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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 = 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 implies cos nx rational.
Equivalently, cos nx irrational implies cos x irrational.

Taking n = 30 and x = 1°, since we know cos 30° = root 3/2 is irrational, it follows that cos 1° is irrational.

sin 1°

Building on the above result, since cos 2x = 1 − 2 sin2x, we have

cos 2nx irrational implies cos 2x irrational implies sin2x irrational implies sin x irrational.

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

tan 1°

The standard addition formula for tangents

tan(a+b) = (tan a + tan b)/(1 - tan a tan b)

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 not equal to 1.)

We know that tan 30° = 1 /root 3 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.


  1. 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.
  2. 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 implies cos 30° rational, a contradiction.
  3. Notice that we do not claim sin x rational implies 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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