×

# Can you prove ?!

Prove that $$cos(1^{o})$$ is irrational

Note by A K
2 years, 8 months ago

Sort by:

The 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. · 2 years, 8 months ago

HAVE U USED INDUCTION · 2 years, 8 months ago

Yes, I have essentially used strong induction on the equation

$$\cos(k^{\circ} + 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ}) - \cos(k^{\circ} - 1^{\circ})$$. · 2 years, 8 months ago