This Demonstration shows Lill's angle trisection. The angle to be trisected is marked with a green arc.

Consider the trigonometric identity for in the form

.

Substitute and to get

.

This equation has the solutions

,

,

.

By Lill's construction,

,

so given (the green arc), we must find such that (i.e. the length of the thick red segment is 0, thus the points and coincide). Then we find . From , we can construct .