Sum The Secants, Get 6 (Posted on 2007-09-12)

	Submitted by K Sengupta
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Substituting h = 2*pi/9, 4*pi/9 and 8*pi/9 in turn, we observe that: cos 3h =-1/2, for each of the foregoing values, and accordingly: 4 cos3 h - 3 cos h = -1/2, giving: 8 cos3 h – 6 cos h + 1 =0 Therefore, cos(2*pi/9), cos(4*pi/9) and cos(8*pi/9) are the roots of the cubic equation 8m3 – 6m + 1 =0; for m = cos h. Hence, their respective reciprocals sec(2*pi/9), sec(4*pi/9) and sec(8*pi/9) must correspond to the roots of the cubic equation: m3 – 6m2 + 8 = 0......(*) Now, the sum of the roots of equation (*) is 6. Consequently, sec(2*pi/9) + sec(4*pi/9) + sec(8*pi/9) = 6

	Subject	Author	Date
	Some Secants Sum Six	Brian Smith	2007-09-16 17:11:57
	Hint	K Sengupta	2007-09-13 03:19:26