## Radical Expression For cos(2pi/7)

We sometimes wish to express the values of trigonometric functions
for some rational multiple of pi as a radical expression. For example,
we might wish to express cos(2pi/7) as a radical function of rational
numbers. However, solutions by radicals are sometimes less helpful
than one might think. To illustrate, note that the values of
cos(k pi/m) can be expressed algebraically as
1 / [ 1 + 1^(1/m) ]^2 \
--- sqrt( ------------------- )
2 \ 1^(1/m) /
but this isn't very useful for computing numerical answers. In
general the values of 4cos^2(k pi/m) are the roots of the polynomial
whose coefficients (with alternating signs) are taken from a diagonal
of Pascal's triangle starting on the mth row. To take the specific
example, the value of 4cos^2(2pi/7) is a root of the 7th diagonal
1
1 1
1 2 1
1 3 3 (1)
1 4 (6) 4 1
1 (5) 10 10 5 1
(1) 6 15 20 15 6 1
so we have the cubic
x^3 - 5x^2 + 6x - 1 = 0
We could just solve this cubic for the values of 4cos^2(k pi/7), but
since those values are all squares, we know that each root x of this
polynomial is a square, so setting x=y^2 and substituting into the
above equation gives
y^6 - 5y^4 + 6y^2 - 1 = (y^3 - y^2 - 2y + 1) (y^3 + y^2 - 2y - 1)
The value of 2cos(2pi/7) is a root of the right-hand cubic, which
gives the result
/2pi\ 1 / / 7+21R(-3) \ 1/3 / 7-21R(-3) \ 1/3 \
cos( --- ) = --- ( ( ---------- ) + ( ----------- ) - 1 )
\ 7 / 6 \ \ 2 / \ 2 / /
where R() signifies square root. So here we have an explicit
expression in radicals, but arguably it still isn't all that helpful
because we need to evaluate the cube root of a complex number.
By the way, for a related bit of trivia, notice that 7 = 2^3 - 1 is
a Mersenne prime. Furthermore, if we write
6cos(pi/7) = 2 - (q)^(1/3) - (q*)^(1/3)
where "*" denotes the complex conjugate we have
q = 28 + 144 sqrt(-3)
whose norm is
(28)^2 + 3(144)^2 = 62992
= (2^4)(2^5 - 1)(2^7 - 1)
so the odd prime divisors are the next two Mersenne primes after
7, i.e., 31 and 127.

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