
A puzzle was sent to me: what is the sum of sec 40�, sec 80�, and sec 160�. I discovered using Plouffe's Inverter (a.k.a. Inverse Symbolic Calculator) that these are the roots of x^{3}6x^{2}+8, and wondered why. I knew that cos 40�, cos 80�, and cos 160� would be the roots of 8x^{3}6x+1 (the reciprocals of the roots of the polynomial with coefficients in reverse order), but I still needed help understanding this puzzle.
It was pointed out to me that these three angles are three of the nine roots of
(1) cos 9x  1 = 0
Then I discovered an easy way to express cos nx in terms of cos x, and, in particular
(2) cos 9x = 9cos x  120cos^{3}x + 432cos^{5}x  576cos^{7}x + 256cos^{9}
Using this fact, equation 1 can be rewritten as
(3) 0 = 1 + 9cos x  120cos^{3}x + 432cos^{5}x  576cos^{7}x + 256cos^{9}x
Equation 3 has solutions including cos 0�, cos 120�, and cos 240�, which are simple rational numbers that can easily be divided out of this polynomial. What remains is a perfect square, because cos 40� = cos 320�, etc. So,
(4) 0 = (cos x  1) (2 cos x + 1)^{2} (8 cos^{3}x  6 cos x + 1)^{2}
Now the mystery is explained; we know why the roots of 8x^{3}6x+1 are cos 40�, cos 80�, and cos 160�. The original puzzle is solved, and better yet, we might have a way to get arithmetic expressions for cosines of these special angles.
According to The Cubic Formula, the cubic, ax³+bx²+cx+d=0, has roots given by
x=cuberoot(B/2 + sqrt(B²/4+A³/27))  cuberoot(B/2 + sqrt(B²/4+A³/27))b/(3a), where
A=c/ab²/(3a²) and B=d/a+2b³/(27a³)bc/(3a²)
This is already a "depressed cubic" (no x² term), so it's fairly easy
to calculate A and B.
a=8, b=0, c=6, and d=1, so A=3/4, and B=1/8
So x = cuberoot(1/16 + sqrt(3)/16)  cuberoot(1/16 + sqrt(3)/16).
Easy, right? Not so fast. There are two square roots of every number, and three cube roots. Since each one appears twice, it looks like there are 3²*2²=36 different ways to calculate x. Luckily, and for reasons I don't fully understand, you can use either of the two square roots (consistently), so there are nine different combinations of cube roots. I just tried each of the nine roots obtained this way in x^{3}6x^{2}+8 to see which three give zero. Here's what I found this way...
Look at the first cube root. The argument (the angle that fixes the direction) of 1/16 + i sqrt(3)/16 is 120�, so its three cube roots (and their arguments) are
0.383022+0.321394i (40�)
0.469846+0.171010i (160�)
0.08682410.492404i (280�)
The second number is 1/16 + i sqrt(3)/16 (argument is 60�) its three cube roots and their arguments are:
0.469846+0.171010i (20�)
0.383022+0.321394i (140�)
0.08682410.492404i (260�)
As you can see, the same three imaginary components of the first cube roots appear as the second cube roots. That is, if you pair cube root (1a) with cube root (2b), the difference is a real number.
So the three roots of this cubic are
0.383022+0.321394i  0.383022+0.321394i = 0.766044222 = cos(40�)
0.469846+0.171010i  0.469846+0.171010i = 0.93969231 = cos(160�)
0.08682410.492404i  0.08682410.492404i = 0.173648189 = cos(80�)
Does this get us any closer to finding an arithmetic expression that yields the cosine of 40�? I don't think so, alas.







Prove that the sum
sec² 1� + sec² 3� + ... + sec² 89�
is an integer.
Remark
A little trial and error will convince you that if you divide the interval [0,90�] into n equal subintervals, and sum the squares of tan of the angle at the center of each of the intervals, then this sum is 2n²n.
This problem is a case in point, where n=45. The sum of squares of tans is 2n²n, which is 4005 when n=45.
Other similar problems would be to prove the sum of
tan² 15� + tan² 45� + tan² 75� = 15,
or that tan² 9� + tan² 27� + tan² 45� + tan² 63� + tan² 81� = 45.
Outline of a Proof
These angles in the range between 0� and 90� are just the solutions in the first quadrant to the equation
cos 180x + 1 = 0
Expanding cos 180x in terms of cos x, we get
cos 180x = 1  16200cos² x + ...
where the "..." represent a whole bunch of other even powers of cos x, with humongous coefficients, the values of which, luckily, don't matter to us!
Plugging this into the first equation above, we get the polynomial equation,
0 = 2  16200cos² x + ...
I'm sure you'll recognize 16200 as half the square of 180  this is important for the general case.
Since sec is the reciprocal of cos, this polynomial equation can be expressed as
0 = 2sec^{180} x  16200sec^{178} x + ...
Let u = sec² x, so now we have
0 = 2u^{90}  16200u^{89} + ...
The sum of the 90 roots of this polynomial is 8100, but we're only looking for the sum of the 45 values of u representing angles in the first quadrant, so we divide that in half, giving us 4050.
To complete the proof, use 4n in place of 180 to find the sum of squares of secants of the n angles in the first quadrant is 2n², and then subtract n to get the sum of squares of tangents, because tan² = sec²  1.
Plouffe's Inverter (formerly Inverse Symbolic Calculator)  turns a number into the formula that generated that number!
Trig Functions of Special Angles
The Cubic Formula  the cubic, ax³+bx²+cx+d=0, has roots given by
x=cuberoot(B/2 + sqrt(B²/4+A³/27))  cuberoot(B/2 + sqrt(B²/4+A³/27))b/(3a), where
A=c/ab²/(3a²) and B=d/a+2b³/(27a³)bc/(3a²)
The webmaster and author of this Math Help site is Graeme McRae.