Matematické Fórum

stuart clark · 31. 08. 2012 20:32

How can I prove $cos 1^{0}$ is Rational or Irrational

check_drummer · 31. 08. 2012 22:12

↑ stuart clark:
Hi, what about trying to prove that this number is even not algebraic...(?)

jarrro · 01. 09. 2012 09:51

↑ check_drummer:I think he thinks one degree

vanok · 01. 09. 2012 17:47

Hi ↑ stuart clark:,
We shall use to solve this exercise properties of polynomials of Chebyshev of the first kind.
For details see:
http://en.wikipedia.org/wiki/Chebyshev_polynomials
We posed $\cos (\theta)=x$ , and we know that T_n is a polynomial with integer coefficients in x.
For example:

$T_0(x) = 1 \,$

$T_1(x) = x \,$

$T_2(x) = 2x^2 - 1 \,$

$T_3(x) = 4x^3 - 3x \,$

$T_4(x) = 8x^4 - 8x^2 + 1 \,$

$T_5(x) = 16x^5 - 20x^3 + 5x \,$

$T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1 \,$

$T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x \,$

$T_8(x) = 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1 \,$

$T_9(x) = 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x. \,$ ...

Let us suppose that $\cos( 1 °)$ is rational, then $\cos( n°)$ is also rational
(n is natural numbers).
So, for exemple $\cos (30 °)$ must be rational.
But this is a contradiction, because $\cos(30°) = \frac {\sqrt 3} 2$
Conclusion: $\cos (1°)$ is irrational.