## Sunday, November 28, 2010

### Puzzle: Show that cos 1°, sin 1°, and tan 1° are irrational numbers.

Let $$\theta=$$1°. Assume otherwise $$\sin(\theta)$$ is rational.
Since $$\cos(2\theta)=1-2 \sin^2(\theta).$$ We know that $$\cos(2^k \theta)$$ are rational.
Now
$$\cos(30\theta) = \cos(32\theta)\cos(2\theta)+\sin(32\theta)\sin(2\theta)$$
The first part $$\cos(32\theta)\cos(2\theta)$$ is rational.
And we have
$$\sin(32\theta)=32sin(\theta)\cos(\theta)\cos(2\theta) \cdots \cos(16\theta)$$
and
$$\sin(2\theta)=2\sin(\theta)\cos(\theta)$$.
Thus for the second part $$\sin(32\theta)\sin(2\theta)$$, we'll get a rational number times $$\cos^2(\theta)=1-\sin^2(\theta)$$, which is also a rational number. Therefore the second part is also rational.
Thus we conclude that $$\cos(30\theta)=\sqrt{3}/2$$ is rational, which contradicts the fact that $$\sqrt{3}$$ is irrational.

To prove that $$\cos(\theta)$$ is irrational, we can use the formula
$$\cos(2\theta) = 2\cos^2(\theta)-1$$
and similar arguments as above.

To prove that $$\tan(\theta)$$ is irrational, we can use the formula
$$\sec^2(\theta) = \tan^2(\theta)+1$$
and similar arguments as above.