In trapezoid ABCD, with sides AB and CD parallel, DAB = 6° and ABC = 42°. Point X on side AB is such that AXD = 78° and CXB = 66°. If AB and CD are 1 inch apart, prove that AD + DX − (BC + CX) = 8 inches.

Dropping a perpendicular (of length 1) from D to AX, and similarly from C to BX, we see that:

- AD = cosec 6°
- DX = cosec 78°
- BC = cosec 42°
- CX = cosec 66°

Thus, we are asked to prove that cosec 6° + cosec 78° − cosec 42° − cosec 66° = 8.

Notice that, for x = 6°, 78°, −42°, −66°, and 30°, sin 5x = ½. We now express sin 5x in terms of sin x.

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}

Setting n = 5, expanding the right-hand side using the binomial theorem, and equating imaginary parts, we obtain

sin 5x | = sin^{5}x − 10 sin^{3}x cos^{2}x + 5 sin x cos^{4}x |

= sin^{5}x − 10 sin^{3}x (1 − sin^{2}x) + 5 sin x (1 − sin^{2}x)^{2}, since sin^{2}x + cos^{2}x = 1 | |

= 16 sin^{5}x − 20 sin^{3}x + 5 sin x |

This result can also be obtained by means of trigonometric identities. Setting s = sin x, it follows that the five distinct real numbers,

sin 6°, sin 78°, −sin 42°, −sin 66°, and sin 30° = ½, (1)^{ }

are roots of the equation 16s^{5} − 20s^{3} + 5s = ½, or, equivalently, of 32s^{5} − 40s^{3} + 10s − 1 = 0. (2)

By the Fundamental Theorem of Algebra, (2) has exactly five roots, up to multiplicity, and hence these must be precisely the *distinct* roots identified in (1).^{ }

Since s = ½ is a root of (2), the equation factorizes:

(2s − 1)(16s^{4} + 8s^{3} − 16s^{2} − 8s + 1) = 0,

yielding the quartic equation whose roots are sin 6°, sin 78°, −sin 42°, and −sin 66°.^{ }

As s = 0 is not a root of this quartic equation, we may divide by s^{4}, and, setting t = 1/s, obtain

t^{4} − 8t^{3} − 16t^{2} + 8t + 16 = 0,

an equation whose roots are cosec 6°, cosec 78°, −cosec 42°, and −cosec 66°.^{ }

By Viète's formulas, the sum of the roots of this equation is 8.

Thus, AD + DX − (BC + CX) = 8 inches.

Source: Original; inspired by a similar problem in M500 Magazine, Issue 210.