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