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Sum The Secants, Get 6 (Posted on 2007-09-12) Difficulty: 2 of 5
Show that:

sec(2π/9) + sec(4π/9) + sec(8π/9) = 6

See The Solution Submitted by K Sengupta    
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Solution Some Secants Sum Six Comment 2 of 2 |

sec(2*pi/9) + sec(4*pi/9) + sec(8*pi/9) = 6

To prove the above equality, the multiangle cosine identites are useful:
cos(2t) = 2*(cos t)^2 - 1
cos(3t) = 4*(cos t)^3 - 3*(cos t)
cos(4t) = 8*(cos t)^4 - 8*(cos t)^2 + 1

If cos(2*pi/9) = x then by using the above identitites and sec t = 1/(cos t), the equality can be rewritten as:
1/x + 1/(2x^2 - 1) + 1/(8x^4 - 8x^2 + 1) = 6

This simplifies to a degree 7 polynomial:
96x^7 - 16x^6 - 152x^5 + 24x^4 + 66x^3 - 10x^2 - 6x + 1 = 0

The polynomial factors as:
(8x^3 - 6x + 1)*(12x^4 - 2x^3 - 10x^2 + 1) = 0

Substituting cos(2*pi/9) = x into the cos(3t) identity yields:
-1/2 = 4x^3 - 3x, which simplifies to 8x^3 - 6x + 1 = 0

Since cos(2*pi/9) is a root of 8x^3 - 6x + 1 = 0 and that polynomial is a factor of the seventh degree

polynomial, it follows that the identity is true.


  Posted by Brian Smith on 2007-09-16 17:11:57
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