# Solution to puzzle 70: One degree

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.

## cos 1°

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 = cos^{n}x − [n(n − 1)/2] cos^{n−2}x sin^{2}x + ...

Given that sin^{2}x = 1 − cos^{2}x, 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.

## sin 1°

Building on the above result, since cos 2x = 1 − 2 sin^{2}x, we have

cos 2nx irrational cos 2x irrational sin^{2}x irrational sin x irrational.

Taking n = 15 and x = 1°, it follows that sin 1° is irrational.

## tan 1°

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.

## Remarks

- The result for cos 1° can also be established through repeated application of the identity cos nx = 2 cos (n−1)x cos x − cos (n−2)x. See the credit, below, for further details.
- Another approach to the result for cos 1° is through repeated application of the identity cos(n° + 1°) + cos(n° − 1°) = 2 cos n° cos 1°, for n = 1, 2, ... , 29. Since cos 0° = 1, we would obtain cos 1° rational cos 30° rational, a contradiction.
- Notice that we do not claim sin x rational sin nx rational; i.e., the analogue of the result we established for cosines. In fact, this analogue does not hold. A simple counter-example would be n = 2, x = 30°.

Source: cos(1 deg) = irrational on wu :: forums

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