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Thread: Addition of secants

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    Addition of secants

    Please let me know how to find the value of this trigonometric expression

    sec(pi/7) + sec(3 * pi/7) + sec(5 * pi/7)
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  2. #2
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    Quote Originally Posted by ravishank View Post
    Please let me know how to find the value of this trigonometric expression

    sec(pi/7) + sec(3 * pi/7) + sec(5 * pi/7)
    \frac{\pi}7,\; \frac{3\pi}7,\; \frac{5\pi}7,\; \pi,\; \frac{9\pi}7,\; \frac{11\pi}7,\; \frac{13\pi}7 are the seven roots of the equation \cos(7x) = -1. Their cosines are the x-coordinates of the seven coloured dots in the attached picture, which are equally spaced around the unit circle. Notice that the blue dots have the same x-coordinates as the red dots.

    There is a formula for \cos(7x) in terms of c=\cos x, namely \cos(7x) = 64c^7-112c^5+56c^3 - 7c. So the equation \cos(7\pi) = -1 can be written as 64c^7-112c^5+56c^3 - 7c +1=0. The seven roots of this equation are \cos(\pi/7),\; \cos(3\pi/7),\; \cos(5\pi/7) (each of which is a repeated root), the seventh root being \cos\pi = -1.

    The next step is to factorise the left side of that equation as 64c^7-112c^5+56c^3 - 7c +1 = (c+1)(8c^3 - 4c^2 - 4c + 1)^2.
    The c+1 factor obviously corresponds to x = \pi (the green dot in the picture) and the three roots of the squared factor correspond to the x-coordinates of the red (or blue) dots. Thus the equation 8c^3 - 4c^2 - 4c + 1 = 0 has roots \cos(\pi/7),\; \cos(3\pi/7),\; \cos(5\pi/7). Since \sec x = 1/\cos x, we can put s = 1/c and conclude that the equation s^3-4s^2-4s+8=0 has roots \sec(\pi/7),\; \sec(3\pi/7),\; \sec(5\pi/7). The sum of the roots of the equation is 4, so that is the answer to the problem.
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