Prove this is a polynomial of degree five with integer coefficients?

This follows pretty much from gianlino's answer in the link.

Note that cos (5t) = 1/2 for t = 12°, 60°, 84°, 132°, 156° all of which are distinct when you consider the value of cos t.

cos (5t) = T_5(cos t), where T_5(x) = P(x) is the Chebyshev Polynomial of the first kind of degree 5.

Since the Chebyshev polynomial is of degree 5, we know that the equation 1/2 = P(x) has the roots cos 12°, cos 60°, cos 84°, cos 132°, cos 156° according to the factor theorem. And by the Fundamental Theorem of Algebra, these are the only roots of the polynomial.

Q(x) = 2P(x) - 1 is a polynomial with the roots cos 12°, cos 60°, cos 84°, cos 132°, cos 156°, thus Q(x) = c(x - cos12°) (x - cos60°) (x - cos84°) (x - cos132°) (x - cos156°) for a constant c

http://en.wikipedia.org/wiki/Chebyshev_polynomial

The Chebyshev polynomial is defined by the recurrence relation given in the above link.

It is easy to see that T_n(x) has integer coefficients (easy to prove by induction) and the leading coefficient is 2^(n-1) (also easy to prove by induction).

Therefore P(x) is a polynomial with integer coefficients with leading coefficient 2^(5-1) = 2^4 = 16. Q(x) as a result has leading coefficient 32 = c.

Therefore Q(x) = 32(x - cos12°) (x - cos60°) (x - cos84°) (x - cos132°) (x - cos156°). Since P(x) has integer coefficients, obviously Q(x) = 2P(x) - 1 = 32(x - cos12°) (x - cos60°) (x - cos84°) (x - cos132°) (x - cos156°) has integer coefficients.

Q.E.D.

First observe that it is comparable to proving the set of polynomials with total style coefficients, using fact the integers have an identical length using fact the full numbers. This only makes the data less difficult. define style(monomial) using fact the sum of the coefficient * exponent. So style(3x^4) = 12, style(1x) = 2 Then define the type(polynomial) using fact the sum of the type(monomial) of the monomials interior the polynomial. style 0 polynomials 0 style a million polynomials a million style 2 polynomials 2, x style 3 polynomials 3, 2x, 1x + a million, 1x^2 style 4 polynomials 4, 3x, 2x + a million, 1x + 3, 2x^2, 1x^2 + a million, 1x^3 you will see that each and each style has a finite style of polynomials, as a manner to affiliate each and each polynomial with a organic style. #0 0, #a million a million, #2 2, #3 x, #4 3, #5 2x, #6 1x+a million, #7 1x^2, etc and it is the definition of a countable set.