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| Sum The Secants, Get 6 (Posted on 2007-09-12) |
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Show that:
sec(2π/9) + sec(4π/9) + sec(8π/9) = 6
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Submitted by K Sengupta
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Rating: 2.0000 (1 votes)
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Solution:
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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
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