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.