Trigonometric Tables Won't Help!

Using the chebyshev polynomial $T_7 = 64x^7 - 112x^5 + 56x^3 - 7x$ , the roots to $T_7 (x) = 1$ are $\cos \frac{ 2n \pi } { 7}$ .

Lets factor out $(x-1)$ from $T_7 (x) - 1$ : $T_7 (x) - 1 = (x-1) (64x^6 + 64x^5 - 48x^4 - 48x^3 + 8x^2 + 8x + 1 )$ .
The zeroes of the second factor are $\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7}, \cos \frac{6\pi}{7}, \cos \frac{8\pi}{7}, \cos \frac{10\pi}{7}, \cos \frac{12\pi}{7}$ .

Taking the mapping $x \rightarrow \frac{1}{x}$ , the factor becomes $x^6 + 8x^5 + 8x^4 - 48x^3 - 48x^2 + 64x + 64$ . The zeroes of this are $\sec \frac{2\pi}{7}, \sec \frac{4\pi}{7}, \sec \frac{6\pi}{7}, \sec \frac{8\pi}{7}, \sec \frac{10\pi}{7}, \sec \frac{12\pi}{7}$ .

Finally, we apply newton's identities and want to calculate $P_4$ .
$P_1 = -8$
$P_2 = (-8) P_1 - 8 \times 2 = 48$
$P_3 = (-8) P_2 - 8 P_1 + 48 \times 3 = -176$
$P_4 = (-8) P_3 - 8 P_2 + 48 P_1 - (-48) \times 4 = 832$

Now, since $\sec \frac{ \pi}{7} = - \sec \frac{ 8 \pi } { 7} = - \sec \frac{ 6 pi } { 7 }$ ,
and $\sec \frac{ 2 \pi }{7} = \sec \frac{ 12 \pi }{ 7}$ ,
and $\sec \frac{ 3 \pi } { 7} = - \sec \frac { 4 \pi } { 7} = - \sec \frac{ 10 pi } { 7 }$ ,

Hence we conclude that

$\sec^4 \frac{2\pi}{7} + \sec^4 \frac{4\pi}{7} + \sec^4 \frac{6\pi}{7} + \sec^4 \frac{8\pi}{7} + \sec \frac{10\pi}{7}^4 + \sec^4 \frac{12\pi}{7} = 2 ( \sec^4 \frac{\pi}{7} + \sec^4 \frac{2\pi}{7} + \sec^4 \frac{3\pi}{7})$

Hence, the expression is $\frac{ 832 } { 2 } = 416$ .

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Note: The solution could be simplified by making the observation about the duplicity of roots first. If so, we can conclude that the polynomial whose roots are $\cos \frac{\pi}{7}, \cos \frac{2\pi}{7}, \cos \frac{3\pi}{7}$ is given by $\sqrt{ \frac{ T_7(x) - 1 } { x-1 } } = 8x^3 - 4x^2 - 4x + 1$ , and then using $x \rightarrow \frac{1}{x}$ and newton identities accordingly. Give it a try!