Talk:Trigonometry/The sine of 15 degrees

Would this section benefit from a geometric derivation? One could take right triangle ABC with the measure of angle ABC to be 90, that of angle ACB to be 15, and that of angle CAB to be 75 degrees. Then let point D be on line segment BC such that angle BAD has a measure of 60 degrees. With some angle chasing, it then follows that triangle ACD is isosceles. We can then use the sine of 30 degrees to calculate the lengths of AD, BC, and CD in terms of the length of AB, and then we can use the Pythagorean Theorem to find the length of AC in terms of the length of AB as well. The sine of 15 would then follow. --Oneequalstwo (discuss • contribs) 13:58, 26 December 2011 (UTC)