Show that cos 1°, sin 1°, and tan 1° are irrational numbers.
We will use appropriate multiple angle formulae to show that, if cos 1°, sin 1°, or tan 1° is rational, then the cosine or tangent of some multiple of 1° must also be rational. By choosing a multiple of 1° for which a suitable trigonometric function is known to be irrational, we can derive a contradiction and thereby establish each result.
De Moivre's theorem states that for any real number x and any integer n,
cos nx + i sin nx = (cos x + i sin x)n
Expanding the right-hand side using the binomial theorem, and equating real parts, we have
cos nx = cosnx − [n(n − 1)/2] cosn−2x sin2x + ...
Given that sin2x = 1 − cos2x, we can thereby express cos nx as a polynomial in cos x, with integer coefficients.
Hence, cos x rational cos nx rational.
Equivalently, cos nx irrational cos x irrational.
Taking n = 30 and x = 1°, since we know cos 30° = /2 is irrational, it follows that cos 1° is irrational.
Building on the above result, since cos 2x = 1 − 2 sin2x, we have
cos 2nx irrational cos 2x irrational sin2x irrational sin x irrational.
Taking n = 15 and x = 1°, it follows that sin 1° is irrational.
The standard addition formula for tangents
tells us that, if tan a and tan b are rational, then tan(a + b) is rational. (Of course, we must also have tan a, tan b, and tan(a + b) defined, so that tan a tan b 1.)
We know that tan 30° = 1 / is irrational.
Since 30° can be built up as a series of binary sums, beginning with 1° and 1°, it follows, by contradiction, that tan 1° is irrational.