361. Prove that the polynomial $(\cos\theta + x\sin\theta)^n - \cos n\theta - x\sin n\theta$ is divisible by $x^2 + 1$.

Solution: Let $f(x) = (\cos\theta + x\sin\theta)^n - \cos n\theta - x\sin n\theta$. But $x^2 + 1 = (x + i)(x - i)$ and $f(i) = \cos n\theta + i\sin n\theta - \cos n\theta - i\sin n\theta = 0$. Similarly, $f(-i) = 0$. And hence, required condition is proved.

362. Prove that the polynomial $x^n\sin\theta - p^{n - 1}x\sin n\theta + p^n\sin(n - 1)\theta$ is divisible by $x^2 - 2px\cos\theta + p^2$.

Solution: Roots of the equation $x^2 - 2px\cos\theta + p^2 = 0$ are $p(\cos\theta\pm\sin\theta)$. Let $f(x) = x^n\sin\theta - p^{n - 1}x\sin n\theta + p^n\sin(n - 1)\theta$, then

$f[p(\cos\theta + i\sin\theta)] = p^n(\cos n\theta + i\sin n\theta)\sin\theta - p^n(\cos\theta + i\sin \theta)\sin n\theta + p^n\sin(n - 1)\theta$. Separating real and imaginary parts

$\cos n\theta\sin\theta - \cos\theta\sin n\theta + \sin(n - 1)\theta = -\sin(n - 1)\theta + \sin(n - 1)\theta = 0$

and $\sin\theta\sin n\theta - \sin\theta\sin n\theta = 0$. Hence, $f(x)$ is divisible by $p(\cos\theta + i\sin\theta)$ and similarly we can prove it for the other root.

363. Find out for what values of $p$ and $q$ the binomial $x^4 + 1$ is divisible by $x^2 + px + q$.

Solution: Let $x^4 + 1 = (x^2 + px + q)(x^2 + p'x + q') = x^4 + (p + p')x^3 + (pp' + q + q')x^2 + (pq' + p'q)x + qq'$ which gives us four equations $p + p' = 0, pp' + q + q' = 0, pq' + p'q = 0$ and $qq' = 1$.

Assuming $p = 0, p' = 0, q + q' = 0, qq' = 1, q^2 = -1, q = \pm i, q' = \mp i$. Consequently, corresponding factorization has form $x^4 + 1 = (x^2 + i)(x^2 - i)$.

Let $q = q', q^2 = 1, q = \pm 1$. First let $q = q' = 1$. Then $pp' = -2, p + p' = 0, p^2 = 2, p = \pm\sqrt{2}, p' = \mp\sqrt{2}$. The corresponding factorization is $x^4 + 1 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$.

Then we assume $q = q' = -1, p + p' = 0, pp' = 2, p = \pm \sqrt{2}i, p' = \mp\sqrt{2}i$. The factorization will be $(x^2 + \sqrt{2}ix - 1)(x^2 - \sqrt{2}ix - 1)$.

364. Find the sum of the $p$th($p\in\mathbb{P}$) power of the roots of the equation $x^n = 1$.

Solution: Let $S = \displaystyle\sum_{k = 1}^{n - 1}x_k^p = \sum_{k = 1}^{n - 1}z^{kp}$ where $z = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$.

Thus, $\displaystyle\sum_{k = 1}^{n - 1}x_k^p = 1 + z^p + z^{2p} + \cdots + z^{(n - 1)p}$ but $z^p = \cos\frac{2p\pi}{n} + i\sin\frac{2p\pi}{n}$. Obviously $z^p = 1$ if and only if $p$ is divisible by $n$, in which case $S = n$. If $z^p\neq 1$, then $S = \frac{z^{np - 1}}{z^p - 1} = 0\because z^{np} = 1$.

365. Let $\epsilon = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}, \;\forall\;n\in P$, and let $A_k = x + y\epsilon^k + z\epsilon^{2k} + \cdots + w\epsilon^{(n - 1)k}, (k = 0, 1, 2, \ldots, n - 1)$ where $x, y, z, \ldots, w$ are $n$ arbitrary complex numbers. Prove that $\displaystyle \sum_{k = 0}^{n - 1}|A_k|^2 = n(|x|^2 + |y|^2 + \ldots + |w|^2)$

Solution: We have $\displaystyle\sum_{k=0}^{n - 1}|A_k|^2 = \sum_{k=0}^{n - 1}A_k\overline{A_k}$.

But $A_k\overline{A_k} = (x + y\epsilon^k + z\epsilon^{2k} + \ldots + w\epsilon^{(n - 1)k})(\overline{x} + \overline{y}\epsilon^{-k} + \overline{z}\epsilon^{-2k} + \cdots + \overline{w}\epsilon^{-(n - 1)k})$

$= (x\overline{x} + y\overline{y} + \cdots + w\overline{w})+ x(\overline{y}\epsilon^{-k} + \overline{z}\epsilon^{-2k} + \cdots + \overline{w}\epsilon^{-(n - 1)k}) + y\epsilon^k(\overline{x} + \overline{x}\epsilon^{-2k} + \cdots + \overline{w}\epsilon^{-(n - 1)k}) + \cdots + w\epsilon^{(n - 1)k}(\overline{x} + \overline{y}\epsilon^{-k} + \cdots + \overline{u}\epsilon^{-(n - 2)k})$

Therefore, $\displaystyle\sum_{k=0}^{n - 1}|A_k|^2 = n(|x|^2 + |y|^2 + \cdots + |w|^2) + x\sum_{k=1}^{n - 1}(\overline{y}\epsilon^{-k} + \overline{z}\epsilon^{-2k} + \cdots + \overline{w}\epsilon^{-(n - 1)k}) + y\sum_{k=0}^{n - 1}(\overline{x}\epsilon^k + \overline{z}\epsilon^{-k} + \cdots + \overline{w}\epsilon^{-(n - 2)k}) + \cdots + w\sum_{k = 0}^{n - 1}(\overline{x}\epsilon^{(n - 1)k}+ \overline{y}\epsilon^{(n - 2)k} + \cdots + \overline{u}\epsilon^k)$

But $\displaystyle\sum_{k=0}^{n - 1}\epsilon^{lk} = 0$ if $l$ is not divisible by $n$ from previous problem. Therefore all the sums in the right vanish and we get $\displaystyle\sum_{k = 0}^{n - 1}|A_k|^2 = n(|x|^2 + |y|^2 + \ldots + |w|^2)$.

366. Prove the identity $x^{2n} - 1 = (x^2 - 1)\displaystyle\prod_{k = 1}^{n - 1}\left(x^2 - 2x\cos\frac{k\pi}{n} + 1\right)$.

Solution: Considering $2n$th root of unity $x_s = \cos\frac{2s\pi}{n} + i\sin\frac{2s\pi}{n}\;\;(s = 1, 2, 3, \ldots, n)$.

Therefore, $x^{2n} - 1 = \displaystyle\prod_{s=1}^{2n}(x - x_s) = \prod_{s=1}^{n - 1}(x - x_s) \prod_{s=n + 1}^{2n - 1}(x - x_s)(x^2 - 1)\;\because x_n = -1, x_{2n} = 1$. But $x_{2n - s} = \overline{x_s}$, consequently,

$x^{2n} - 1 = (x^2 - 1)\displaystyle\prod_{s = 1}^{n - 1}(x - x_s)(x - \overline{x_s}) = (x^2 - 1)\prod_{s = 1}^{n - 1}(x^2 - 2x\cos\frac{s\pi}{n} + 1)$.

367. Prove the identity $x^{2n + 1} - 1 = (x - 1)\displaystyle\prod_{k = 1}^n\left(x^2 - 2x\cos\frac{2k\pi}{2n + 1} + 1\right)$.

Solution: Considering $2n + 1$th root of unity $x_s = \cos\frac{2(2s + 1)\pi}{2n + 1} + i\sin\frac{(2s + 1)\pi}{2n + 1}\;\;(s = 1, 2, 3, \ldots, n)$.

Therefore $x^{2n + 1} - 1 = \displaystyle\prod_{s = 1}^{2n + 1}(x - x_s)$. However, $x_{2n + 1} = 1$, therefore

$x^{2n + 1} - 1 = \displaystyle(x - 1)\prod_{s=1}^{2n}(x - x_s)$, but $x_{2n - s} = \displaystyle\overline{x_s} \Rightarrow x^{2n + 1} - 1 = (x - 1)\prod_{x=1}^{n}(x - x_s)(x - \overline{x_s}) = (x + 1)\displaystyle\prod_{k = 1}^n\left(x^2 - 2x\cos\frac{2k\pi}{2n + 1} + 1\right)$.

368. Prove the identity $x^{2n + 1} + 1 = (x + 1)\displaystyle\prod_{k = 1}^n\left(x^2 + 2x\cos\frac{2k\pi}{2n + 1} + 1\right)$.

Solution: Considering $2n + 1$th root of $-1, x_s = -\cos\frac{2(2s + 1)\pi}{2n + 1} + i\sin\frac{(2s + 1)\pi}{2n + 1}\;\;(s = 1, 2, 3, \ldots, n)$. Therefore $x^{2n + 1} + 1 = \displaystyle\prod_{s = 1}^{2n + 1}(x - x_s)$. However, $x_{2n + 1} = -1$, therefore

$x^{2n + 1} + 1 = \displaystyle(x + 1)\prod_{s=1}^{2n}(x - x_s)$, but $x_{2n - s} = \displaystyle\overline{x_s} \Rightarrow x^{2n + 1} + 1 = (x + 1)\prod_{x=1}^{n}(x - x_s)(x - \overline{x_s}) = (x + 1)\displaystyle\prod_{k = 1}^n\left(x^2 + 2x\cos\frac{2k\pi}{2n + 1} + 1\right)$.

369. Prove the identity $x^{2n} + 1 = \displaystyle\prod_{k = 0}^{n - 1}\left(x^2 - 2x\cos\frac{(2k + 1)\pi}{2n} + 1\right)$.

Solution: This problem can be solved like previous problem.

370. If $n$ is even, then prove the identity $\sin\frac{\pi}{2n}\sin\frac{2\pi}{2n}\ldots \sin\frac{(n - 1)\pi}{2n} = \frac{\sqrt{n}}{2^{n - 1}}$.

Solution: We have proven that $x^{2n} - 1 = (x^2 - 1)\displaystyle\prod_{k = 1}^{n - 1}\left(x^2 - 2x\cos\frac{k\pi}{n} + 1\right)$

$\Rightarrow x^{2n - 2} + x^{2n - 4} + \cdots + x^2 + 1 = \displaystyle\prod_{k = 1}^{n - 1}\left(x^2 - 2x\cos\frac{k\pi}{n} + 1\right)$

Putting $x = 1$, we have $n = \displaystyle\prod_{k = 1}^{n - 1}\left(2 - 2\cos\frac{k\pi}{n}\right) = \prod_{k=1}^{n - 1}4\sin^2\frac{k\pi}{2n} = 2^{2(n - 1)}\sin^2\frac{\pi}{2n}\sin^2\frac{2\pi}{2n}\cdots\sin^2\frac{(n - 1)\pi}{2n}$

$\Rightarrow \sin\frac{\pi}{2n}\sin\frac{2\pi}{2n}\ldots \sin\frac{(n - 1)\pi}{2n} = \frac{\sqrt{n}}{2^{n - 1}}$.