371. If $n$ is even, then prove the identity $\cos\frac{2\pi}{2n + 1}\cos\frac{4\pi}{2n + 1}\ldots\cos\frac{2n\pi}{2n + 1} = \frac{(-1)^{n/2}}{2^n}$.

Solution: This problem can be solved like previous problem.

372. Prove that if $\cos\alpha + i\sin\alpha$ is the solution of the equation $x^n + p_1x^{n - 1} + \cdots + p_n = 0$, then $p_1\sin\alpha + p_2\sin2\alpha + \cdots + p_n\sin n\alpha = 0(p1, p2, \ldots, p_n\text{ are real})$.

Solution: Since $\cos\alpha + i\sin\alpha$ is the root of the given equation, we have $\displaystyle\sum_{k = 0}^np_k(\cos\alpha + i\sin\alpha)^{n - k} = 0\;\;(p_0 = 1)$

$\Rightarrow \displaystyle(\cos\alpha + i\sin\alpha)^n\sum_{k=0}^np_k(\cos\alpha + i\sin\alpha)^{-k} = 0 \Rightarrow \sum_{k= 0}^np_k(\cos\alpha k - i\sin\alpha k) = 0$.

Hence, $\displaystyle\sum_{k= 0}^np_k\sin\alpha k = p_1\sin\alpha + p_2\sin2\alpha + \cdots + p_n\sin n\alpha = 0$.

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

Solution: The roots of the equation $x^7 = 1$ are $\cos\frac{2k\pi}{7} + i\sin\frac{2k\pi}{7}\;\;(k = 0, 1, 2, \ldots, 6)$.

Therefore, the roots of the equation $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0$ will be $x_k = \cos\frac{2k\pi}{7} + i\sin\frac{2k\pi}{7}\;\;(k = 1, 2, 3, \ldots, 6)$.

Putting $x + \frac{1}{x} = y$, then $x^2 + \frac{1}{x^2} = y^2 - 2$ and $x^3 + \frac{1}{x^3} = y^3 - 3y$. Rewriting the above equation $\left(x^3 + \frac{1}{x^3}\right) + \left(x^2 + \frac{1}{x^2}\right) + \left(x + \frac{1}{x}\right) + 1 = 0$.

Clearly, $x_1 = \overline{x_6}, x_2 = \overline{x_5}, x_3 = \overline{x_4}, x_k + \frac{1}{x_k} = x_k + \overline{x_k} = 2\cos\frac{2k\pi}{7}$.

Hence we can say that quantities $2\cos\frac{2\pi}{7}, 2\cos\frac{4\pi}{7}, 2\cos\frac{6\pi}{7}$ are the rootss of the equation $y^3 + y^2 - 2y - 1 = 0$. Let the roots of the cubic equation $x^3 - ax^2 + bx - c = 0$ be $\alpha, \beta, \gamma$. Then $\alpha + \beta + \gamma = a, \alpha\beta + \beta\gamma + \gamma\alpha = b, \alpha\beta\gamma = c$.

Let the equation, whose roots are $\sqrt[3]{\alpha}, \sqrt[3]{\beta}, \sqrt[3]{\gamma}$, be $x^3 - Ax^2 + Bx - C = 0$. Then, $\sqrt[3]{\alpha} + \sqrt[3]{\beta} + \sqrt[3]{\gamma} = A, \sqrt[3]{\alpha\beta} + \sqrt[3]{\beta\gamma} + \sqrt[3]{\gamma\alpha} = B, \sqrt[3]{\alpha\beta\gamma} = C$.

We know that $(m + p + q)^3 = m^3 + p^3 + q^3 + 3(m + p + q)(mp + mq + pq) - 3mpq$. Substituting $\sqrt[3]{\alpha}, \sqrt[3]{\beta}, \sqrt[3]{\gamma}$ and $\sqrt[3]{\alpha\beta}, \sqrt[3]{\beta\gamma}, \sqrt[3]{\gamma\alpha}$ for $m, p, q$ we obtain

$A^3 = a + 3AB - 3C, B^3 = b + 3BCA - 3C^2$. In our case, $a = -1, b = -2, c = 1, C = 1$. Hence, $A^3 = 3AB - 4, B^3 = 3AB - 5$. Multiplying these equations and putting $AB = z$, we find

$z^3 - 9z^2 + 27z - 20 = 0 \Rightarrow (z - 3)^3 + 7 = 0 \Rightarrow z = 3 - \sqrt[3]{7}$ But $A^3 = 3z - 4 \Rightarrow A = \sqrt[3]{5 - 3\sqrt[3]{7}}$ and hence

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

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

Solution: This problem can be solved like previous problem.

375. Let $A = x_1 + x_2\omega + x_3\omega^2, B = x_1 + x_2\omega^2 + x_3\omega,$ where $\omega, \omega^2$ are complex roots of unity and $x_1, x_2, x_3$ are roots of the cubic equation $x^3 + px + q = 0$. Prove that $A^3$ and $B^3$ are the roots of the quadratic equation $x^2 + 27qx - 27p^3 = 0$.

Solution: Squaring the first trimonial, $A^2 = (x_1^2 + 2x_2x_3) + (x_3^2 + 2x_1x_2)\omega + (x_2^2 + 2x_1x_3)\omega^2$.

Then $A^3 = (x_1^2 + x_2^2 + x_3^2 + 6x_1x_2x_3) + (3x_1^2x_2 + 3x_2^2x_1 + 3x_2^2x_3)\omega + (3x_1^2x_3 + 3x_2^2x_1 + 3x_3^2x_2)\omega^2$

Putting $3\alpha = 3x_1^2x_2 + 3x_2^2x_1 + 3x_2^2x_3$ and $3\beta = 3x_1^2x_3 + 3x_2^2x_1 + 3x_3^2x_2$.

Now $x_1^3 + x_2^3 + x_3^3 = -(px_1 + q) - (px_2 + q) - (px_3 + q) = -3q$ since $x_1 + x_2 + x_3 = 0$. Moreover, $x_1x_2x_3 = -q$, therefore

$A^3 = -9q + 3\alpha\omega + 3\beta\omega^2$, we also find $B^3 = -9q + 3\alpha\omega^2 + 3\beta\omega$. Hence, $A^3 + B^3 = -18q - 3\alpha - 3\beta = -27q$, and similarly, $A^3B^3 = -27p^3$.

376. Solve the equation $\frac{(5x^4 + 10x^2 + 1)(5a^4 + 10a^2 + 1)}{(x^4 + 10x^2 +1)(a^4 + 10a^2 + 5)} = ax$.

Solution: Let $f(x) = \frac{5x^4 + 10x^2 + 1}{x^4 + 10x^2 + 5}$ then the equation takes the form $f(x).f(a) = ax$.

$x - f(x) = \frac{(x - 1)^5}{x^4 + 10x^2 + 5}$ and $x + f(x) = \frac{(x + 1)^5}{x^4 + 10x^2 + 5}$. Dividing,

$\frac{x - f(x)}{x + f(x)} = \left(\frac{x - 1}{x + 1}\right)^5$. Let $\frac{x - 1}{x + 1} = y$ and $\frac{a - 1}{a + 1} = b$.

$\Rightarrow x - f(x) = y^5x + y^5f(x), x(1 - y^5) = f(x)(1 + y^5)\Rightarrow \frac{f(x)}{x} = \frac{1 - y^5}{1 + y^5}$.

Similarly, $\frac{f(a)}{a} = \frac{1 - b^5}{1 + b^5}$. So we can write the equation as $\frac{1 - y^5}{1 + y^5} = \frac{1 + b^5}{1 - b^5}\Rightarrow y^5 = -b^5$.

The last equation has five roots. $y_k = -b\epsilon^k$, where $\epsilon = \cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}$.

But $x = \frac{1 + y}{1 - y}\Rightarrow x_k = \frac{(a + 1) - (a - 1)\epsilon^k}{(a + 1) + (a - 1)\epsilon^k} = \frac{\cos\frac{k\pi}{5} - ia\sin\frac{k\pi}{5}}{a\cos\frac{k\pi}{5} - i\sin\frac{k\pi}{5}}$.

377. Find the magnitude of the sum $S = {}^nC_1 - 3.{}^nC_3 + 3^2.{}^nC_5 - 3^3.{}^nC_7 + \cdots$.

Solution: $\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^n = \cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3}$

Further $\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^n = \frac{(-1)^n}{2^n}(1 - i\sqrt{3})^n = \frac{(-1)^n}{2^n}[1 + {}^nC_1(-i\sqrt{3}) + {}^nC_2(-i\sqrt{3})^2 + {}^nC_3(-i\sqrt{3})^3 + \cdots]$

$= \frac{(-1)^n}{2^n}[1 - 3.{}^nC_2 + \cdots] -i\sqrt{3}[{}^nC_1 -3.{}^nC_3 + 3^2.{}^nC_5 - 3^3.{}^nC_7 + \cdots]$

Equating coefficient of $i$ in both the equations, $S = (-1)^{n + 1}\frac{2^n}{\sqrt{3}}\sin\frac{2n\pi}{3}$.

378. Find the magnitude of the follwing sums: $$\sigma = 1 - {}^nC_2 + {}^nC_4 - {}^nC_6 + \cdots$$ $$\sigma' = {}^nC_1 - {}^nC_3 + {}^nC_5 - {}^nC_7 + \cdots$$

Solution: We have $(1 + i)^n = 1 + {}^nC_1 i + {}^nC_2 i^2 + {}^nC_3 i^3 + \cdots = 1 + {}^nC_1 i - {}^nC_2 - {}^mC_3 i + \cdots$

But $1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

Therefore, $\sigma = 1 - {}^nC_2 + {}^nC_4 - {}^nC_6 + \cdots = 2^{\frac{n}{2}}\cos\frac{n\pi}{4}$,

$\sigma' = {}^nC_1 - {}^nC_3 + {}^nC_5 - {}^nC_7 + \cdots = 2^{\frac{n}{2}}\sin\frac{n\pi}{4}$.

Hence, if $n = 0(\mod 4)$ i.e. $n = 4m\;\forall\;m\in\mathbb{I}$, then $\sigma = (-1)^m2^{2m}, \sigma' = 0$. If $n = 4m + 1$, then $\sigma = \sigma' = (-1)^m2^{2m}$, for $n = 4m + 2, \sigma = 0, \sigma' = (-1)2^{2m + 1}$ and for $n = 4m + 3, \sigma = (-1)^{m + 1}2^{2m + 1}, \sigma' = (-1)^m2^{2m + 1}$.