0

Prove or disprove the following :

$1.\sin ^{-1} 1 $ is algebraic over $\mathbb Q$

$2.\cos (\frac{\pi}{17})$ is algebraic over $\mathbb Q$

As suggested by @Andre ,for the 2nd one

$(\cos (\frac{\pi}{17})+i\sin (\frac{\pi}{17}))^{17}=-1$ (De Moivre's Theorem) Expanding the L.H.S binomially and separating the real and imaginary parts and equating them to the R.H.S. we get a polynomial whose root is $\cos (\frac{\pi}{17})$

Is the above solution correct?

How should I proceed here? I am not getting any hints to work this one out

Community
  • 1
Learnmore
  • 31,675
  • 10
  • 110
  • 250
  • 5
    You probably know what $\sin^{-1} 1$ is (I assume that you mean $\arcsin 1$). For the second, think about $(\cos(\pi/17)\pm i\sin(\pi/17))^{17}$. – André Nicolas Apr 28 '16 at 16:59
  • @AndréNicolas;I did the second one ;please check – Learnmore Apr 28 '16 at 17:09
  • 1
    Yes, your method for the second is correct. But you need to add some detail. When we expand, the real part gives $\cos^{17} \theta-\binom{17}{2}\cos^{15}\theta\sin^2\theta+\cdots=-1$. There will be various even powers of $\sin\theta$. Replace $\sin^2\theta$ everywhere by $1-\cos^2\theta$. – André Nicolas Apr 28 '16 at 17:16
  • Alternately, use the hint to show that $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$ are both algebraic. Then use the fact that $\cos\theta$ is half the sum of these, and the sum of two algebraics is algebraic. The disadvantage of this method is that we have to use the non-trivial result that the sum of two algebraics is algebraic. Your suggested method is "closer to the ground." – André Nicolas Apr 28 '16 at 17:25
  • 1
    For an elementary solution see here. – Dietrich Burde Apr 28 '16 at 17:45
  • Thank you very much @AndréNicolas – Learnmore Apr 29 '16 at 15:00
  • @AndréNicolas; you did not say what to do with $\sin ^{-1} 1$ – Learnmore Apr 29 '16 at 15:01
  • 1
    @learnmore: You are welcome. For the first question, we have $\sin^{-1}(1)=\pi/2$. It is a standard result that $\pi$ is transcendental (not algebraic). From this it follows that $\pi/2$ is not algebraic. You would not be expected to prove that $\pi$ is transcendental, the proof is difficult. It is enough to quote the result, which is due to Lindemann. – André Nicolas Apr 29 '16 at 15:04
  • Okay I got that @AndréNicolas – Learnmore Apr 29 '16 at 15:09

0 Answers0