19 problems found
Solution:
Solution:
Let \(\omega = \e^{2\pi {\rm i}/n}\), where \(n\) is a positive integer. Show that, for any complex number \(z\), \[ (z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,. \] The points \(X_0, X_1, \ldots\, X_{n-1}\) lie on a circle with centre \(O\) and radius 1, and are the vertices of a regular polygon.
Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).
Show that \((z-\e^{i\theta})(z-\e^{-i\theta})=z^2 -2z\cos\theta +1\,\). Write down the \((2n)\)th roots of \(-1\) in the form \(\e^{i\theta}\), where \(-\pi <\theta \le \pi\), and deduce that \[ z^{2n} +1 = \prod_{k=1}^n \left(z^2-2z \cos\left( \tfrac{(2k-1)\pi}{2n}\right) +1\right) \,. \] Here, \(n\) is a positive integer, and the \(\prod\) notation denotes the product.
Solution: \begin{align*} && (z-e^{i \theta})(z-e^{-i\theta}) &= z^2 - (e^{i\theta}+e^{-i\theta})z + 1 \\ &&&= z^2-2\cos \theta z + 1 \end{align*} The \(2n\)th roots of \(-1\) are \(e^{\frac{i (2k+1)\pi}{2n}}, k \in \{-n, \cdots, n-1 \}\) or \(e^{\frac{i k \pi}{2n}}, k \in \{-2n+1, -2n+3, \cdots, 2n-3, 2n-1 \}\) \begin{align*} && z^{2n}+1 &= (z-e^{-i(2n-1)/2n})\cdot (z-e^{-i(2n-3)/2n})\cdots (z-e^{i(2n-3)/2n})\cdot (z-e^{i(2n-1)/2n}) \\ &&&= \prod_{k=1}^n \left (z - e^{i \frac{2k-1}{2n}\pi} \right)\left (z - e^{-i \frac{2k-1}{2n}\pi} \right)\\ &&&= \prod_{k=1}^n \left (z^2 - 2z \cos \left ( \frac{(2k-1)\pi}{2n}\right) + 1 \right) \end{align*}
For any given positive integer \(n\), a number \(a\) (which may be complex) is said to be a primitive \(n\)th root of unity if \(a^n=1\) and there is no integer \(m\) such that \(0 < m < n\) and \(a^m = 1\). Write down the two primitive 4th roots of unity. Let \({\rm C}_n(x)\) be the polynomial such that the roots of the equation \({\rm C}_n(x)=0\) are the primitive \(n\)th roots of unity, the coefficient of the highest power of \(x\) is one and the equation has no repeated roots. Show that \({\rm C}_4(x) = x^2+1\,\).
Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.
Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]
Solution: \begin{align*} && LHS &= (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) \\ &&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\ &&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\ &&&= RHS \end{align*} Therefore we can see \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta\). \begin{align*} && (5-i)^2(1+i) &= (24-10i)(1+i) \\ &&&= (24+10) + i(24-10) \\ &&&= 34+14i \\ \Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\ \Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\ \Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\ \end{align*} Consider \((1+i)(4-i)^3(20-i)\) \begin{align*} && (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\ &&&= 1985+i \\ \Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right) \end{align*}
By considering the solutions of the equation \(z^n-1=0\), or otherwise, show that \[(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},\] where \(z\) is any complex number and \(\omega={\rm e}^{2\pi i/n}\). Let \(A_1,A_2,A_3,\dots,A_n\) be points equally spaced around a circle of radius \(r\) centred at \(O\) (so that they are the vertices of a regular \(n\)-sided polygon). Show that \[\overrightarrow{OA_1}+\overrightarrow{OA_2}+\overrightarrow{OA_3} +\dots+\overrightarrow{OA_n}=\mathbf0.\] Deduce, or prove otherwise, that \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]
If $$ z^{4}+z^{3}+z^{2}+z+1=0\tag{*} $$ and \(u=z+z^{-1}\), find the possible values of \(u\). Hence find the possible values of \(z\). [Do not try to simplify your answers.] Show that, if \(z\) satisfies \((*)\), then \[z^{5}-1=0.\] Hence write the solutions of \((*)\) in the form \(z=r(\cos\theta+i\sin\theta)\) for suitable real \(r\) and \(\theta\). Deduce that \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]
Solution: \begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)
Show, using de Moivre's theorem, or otherwise, that \[ \tan7\theta=\frac{t(t^{6}-21t^{4}+35t^{2}-7)}{7t^{6}-35t^{4}+21t^{2}-1}\,, \] where \(t=\tan\theta.\)
By applying de Moivre's theorem to \(\cos5\theta+\mathrm{i}\sin5\theta,\) expanding the result using the binomial theorem, and then equating imaginary parts, show that \[ \sin5\theta=\sin\theta\left(16\cos^{4}\theta-12\cos^{2}\theta+1\right). \] Use this identity to evaluate \(\cos^{2}\frac{1}{5}\pi\), and deduce that \(\cos\frac{1}{5}\pi=\frac{1}{4}(1+\sqrt{5}).\)
Solution: \begin{align*} && (\cos \theta + i \sin \theta)^n &= \cos n \theta + i \sin n \theta \\ n = 5: && \cos 5 \theta + i \sin 5 \theta &= (\cos \theta + i \sin \theta)^5 \\ \textrm{Im}: && \sin 5 \theta &= \binom{5}{1}\cos^4 \theta \sin \theta + \binom{5}{3} \cos^2 \theta (- \sin^3 \theta) + \binom{5}{5} \sin^5 \theta \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta \sin^2 \theta+\sin^4 \theta) \\ &&&= \sin \theta (5\cos^4 \theta-10\cos^2 \theta (1-\cos^2 \theta)+(1-\cos^2 \theta)^2) \\ &&&= \sin \theta((5+10+1)\cos^4 \theta +(-10-2)\cos^2 \theta + 1) \\ &&&= \sin \theta(16\cos^4 \theta -12\cos^2 \theta + 1) \\ \end{align*} Suppose \(\theta= \frac{\pi}{5}\), then \(\sin 5 \theta = 0, \sin \theta \neq 0\), therefore if \(c = \cos \theta\) we must have \begin{align*} && 0 &= 16c^4-12c^2+1 \\ \Rightarrow && c^2 &= \frac{3 \pm \sqrt{5}}{8} \\ &&&= \frac{6\pm 2\sqrt{5}}{16} \\ &&&= \frac{(1 \pm \sqrt{5})^2}{16} \\ \Rightarrow && c &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*} Since \(c > 0\) we either have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\) or \(\cos \frac15 \pi = \frac{\sqrt{5}-1}4\), however \(\sqrt{5}-1 < 1.5\) and so \(\frac{\sqrt{5}-1}{4} < \frac12 = \cos \frac13 \pi\) we must have \(\cos \frac15 \pi = \frac{1+\sqrt{5}}4\)
If \(u\) and \(v\) are the two roots of \(z^{2}+az+b=0,\) show that \(a=-u-v\) and \(b=uv.\) Let \(\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).\) Show that \(\alpha\) is a root of \(z^{6}-1=0\) and express the roots in terms of \(\alpha.\) The number \(\alpha+\alpha^{2}+\alpha^{4}\) is a root of a quadratic equation \[ z^{2}+Az+B=0 \] where \(A\) and \(B\) are real. By guessing the other root, or otherwise, find the numerical values of \(A\) and \(B\). Show that \[ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2}, \] and evaluate \[ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}, \] making it clear how you determine the sign of your answer.
Solution: \begin{align*} 0 &= z^2+az+b \\ &= (z-u)(z-v) \\ &= z^2-(u+v)z+uv \end{align*} Therefore by comparing coefficients, \(a = -u-v\) and \(b = uv\). Suppose \(\alpha = \cos(2\pi/7) + i \sin (2\pi/7)\), then by De Moivre, \(\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1\), ie \(\alpha^7-1 = 0\). Notice that \((\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1\) and \begin{align*} P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6) \\ &= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\ &= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\ &= 2 \end{align*} Therefore it is a root of \(x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}\) Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$ And \(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2\) since it is positive it is \(\frac{\sqrt{7}}{2}\)
Let \[ \mathrm{C}_{n}(\theta)=\sum_{k=0}^{n}\cos k\theta \] and let \[ \mathrm{S}_{n}(\theta)=\sum_{k=0}^{n}\sin k\theta, \] where \(n\) is a positive integer and \(0<\theta<2\pi.\) Show that \[ \mathrm{C}_{n}(\theta)=\frac{\cos(\tfrac{1}{2}n\theta)\sin\left(\frac{1}{2}(n+1)\theta\right)}{\sin(\frac{1}{2}\theta)}, \] and obtain the corresponding expression for \(\mathrm{S}_{n}(\theta)\). Hence, or otherwise, show that for \(0<\theta<2\pi,\) \[ \left|\mathrm{C}_{n}(\theta)-\frac{1}{2}\right|\leqslant\frac{1}{2\sin(\frac{1}{2}\theta)}. \]
Solution: \begin{align*} && C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\ &&&= \textrm{Re} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\ \\ && S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\ &&&= \textrm{Im} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\ \\ && C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\ &&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (2n+1)\theta/2\right)}{2 \sin (\theta/2)} \leqslant\frac{1}{2 \sin (\theta/2)} \\ \end{align*}
Show that \[ \sin(2n+1)\theta=\sin^{2n+1}\theta\sum_{r=0}^{n}(-1)^{n-r}\binom{2n+1}{2r}\cot^{2r}\theta, \] where \(n\) is a positive integer. Deduce that the equation \[ \sum_{r=0}^{n}(-1)^{r}\binom{2n+1}{2r}x^{r}=0 \] has roots \(\cot^{2}(k\pi/(2n+1))\) for \(k=1,2,\ldots,n\). Show that
Let \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}.\) Show that \(1+\omega+\omega^{2}=0\) and calculate the modulus and argument of \(1+\omega^{2}.\) Let \(n\) be a positive integer. By evaluating \((1+\omega^{r})^{n}\) in two ways, taking \(r=1,2\) and \(3\), or otherwise, prove that \[ \binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots+\binom{n}{k}=\frac{1}{3}\left(2^{n}+2\cos\left(\frac{n\pi}{3}\right)\right), \] where \(k\) is the largest multiple of \(3\) less than or equal to \(n\). Without using a calculator, evaluate \[ \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} \] and \[ \binom{24}{2}+\binom{24}{5}+\cdots+\binom{24}{23}\,. \] {[}\(2^{25}=33554432.\){]}
Solution: Since \(\omega^3 = 1\) and \(\omega \neq 1\) we must have that \((\omega-1)(1 + \omega + \omega^2) = 0\) but by dividing by \(\omega - 1\) we obtain the desired result. \(1+\omega^2 = -\omega\) so \(|1 + \omega^2| = |-\omega| = 1\) and \(\arg ( 1 + \omega^2) = \arg(-\omega) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}\) \begin{align*} && (1 + 1)^n &= \sum_{k=0}^n \binom{n}{k}\\ && (1+ \omega)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{k} \\ && (1+ \omega^2)^n &= \sum_{k=0}^n \binom{n}{k} \omega^{2k} \\ \Rightarrow && 2^n+(-\omega^2)^n + (-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n (1+1+1)\binom{n}{k} + \sum_{k=0, k \equiv 1 \pmod{3}}^n (1 + \omega + \omega^2) \binom{n}{k} + \sum_{k=0, k \equiv 2 \pmod{3}}^n (1 + \omega^2 + \omega) \binom{n}{k} \\ \Rightarrow && 2^n +((-\omega)^n)^{-1}+(-\omega)^n &= \sum_{k=0, k \equiv 0 \pmod{3}}^n \binom{n}{k} \end{align*} \(2^n +((-\omega)^n)^{-1}+(-\omega)^n = 2^n + 2 \textrm{Re}(-\omega^n) = 2^n + 2 \cos \frac{n\pi}{3}\) Therefore our answer follows. \begin{align*} \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} &= \frac13 \l 2^{25} + 2\cos (\frac{25 \pi}{3}) \r \\ &= \frac13 \l 2^{25} + 2 \cos \frac{\pi}{3} \r \\ &= \frac13 \l 2^{25} + 1 \r \\ &= \frac13 \l (4096 \cdot 4096 \cdot 2) + 1 \r \\ &= 11\,184\,811 \end{align*} Notice that \(S_2 = \binom{24}{2} + \cdots +\binom{24}{23} = \binom{24}{1} + \cdots + \binom{24}{22} = S_1\) and \(S_0 = \binom{24}0 + \cdots + \binom{24}{21} = \frac13 \l 2^{24} + 2 \r\) Therefore since \(S_0 + 2 \cdot S_2 = 2^{24}\) we must have \begin{align*} S_2 &= \frac12 \l 2^{24} - \frac13 \l 2^{24} + 2 \r \r \\ &= \frac13 \l 2^{24} - 1 \r \\ &= \frac13 \l 16777216- 1 \r \\ &= \frac13 \cdot 16777215 \\ &= 5\,592\,405 \end{align*}
Show, using de Moivre's theorem, or otherwise, that \[ \tan9\theta=\frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)},\qquad\mbox{ where }t=\tan\theta. \] By considering the equation \(\tan9\theta=0,\) or otherwise, obtain a cubic equation with integer coefficients whose roots are \[ \tan^{2}\left(\frac{\pi}{9}\right),\qquad\tan^{2}\left(\frac{2\pi}{9}\right)\qquad\mbox{ and }\qquad\tan^{2}\left(\frac{4\pi}{9}\right). \] Deduce the value of \[ \tan\left(\frac{\pi}{9}\right)\tan\left(\frac{2\pi}{9}\right)\tan\left(\frac{4\pi}{9}\right). \] Show that \[ \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right)=33273. \]
Solution: Writing \(c = \cos \theta, s = \sin \theta\) then de Moivre states that: \begin{align*} && \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\ &&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\ &&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\ \Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\ &&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\ &&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)} \end{align*} If we consider \(\tan 9\theta = 0\) it will have the roots \(\theta = \frac{n \pi}{9}, n \in \mathbb{Z}\), in particular, the numerator of our fraction for \(\tan 9 \theta\) will be zero for \(t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}\). It's worth noting all other values of \(\theta\) will repeat these values. Also note that \(0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}\) are the roots of \(t\) and \(t^2-3\) respectively. Therefore the other values are the roots of our sextic. However, also note that \(\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}\) and similar, therefore we can notice that all the roots in pairs can be mapped to \(\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}\) and \(\tan \frac{4 \pi}{9}\) and all those values are squared, so the roots of: \(x^3 - 33x^2+27x-3\) will be \(\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}\) and \(\tan^2 \frac{4 \pi}{9}\). The product of the roots will be \(3\), so \begin{align*} && \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\ \Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\ \underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \sqrt{3} \\ \end{align*} Notice that \(x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))\) Therefore \begin{align*} \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\ &= 33\,273 \end{align*}