# Solution to puzzle 111 (Six tangents)

Show that tan (/13) · tan (2/13) · tan (3/13) · tan (4/13) · tan (5/13) · tan (6/13) = .

The equation sin 13x = 0 has solutions x = n/13, where n is an integer.

We will equate imaginary parts of cos 13x + i sin 13x = (cos x + i sin x)13.
Letting C = cos x and S = sin x, and expanding the right-hand side using the binomial theorem, we obtain

sin 13x = 13C12S − 286C10S3 + 1287C8S5 − 1716C6S7 + 715C4S9 − 78C2S11 + S13.

If sin 13x = 0, then, dividing by C13, and letting T = tan x = S/C, we have

T13 − 78T11 + 715T9 − 1716T7 + 1287T5 − 286T3 + 13T = 0.

This equation has 13 roots:
T = 0, tan (/13), tan (2/13), ... , tan (12/13), or, equivalently,
T = 0, ± tan (/13), ± tan (2/13), ... , ± tan (6/13).

Hence the 12 roots of the equation T12 − 78T10 + ... + 13 = 0 are
T = ± tan (/13), ± tan (2/13), ... , ± tan (6/13).

But the product of these 12 roots, using Viète's formulas, is simply (−1)12 · 13 / 1 = 13.

Hence tan2(/13) · tan2(2/13) · tan2(3/13) · tan2(4/13) · tan2(5/13) · tan2(6/13) = 13.

Since tan (/13), tan (2/13), ... , tan (6/13) are all positive, we conclude that

tan (/13) · tan (2/13) · tan (3/13) · tan (4/13) · tan (5/13) · tan (6/13) = .

Source: M500 Magazine, Issue 195