321. If $(1 + x)^n = p_0 + p_1x + p_2x^2 + \ldots + p_nx^n$, show that $p_0 - p_2 + p_4 - \ldots = 2^{n/2}\cos\frac{n\pi}{4}$ and $p_1 - p_3 + p_5 - \ldots = 2^{n/2}\sin\frac{n\pi}{4}$.

Solution: Given, $(1 + x)^n = p_0 + p_1x + p_2x^2 + \cdots + p_nx^n$. Putting $x = i$, we get $(1 + i)^n = p_0 + p_1i + p_2i^2 + \cdots + p_ni^n$

$= (p_0 - p_2 + p_4 - \cdots) + i(p_1 - p_3 + p_5 - \cdots)\Rightarrow \left[\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\right]^n = (p_0 - p_2 + p_4 - \cdots) + i(p_1 - p_3 + p_5 - \cdots)$

Equating real and imaginary parts, we have $p_0 - p_2 + p_4 \cdots = 2^{\frac{n}{2}}\cos\frac{n\pi}{4}$ and $p_1 - p_3 + p_5 - \cdots = 2^{\frac{n}{2}}\sin\frac{n\pi}{4}$.

322. If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + \ldots + a_{2n}x^{2n}$, show that $a_0 + a_3 + a_6 + \ldots = \frac{1}{3}\left(1 + 2^{n + 1}\cos\frac{n\pi}{3}\right)$.

Solution: Given, $(1 - x + x^2)^n = a_0 + a_1 + a_2x^2 + \cdots a_{2n}x^{2n}$. Putting $x = 1, \omega$ and $\omega^2$, we get

$1 = a_0 + a_1 + a_2 + \cdots + a_{2n}, (-2\omega)^n = a_0 + a_1\omega + a_2\omega^2 + \cdots + a_{2n}\omega^{2n}, (-2\omega^2)^n = a_0 + a_1\omega^2 + a_2\omega^4 + \cdots + a_{2n}\omega^{4n}$

Adding these we get, $3(a_0 + a_3 + a_6 + \cdots) = 1 + (-2)^n(\omega^n + \omega^{2n})$. Now $\omega = \frac{-1 + \sqrt{3}i}{2} = \left(\cos\frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right)$

$\omega^n = \cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3}$. Now $\omega^2 = \frac{-1 - \sqrt{3}i}{2} = \left(\cos\frac{2\pi}{3} - i\sin \frac{2\pi}{3}\right)\therefore \omega^n + \omega^{2n} = 2\cos\frac{2n\pi}{3} = 2\cos\left(n\pi - \frac{n\pi}{3}\right)$

$= 2(-1)^n\cos\frac{n\pi}{3}$. Thus, $3(a_0 + a_3 + a_6 + \cdots) = 1 + (-2)^n2(-1)^n\cos\frac{n\pi}{3} = 1 +2^{n + 1}\cos\frac{n\pi}{3}$. $\Rightarrow a_0 + a_3 + a_6 + \cdots = \frac{1}{3}\left(1 + 2^{n+1}\cos\frac{n\pi}{3}\right)$.

323. If $n$ is a positive integer and $(1 + x)^n = c_0 + c_1x + c_2x^2 + \ldots + c_nx^n$, show that $c_0 + c_4 + c_8 + \ldots = 2^{n - 2} + 2^{n/2 - 1}\cos\frac{n\pi}{4}$.

Solution: Given, $(1 + x)^n = c_0 + c_1x + c_2x^2 + \cdots + c_nx^n$. Putting $x = 1$ and $x = -1$, we get $2^n = c_0 + c_1 + c_2 + \cdots + c_n$

and $0 = c_0 - c_1 + c_2 - \cdots + (-1)^nc_n$. Adding these two, we get $2^n = 2(c_0 + c_2 + c_4 + \cdots)$ or $c_0 + c_2 + c_4 + \cdots = 2^{n - 1}$

Putting $x = i$, we get $(1 + i)^n = c_0 + c_1i + c_2i^2 + c_3i^3 + \cdots + c_ni^n\Rightarrow \left[\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right]^n = (c_0 -c_2 + c_4 - \cdots) + i(c_1 - c_3 + \cdots)$

$\Rightarrow 2^{\frac{n}{2}}\left(\cos\frac{n\pi}{4}+i\sin\frac{i\pi}{4}\right) = (c_0 -c_2 + c_4 - \cdots) + i(c_1 - c_3 + \cdots)$

Equating real parts, we get $c_0 - c_2 + c_4 - \cdots = 2^{\frac{n}{2}}\cos\frac{n\pi}{4}$. Adding this result with the one obtained previously, we have $2[c_0 + c_4 + c_8 + \cdots] = 2^{n - 1} + 2^{\frac{n}{2}}\cos\frac{n\pi}{4}$.

324. Solve the equation $z^8 + 1= 0$ and deduce that $\cos4\theta = 8\left(\cos\theta - \cos\frac{\pi}{8}\right)\left(\cos\theta - \cos\frac{3\pi}{8}\right)$ $\left(\cos\theta - \cos\frac{5\pi}{8}\right)\left(\cos\theta - \cos\frac{7\pi}{8}\right)$.

Solution: $z^8 + 1 = 0 \Rightarrow z^8 = -1 = \cos\pi + i \sin\pi\therefore z = (\cos\pi + i \sin\pi)^{\frac{1}{8}} = \cos\frac{2r\pi + \pi}{8} + i \sin\frac{2r\pi + \pi}{8}, r = 0, 1, 2, \ldots, 7$

$\therefore z = \cos\frac{\pi}{8} \pm \sin\frac{\pi}{8}, \cos\frac{3\pi}{8} \pm \sin\frac{3\pi}{8}, \cos\frac{5\pi}{8} \pm \sin\frac{5\pi}{8}, \cos\frac{7\pi}{8} \pm \sin\frac{7\pi}{8}$

Now, quadratic equation whose roots are $\cos\frac{\pi}{8} \pm \sin\frac{\pi}{8},$ is $z^2 - 2\cos\frac{\pi}{8}z + 1 = 0$. Similarly, we can find the quadratic equations for remaining three pairs of roots. Thus,

$z^8 + 1 = \left(z^2 - 2\cos\frac{\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{3\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{5\pi}{8}z + 1\right)\left(z^2 - 2\cos\frac{7\pi}{8}z + 1\right)$

Dividing both sides by $z^4$, we get $z^4 + \frac{1}{z^4} = \left(z + \frac{1}{z} - 2\cos\frac{\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{3\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{5\pi}{8}\right)\left(z + \frac{1}{z} - 2\cos\frac{7\pi}{8}\right)$

Putting $z = \cos\theta + i\sin\theta$, so that $z^n + \frac{1}{z^n} = 2\cos n\theta$, we get $2\cos 4\theta = 2\left(\cos \theta - \cos\frac{\pi}{8}\right)2\left(\cos \theta - \cos\frac{3\pi}{8}\right)2\left(\cos \theta - \cos\frac{5\pi}{8}\right)2\left(\cos \theta - \cos\frac{5\pi}{8}\right)$

$\therefore \cos 4\theta = 8\left(\cos \theta - \cos\frac{\pi}{8}\right)\left(\cos \theta - \cos\frac{3\pi}{8}\right)\left(\cos \theta - \cos\frac{5\pi}{8}\right)\left(\cos \theta - \cos\frac{7\pi}{8}\right)$

325. Prove that the roots of the equation $8x^3 - 4x^2 - 4x + 1= 0$ are $\cos\frac{\pi}{7}, \cos\frac{3\pi}{7}, \cos\frac{5\pi}{7}$.

Solution: Let $z = \cos\theta + i \sin\theta$, then $z^7 = \cos 7\theta + i \sin 7\theta$. If $\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$ then $z^7 = \cos 7\theta + i \sin 7\theta = 1$ or $z^7 + 1 =0$

Thus, $z = \cos\theta + i \sin\theta$, where $\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$ are the roots of the equation. Also, when $\theta = \pi, z = -1$. Now, $z^7 + 1 = 0 \Rightarrow (z + 1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) = 0$

Root of equation $z + 1 = 0$ is $\cos \theta + i \sin \theta$, where $\theta = \pi$ Roots of equation $z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0\;\;\;\;(1)$ are $\cos \theta + i \sin \theta,$ where $\theta = \frac{\pi}{7}, \frac{3\pi}{7}, \frac{5\pi}{7}, \frac{7\pi}{7}, \frac{9\pi}{7}, \frac{11\pi}{7}, \frac{13\pi}{7}$

Let $x = \cos \theta$, then $z + \frac{1}{z} = \cos \theta + i \sin \theta + \frac{1}{\cos \theta + i \sin \theta} = 2\cos\theta = 2x$ But $\cos\left(\frac{13\pi}{7}\right) = cos\left(2\pi - \frac{\pi}{7}\right) = \cos\frac{\pi}{7}, \cos\frac{11\pi}{7} = \cos\frac{3\pi}{7}, \cos\frac{9\pi}{7} = \cos\frac{5\pi}{7}$

Dividing (1) by $z^3$, we get $z^3 - z^2 + z - 1 + \frac{1}{z} - \frac{1}{z^2} + \frac{1}{z^3} = 0$ $\left(z^3 + \frac{1}{z^3}\right) - \left(z^2 + \frac{1}{z^2}\right) + \left(z + \frac{1}{z}\right) - 1 = 0$

$\left(z + \frac{1}{z}\right)^3 - 3z.\frac{1}{z}\left(z + \frac{1}{z}\right) - \left[\left(z + \frac{1}{z}\right)^2 - 2z.\frac{1}{z}\right] + z + \frac{1}{z} - 1 = 0$ $\Rightarrow 8x^3 - 4x^2 -4x + 1 = 0$. Roots of this equation are $\cos \frac{\pi}{7}, \cos \frac{3\pi}{7}$ and $\cos \frac{5\pi}{7}$.

326. Solve the equation $z^{10} - 1 = 0$ and deduce that $\sin5\theta = 5\sin\theta\left(1 - \frac{\sin\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin\theta}{\sin^2\frac{2\pi}{5}}\right)$.

Solution: Given, $z^{10} - 1 = 0 \Rightarrow z^{10} = 1 = \cos 0 + i \sin 0\therefore z = (\cos 0 + i \sin 0)^{\frac{1}{10}} = \cos\frac{2r\pi}{10} + i \sin\frac{2r\pi}{10}$ $= \pm 1, \cos\frac{\pi}{5} \pm i\sin\frac{\pi}{5}, \cos\frac{2\pi}{5} \pm i\sin\frac{2\pi}{5}, \cos\frac{3\pi}{5} \pm i\sin\frac{3\pi}{5}, \cos\frac{4\pi}{5} \pm i\sin\frac{4\pi}{5}$

Quadratic equation whose roots are $\pm 1$ is $z^2 - 1 = 0$. And quadratic equation whose roots are $\cos\frac{\pi}{5} \pm \sin\frac{\pi}{5}$ is $z^2 - 2\cos\frac{\pi}{5}z + 1 = 0$. Thus, $z^{10} - 1 = (z^2 - 1)\left(z^2 - 2\cos\frac{\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{2\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{3\pi}{5}z + 1\right)\left(z^2 - 2\cos\frac{4\pi}{5}z + 1\right)$

Dividing both sides by $z^5$, we get $z^5 - \frac{1}{z^5} = \left(z - \frac{1}{z}\right)\left(z + \frac{1}{z} - 2\cos\frac{\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{2\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{3\pi}{5}\right)\left(z + \frac{1}{z} - 2\cos\frac{4\pi}{5}\right)$

Putting $z = \cos \theta + i \sin \theta$ in the above equation, so that $z^5 - \frac{1}{z^5} = 2i\sin 5\theta$, we get $2i\sin 5\theta = 2i\sin\theta.2\left(\cos \theta - \cos\frac{\pi}{5}\right)2\left(\cos \theta - \cos\frac{2\pi}{5}\right)2\left(\cos \theta - \cos\frac{3\pi}{5}\right)2\left(\cos \theta - \cos\frac{4\pi}{5}\right)$ $\therefore \sin 5\theta = 16 \sin \theta\left(\cos \theta - \cos\frac{\pi}{5}\right)\left(\cos \theta - \cos\frac{2\pi}{5}\right)\left(\cos \theta - \cos\frac{3\pi}{5}\right)\left(\cos \theta - \cos\frac{4\pi}{5}\right)$

$= 16\sin \theta\left(\cos\theta -\cos\frac{\pi}{5}\right)\left(\cos\theta +\cos\frac{\pi}{5}\right)\left(\cos\theta -\cos\frac{2\pi}{5}\right)\left(\cos\theta +\cos\frac{2\pi}{5}\right)$ $= 16\sin \theta\left(\cos^2\theta - \cos^2\frac{\pi}{5}\right)\left(\cos^2\theta - \cos^2\frac{2\pi}{5}\right)$ $= 16\sin \theta\left(\sin^2\frac{\pi}{5} - \sin^2\theta\right)\left(\sin^2\frac{2\pi}{5} - \sin^2\theta\right)$

$= 16\sin\theta\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$ $= 16\sin\theta\sin^2{36^\circ}\sin^2{72^\circ}\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$

$= 16\sin\theta\left(\frac{\sqrt{10 - 2\sqrt{5}}}{4}\right)^2\left(\frac{\sqrt{10 + 2\sqrt{5}}}{4}\right)^2\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^2\theta}{\sin^2\frac{2\pi}{5}}\right)$ Thus, $\sin5\theta = 5\sin\theta\left(1 - \frac{\sin^2\theta}{\sin^2\frac{\pi}{5}}\right)\left(1 - \frac{\sin^\theta}{\sin^2\frac{2\pi}{5}}\right)$.

327. Solve the equation $x^7 +1 = 0$ and deduce that $\cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi}{7} = -\frac{1}{8}$.

Solution: Given, $x^7 + 1 = 0$ or $x^7 = -1 = \cos\pi + i \sin\pi$ $\therefore x = \left(\cos\pi + i \sin\pi\right)^{\frac{1}{7}} = \cos\frac{2r\pi + \pi}{7} + i \sin \frac{2r\pi + \pi}{7}, r = 0, 1, 2, \ldots, 6$

$= \cos \frac{\pi}{7} \pm i \sin\frac{\pi}{7}, \cos \frac{2\pi}{7} \pm i \sin\frac{2\pi}{7}, \cos \frac{3\pi}{7} \pm i \sin\frac{3\pi}{7}, \cos \pi + i \sin \pi(= -1)$

$x^7 + 1 = (x + 1)\left(x^2 - 2\cos\frac{\pi}{7}x + 1\right)\left(x^2 - 2\cos\frac{2\pi}{7}x + 1\right)\left(x^2 - 2\cos\frac{3\pi}{7}x + 1\right)$. Putting $x = i$, we get

$i^7 + 1 = (1 + i)\left(-2i\cos\frac{\pi}{7}\right)\left(-2i\cos\frac{2\pi}{7}\right)\left(-2i\cos\frac{3\pi}{7}\right)$

$1 - i = 8(1 + i)\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = -8(1 - i)\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7}$ $\therefore \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{3\pi}{7} = -\frac{1}{8}$.

328. Form the equation whose roots are $\cot^2\frac{\pi}{2n + 1}, \cot^2\frac{2\pi}{2n + 1}, \ldots, \cot^2\frac{n\pi}{2n + 1}$, and hence find the value of $\cot^2\frac{\pi}{2n + 1} + \cot^2\frac{2\pi}{2n + 1} + \ldots + \cot^2\frac{n\pi}{2n + 1}$.

Solution: $(\cos\alpha + i \sin\alpha)^n = \cos^n\alpha + i.{^nC_1}\cos^{n-1}\alpha \sin\alpha + i^2.{^nC_2}\cos^{n-2}\alpha \sin^2\alpha + \cdots + i^n.{^nCn}\sin^n\alpha$

$\Rightarrow \cos n\alpha + i \sin n\alpha = (\cos^n\alpha - {^nC_2}\cos^{n-2}\alpha \sin^2\alpha) + i({^nC_1}\cos^{n-1}\alpha \sin\alpha)$. Equating imaginary parts, we get

$\therefore \sin n\alpha = {^nC_1}\cos^{n-1}\alpha \sin\alpha - {}^nC_3\cos^{n-3}\alpha \sin^3\alpha + \cdots$ $\therefore \sin (2n+1)\alpha = {}^{2n+1}C_1\cos^{2n}\alpha \sin\alpha - {}^{2n+1}C_3\cos^{2n-2}\alpha \sin^3\alpha + \cdots$

$\Rightarrow \sin (2n+1)\alpha = \sin^{2n+1}\alpha[{^{2n+1}C_1\cot^{2n}}\alpha - {}^{2n+1}C_3\cot^{2n-2}\alpha + \cdots]$ $\text{when} \alpha = \frac{\pi}{2n+1}, \frac{2\pi}{2n+1}, \cdots, \frac{n\pi}{2n+1}, \sin(2n+1)\alpha = 0$

$\therefore \cot^2\frac{\pi}{2n+1}, \cot^2\frac{2\pi}{2n+1}, \cdots, \cot^2\frac{n\pi}{2n+1}$ are the roots of the equation. From the second term coefficient we get sum of roots in a polynomial.

$\therefore \cot^2\frac{\pi}{2n+1}+ \cot^2\frac{2\pi}{2n+1}+ \cdots+ \cot^2\frac{n\pi}{2n+1} = \frac{{}^{2n+1}C_3}{{}^{2n+1}C_1}$.

329. If $\theta \neq k\pi$, show that $\cos\theta\sin\theta + \cos^2\theta\sin2\theta + \ldots + \cos^n\theta\sin n\theta = \cot\theta(1 - \cos^n\theta\cos n\theta)$.

Solution: Let $C = \cos\theta \cos\theta + \cos^2\theta \cos 2\theta + \cdots + cos^n\theta \cos n\theta$ and $S = \cos\theta \sin\theta + \cos^2\theta \sin 2\theta + \cdots + cos^n\theta \sin n\theta$

Now, $C + iS = \cos\theta(cos\theta + i \sin\theta) + \cos^2\theta(\cos2\theta + i \sin 2\theta) + \cdots + \cos^n\theta(\cos n\theta + i \sin n\theta)$ $= \cos\theta.e^{i\theta} + \cos^2\theta.e^{2i\theta} + \cdots + \cos^n\theta.e^{ni\theta}$ $= x + x^2 + \cdots + x^n$, where $x = \cos\theta e^{i\theta}= \frac{x(x^n - 1)}{x - 1} = \frac{\cos\theta e^{i\theta}(\cos^n\theta e^{in\theta} - 1)}{\cos\theta e^{i\theta} - 1}$

$= \frac{\cos\theta[\cos^n\theta(\cos n\theta + i \sin n\theta) - 1]}{\cos\theta - e^{-i\theta}} = \frac{\cos\theta[(\cos^n\theta\cos n\theta -1) + i cos^n\theta\sin n\theta]}{\cos\theta - (\cos\theta - i\sin\theta)}$

$= -i \cot\theta(\cos^n\theta \cos n\theta - 1) + i \cos^n\theta \sin n\theta$

Equating imaginary parts, we get $S = -\cot\theta(\cos^n\theta \cos n\theta - 1) = \cot\theta(1 - \cos^n\theta\cos n\theta)$.

330. Show that $-3 - 4i = 5e^{i(\pi + \tan^{-1}4/3)}$.

Solution: L.H.S. $= -3 -4i = 5\left(-\frac{3}{5} - i\frac{4}{5}\right) = 5\left(\cos\left(\pi + \tan^{-1}\frac{4}{5}\right) +i \sin\left(\pi + \tan^{-1}\frac{4}{5}\right)\right)$

$= 5e^{i\left(\pi + \tan^{-1}\frac{4}{5}\right)} =$ R.H.S.