If P = sin frac{2 pi}{7} + sin frac{4 pi}{7} | vedclass

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(C) Let $z = e^{i \frac{2 \pi}{7}}$. Then $z^7 = 1$. Consider the sum $S = z + z^2 + z^4 = (\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8

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$2$

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(C) Let $z = e^{i \frac{2 \pi}{7}}$. Then $z^7 = 1$. Consider the sum $S = z + z^2 + z^4 = (\cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}) + i(\sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7}) = Q + iP$. We know that $z, z^2, z^4$ are roots of the equation $x^3 + x^2 - 2x - 1 = 0$ is not correct here,rather they are roots of $x^3 + x^2 - 2x - 1 = 0$ is for $2\pi/7$ related sums. Actually,for $	heta = \frac{2 \pi}{7}$,the sum $S = Q + iP$. Using the property of roots of unity,$Q + iP = \sum_{k=0}^{2} e^{i \frac{2^{k+1} \pi}{7}}$. It is a known result that for $S = \sum_{k=0}^{2} e^{i \frac{2^k \cdot 2 \pi}{7}}$,the magnitude $|S|^2 = P^2 + Q^2$. For the sum of cosines $Q = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} = -\frac{1}{2}$. For the sum of sines $P = \sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7} = \frac{\sqrt{7}}{2}$. Then $P^2 + Q^2 = (\frac{\sqrt{7}}{2})^2 + (-\frac{1}{2})^2 = \frac{7}{4} + \frac{1}{4} = \frac{8}{4} = 2$. Since $P^2 + Q^2 = 2 = (\sqrt{2})^2$,the point $(P, Q)$ lies on a circle with radius $\sqrt{2}$. Wait,checking the options provided,if the question implies $P^2+Q^2=R^2$,then $R = \sqrt{2}$. Given the options,there might be a typo in the question or options. However,based on standard trigonometric identities,the radius is $\sqrt{2}$.

If $P = \sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} + \sin \frac{8 \pi}{7}$ and $Q = \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7}$,then the point $(P, Q)$ lies on the circle of radius

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