my instructor had mentioned Euler's identity in
conjunction with the problem, but I had not yet figured out how to use it.
Thanks also for the references.
cameron
Samuel
You may try to use the fact that sin (30 * 1 degree)= sin (30 degrees) = 0.5
and together with
the de Movire theorem in complex to solve the problem. (You may find out the
polynomial that you needed, but the computation will be complicated)
Kwok Y F
Phillip Cameron Sisson wrote in message
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