27. Complex Numbers Problems Part 6

1. $x^{2n} + 1= (x + 1)\sum_{k = 0}^{k = n - 1}\left(x^2 - 2xcos\frac{(2k + 1)\pi}{2n} + 1\right)$
2. $sin \frac{\pi}{2n} sin \frac{2\pi}{2n} ... sin \frac{(n - 1)\pi}{2n} = \frac{\sqrt{n}}{2^{n - 1}}$

if $$n$$ is even.

3. $cos \frac{2\pi}{2n + 1} cos \frac{4\pi}{2n + 1} .. cos \frac{2n\pi}{2n + 1} = \frac{(-1)^\frac{n}{2}}{2^n}$

if $$n$$ is even.

4. Prove that if $$cos\alpha + i sin\alpha$$ is the solution of the equation $$x^n + p_1x^{n - 1} + ... + p_n = 0$$ then $$p_1 sin\alpha + p_2 sin2\alpha + ... + p_n sinn\alpha = 0$$ where $$p_1, p_2, ..., p_n$$ are real.

5. Prove that

$\sqrt[3]{cos \frac{2\pi}{7}} + \sqrt[3]{cos \frac{4\pi}{7}} + \sqrt[3]{cos \frac{8\pi}{7}} = \sqrt[3]{\frac{1}{2}(5 - 3\sqrt[3]{7})}$
6. Prove that

$\sqrt[3]{cos \frac{2\pi}{9}} + \sqrt[3]{cos \frac{4\pi}{9}} + \sqrt[3]{cos \frac{8\pi}{9}} = \sqrt[3]{\frac{1}{2}(3\sqrt[3]{9} - 6)}$