Here is an elementary solution using strong induction and contradiction.

Assume that \(\cos 1^\circ \) is rational. Then \( \cos2^\circ=2\cos^2 2^\circ-1\) is rational. Note that \(\cos(n+1)^\circ+\cos(n-1)^\circ=2\cos 1^\circ\cos n^\circ\).

Hence by strong induction \(\cos n^\circ\) is rational for all integers \(n\ge 1\). But this is clearly a contradiction as \(\cos 30^\circ\) is trivially irrational. Thus we conclude that \(\cos 1^\circ\) is irrational.
–
Jack Frost
·
1 year, 10 months ago

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TopNewestHere is an elementary solution using strong induction and contradiction.

Assume that \(\cos 1^\circ \) is rational. Then \( \cos2^\circ=2\cos^2 2^\circ-1\) is rational. Note that \(\cos(n+1)^\circ+\cos(n-1)^\circ=2\cos 1^\circ\cos n^\circ\).

Hence by strong induction \(\cos n^\circ\) is rational for all integers \(n\ge 1\). But this is clearly a contradiction as \(\cos 30^\circ\) is trivially irrational. Thus we conclude that \(\cos 1^\circ\) is irrational. – Jack Frost · 1 year, 10 months ago

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Hint:Use Chebyshev Polynomials. – Calvin Lin Staff · 1 year, 11 months agoLog in to reply