Using the method of proof by contradiction, assume that \(\cos(1^{\circ})\) is rational.

Then \(\cos(2^{\circ}) = 2*\cos^{2}(1^{\circ}) - 1\) must also be rational.

Now \(\cos(k^{\circ} + 1^{\circ}) + \cos(k^{\circ} - 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ})\).

Using this equation, now that we have, by assumption, \(\cos(1^{\circ})\) and hence \(\cos(2^{\circ})\) as rational, (in addition to the fact that \(\cos(0^{\circ}) = 1\) is rational), we see that in turn \(\cos(n^{\circ})\) is rational for each successive integer. But \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\) is clearly irrational, thus implying that the original assumption, i.e., that \(\cos(1^{\circ})\) is rational, is in fact false.
–
Brian Charlesworth
·
2 years, 2 months ago

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TopNewestThe following is the standard approach.

Using the method of proof by contradiction, assume that \(\cos(1^{\circ})\) is rational.

Then \(\cos(2^{\circ}) = 2*\cos^{2}(1^{\circ}) - 1\) must also be rational.

Now \(\cos(k^{\circ} + 1^{\circ}) + \cos(k^{\circ} - 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ})\).

Using this equation, now that we have, by assumption, \(\cos(1^{\circ})\) and hence \(\cos(2^{\circ})\) as rational, (in addition to the fact that \(\cos(0^{\circ}) = 1\) is rational), we see that in turn \(\cos(n^{\circ})\) is rational for each successive integer. But \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\) is clearly irrational, thus implying that the original assumption, i.e., that \(\cos(1^{\circ})\) is rational, is in fact false. – Brian Charlesworth · 2 years, 2 months ago

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HAVE U USED INDUCTION – Vishwesh Agrawal · 2 years, 2 months ago

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strong induction on the equation

Yes, I have essentially used\(\cos(k^{\circ} + 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ}) - \cos(k^{\circ} - 1^{\circ})\). – Brian Charlesworth · 2 years, 2 months ago

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