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

sec(pi/7) + sec(3 * pi/7) + sec(5 * pi/7)

2. Originally Posted by ravishank
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.