Special Angles, Part 2

Expressing cos 9x as a 9th order polynomial in cos x

A puzzle was sent to me: what is the sum of sec 40º, sec 80º, and sec 160º.  I discovered using Plouffe's Inverter (a.k.a. Inverse Symbolic Calculator) that these are the roots of x³-6x²+8, and wondered why.  I knew that cos 40º, cos 80º, and cos 160º would be the roots of 8x³-6x+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 - 120cos³x + 432cos⁵x - 576cos⁷x + 256cos⁹  

Using this fact, equation 1 can be rewritten as

(3)   0 = -1 + 9cos x - 120cos³x + 432cos⁵x - 576cos⁷x + 256cos⁹x  

Equation 3 has solutions including cos 0º, cos 120º, and cos 240º, which are simple rational numbers that can easily be divided out of this polynomial.  What remains is a perfect square, because cos 40º = cos 320º, etc.  So,

(4)   0 = (cos x - 1) (2 cos x + 1)² (8 cos³x - 6 cos x + 1)²

Now the mystery is explained; we know why the roots of 8x³-6x+1 are cos 40º, cos 80º, and cos 160º.  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/a-b²/(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 x³-6x²+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 120º, so its three cube roots (and their arguments) are

0.383022+0.321394i   (40º)
-0.469846+0.171010i   (160º)
0.0868241-0.492404i   (280º)

The second number is 1/16 + i sqrt(3)/16  (argument is 60º)  its three cube roots and their arguments are:

0.469846+0.171010i   (20º)
-0.383022+0.321394i   (140º)
-0.0868241-0.492404i   (260º)

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(40º)
-0.469846+0.171010i  -  0.469846+0.171010i  =  -0.93969231  =  cos(160º)
0.0868241-0.492404i  -  -0.0868241-0.492404i  =  0.173648189  =  cos(80º)

Does this get us any closer to finding an arithmetic expression that yields the cosine of 40º?  I don't think so, alas.

Summary of the Cubic Formula

Another Puzzle

Prove that the sum

sec² 1º + sec² 3º + ... + sec² 89º

is an integer.

Remark

A little trial and error will convince you that if you divide the interval [0,90º] into n equal sub-intervals, and sum the squares of tan of the angle at the center of each of the intervals, then this sum is 2n²-n.

This problem is a case in point, where n=45. The sum of squares of tans is 2n²-n, which is 4005 when n=45.

Other similar problems would be to prove the sum of

tan² 15º + tan² 45º + tan² 75º = 15,

or that tan² 9º + tan² 27º + tan² 45º + tan² 63º + tan² 81º = 45. 

Outline of a Proof

These angles in the range between 0º and 90º 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 = 2sec¹⁸⁰ x - 16200sec¹⁷⁸ x + ... 

Let u = sec² x, so now we have 

0 = 2u⁹⁰ - 16200u⁸⁹ + ... 

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.

Internet References

Plouffe's Inverter (formerly Inverse Symbolic Calculator) - turns a number into the formula that generated that number!

Related Pages in this website

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/a-b²/(3a²)  and  B=d/a+2b³/(27a³)-bc/(3a²)

The webmaster and author of the Math Help site is Graeme McRae.
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