Sin, Cos, Taylor Series, and Commensurability

Question

Through Taylor series, is it possible to say that the cos(x) or sin(x) of any value other than the nice ones {(pi/3), (pi/6), and finally the ((z*pi)/2)} are non commensurable with a 1 unit measure? If yes, prove (although that should go without saying).

Answer 1

A Google search for "irrationality of cosine" turns up a number of relevant links, including these:

http://people.math.sc.edu/filaseta/gradcourses/Math785/Math785Notes2.pdf

This one proves that The number π is irrational and For any rational number α $\ne$ 0, cos α is irrational.

https://en.wikipedia.org/wiki/Niven's_theorem

This says:

In mathematics, Niven's theorem, named after Ivan Niven, states that the only rational values of θ in the interval 0 ≤ θ ≤ 90 for which the sine of θ degrees is also a rational number are:[1]

${\displaystyle {\begin{aligned}\sin 0^{\circ }&=0,\\[10pt]\sin 30^{\circ }&={\frac {1}{2}},\\[10pt]\sin 90^{\circ }&=1.\end{aligned}}} {\begin{aligned}\sin 0^{\circ }&=0,\\[10pt]\sin 30^{\circ }&={\frac {1}{2}},\\[10pt]\sin 90^{\circ }&=1.\end{aligned}}$

In radians, one would require that 0 ≤ x ≤ π/2, that x/π be rational, and that sin x be rational. The conclusion is then that the only such values are sin 0 = 0, sin π/6 = 1/2, and sin π/2 = 1.

The theorem appears as Corollary 3.12 in Niven's book on irrational numbers.[2]

The theorem extends to the other trigonometric functions as well.[2] For rational values of θ, the only rational values of the sine or cosine are 0, ±1/2, and ±1; the only rational values of the secant or cosecant are ±1 and ±2; and the only rational values of the tangent or cotangent are 0 and ±1.