| Newsgroups: net.math
Path: decwrl!decvax!ucbvax!ucbtopaz!gbergman
Subject: Re: math identity
Posted: Mon Jul 30 19:50:46 1984
To prove the asserted identity
tan(3*pi/11) + 4*sin(2*pi/11) = sqrt(11)
recall the exponential definitions of the trigonometric functions:
sin x = (e^ix - e^-ix)/2i
cos x = (e^ix + e^-ix)/2
tan x = sin x / cos x
If we set
w = e^(pi*i/11)
then the desired equation becomes
(w^3 - w^-3) (w^2 - w^-2) __
------------- + 4----------- = \/11 .
i(w^3 + w^-3) 2i
If we clear denominators, square both sides to get rid of the
square root, expand, and bring everything to one side we get
10 8 6 4 2 -2 -4 -8 -10
4w + 4w + 4w + 4w + 4w + 4 + 4w + 4w + 4w + 4w = 0.
Now the thing to note about w = e^(pi*i/11) is that
w^22 = e^(2pi*i) = 1, i.e. w is a root of x^22 - 1 = 0; but it is
not one of the "obvious" roots +-1, hence it is a root of
(x^22 - 1)/(x^2 - 1), which is x^20 + x^18 + ... + x^2 + 1.
Substituting w for x, and multiplying by 4w^-10 gives precisely the
above displayed polynomial equation. Hence that equation is true. Now
tracing our steps backwards, we find that the trigonometric function
must be a square root of 11; checking its sign, we see that it
is the positive square root.
Remarks: If we now write w = e^x*i for arbitrary x, we see that
the left-hand side of the above displayed formula equals
4 sin(11x)/sin(x). Tracing through our computations, we get the
general formula
[11 - (tan(3x) + 4sin(2x))^2][cos(3x)]^2 = sin(11x)/sin(x).
To get from this general formula the specific result asserted,
note that x = pi/11 makes the right hand side zero, hence must make
the first factor on the left 0.
For those who know some Galois theory: The fact that the
square root of 11 can be expressed in terms of the 11th roots of
unity, and hence in terms of the trigonometric
functions of pi/11 -- and the analogous fact with any odd prime in
place of "11" -- is a consequence of the calculation given in the last
paragraph of Ch.VIII, sec.3 (p.208) of Lang's _A_l_g_e_b_r_a (Addison-Wesley
1971 -- I don't know what the page reference will be in the new edition
coming out Aug.1). Lang should also have noted the connection with the
"Example" at the bottom of p.134, since what he is actually doing on
p.208 is displaying the discriminant of the cyclotomic extension in
question.
George Bergman
Math, UC Berkeley 94720 USA
...!ucbvax!ucbcartan!gbergman
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