sin1∘\sin 1^\circ is irrational but how do I prove it in a slick way? And tan(1∘)\tan(1^\circ) is …..

In the book 101 problems in Trigonometry, Prof. Titu Andreescu and Prof. Feng asks for the proof the fact that cos1 is irrational and he proves it. The proof proceeds by contradiction and using the strong induction principle. (Problem on Pg:84, 70 in typeset; solution on Pg:126, 111 in Typeset). However, for Completeness, I’ll append it here:

Proof of irrationality of cos(1)

Assume for the sake of contradiction, that cos(1) is rational. Since, cos(2)=2cos2(1)1 we have that, cos(2) is also rational. Note that, we also have cos(n+1)+cos(n1)=2cos(1)cos(n) By Strong induction principle, this shows that cos(n) is rational for all integers n1. But, this is clearly false, as for instance, cos(30)=32 is irrational, reaching a contradiction.

But, as my title suggests, sin(1) is irrational, (look at the following image for its value!) Sine 1 degree

Is there a proof as short as the above proof or can any of you help me with a proof that bypasses actual evaluation of the above value?

Image Courtesy: http://www.efnet-math.org/Meta/sine1.htm
This link explains how to evaluate this value.

My next question is

Is tan(1) rational and is there a short proof that asserts or refutes its rationality?

P.S.: This is not a homework question.

Answer

If it’s a slick proof you want, nothing beats this proof that the only cases where both r and cos(rπ) are rational are where cos(rπ) is 1, 1/2, 0, 1/2 or 1.

If r=m/n is rational, eiπr and eiπr are roots of z2n1, so they are algebraic integers. Therefore 2cos(rπ)=eiπr+eiπr is an algebraic integer. But the only algebraic integers that are rational numbers are the ordinary integers. So 2cos(rπ) must be an integer, and of course the only integers in the interval [2,2] are 2,1,0,1,2.

Attribution
Source : Link , Question Author : Community , Answer Author :
Robert Israel

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