Problems

Solution: \begin{align*} && LHS &= 4\sin\theta \sin(\tfrac13\pi-\theta) \sin (\tfrac13\pi+\theta) \\ &&&= 4 \sin \theta \left (\tfrac{\sqrt{3}}{2}\cos \theta - \tfrac12 \sin \theta \right)\left (\tfrac{\sqrt{3}}{2}\cos \theta + \tfrac12 \sin \theta \right) \\ &&&= 4 \sin \theta \left (\tfrac{3}{4}\cos^2 \theta - \tfrac14 \sin^2 \theta \right) \\ &&&= 3\sin \theta - 4\sin^3 \theta \\ &&&= \cos 3 \theta = RHS \end{align*}

1. \(\,\) \begin{align*} && 3 \cos 3 \theta &= \sin 3 \theta \left (\cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \right) \\ \Rightarrow && 3 \cot 3\theta &= \cot \theta - \cot (\tfrac13\pi - \theta) + \cot (\tfrac13\pi + \theta) \\ \theta = \tfrac{\pi}{9}: && 3\cot \frac{\pi}{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \\ \Rightarrow && \sqrt{3} &= \cot \tfrac{\pi}{9} - \cot \tfrac{2}{9}\pi + \cot \tfrac49 \pi \end{align*}
2. \(\,\) \begin{align*} \theta = \tfrac16\pi - \phi && \sin(\tfrac12\pi - 3\phi) &= 4\sin(\tfrac16\pi - \phi)\sin(\phi+\tfrac16\pi)\sin(\tfrac12\pi - \phi) \\ \Rightarrow && \cos 3\phi &= 4\cos(\phi - \tfrac13\pi)\cos(\tfrac13\pi - \phi)\cos\phi \\ \Rightarrow && \cot 3\theta &= \cot \theta\cot(\phi - \tfrac13\pi)\cot(\tfrac13\pi - \phi) \tag{dividing by (\(*\))} \\ \\ \frac{\d}{\d \theta}:&& -\csc^2 3\phi &= \cot3\phi \left (-\csc^2 \phi\tan \phi+\csc^2 (\tfrac13\pi - \phi) \tan (\tfrac13\pi - \phi) -\csc^2(\phi - \tfrac13\pi)\tan (\phi - \tfrac13\pi) \right) \\ \Rightarrow && \csc^23\phi\tan3\phi & = 2( \csc2\phi- \csc(\tfrac{2}{3}\pi - 2\phi)+\csc(\phi - \tfrac23\pi)) \\ \phi = \frac{1}{18}\pi: && 4\sqrt{3} &= 2(\csc \tfrac{1}{9}\pi - \csc \tfrac59\pi + \csc \tfrac79 \pi) \\ \end{align*} and the result follows.

Solution:

1. Note that \(\cos \theta = \sin (90^\circ - \theta)\) so \begin{align*} && \sin(90^\circ - \theta) &= \sin 4 \theta\\ && 90^\circ - \theta &= 4\theta +360^{\circ}k \\ && 90^\circ + \theta &= 4\theta +360^{\circ}k \\ \Rightarrow && 5\theta &= 90^\circ, 450^\circ, 810^\circ, \cdots \\ && 3 \theta &= 90^\circ, 450^\circ, \cdots \\ \Rightarrow && \theta &= 18^\circ, 90^\circ, 162^\circ, \ldots \\ && \theta &= 30^\circ, 150^\circ, \ldots \end{align*} Therefore \(\theta = 8^\circ, 30^\circ, 90^\circ, 150^\circ, 162^\circ\). Note also that: \begin{align*} && 0 &= \sin 4 \theta - \cos \theta \\ &&&= 2 \sin 2 \theta \cos 2 \theta- \cos \theta \\ &&&= 4 \sin \theta \cos \theta \cos 2 \theta - \cos \theta \\ &&&= \cos \theta \left (4 \sin \theta (1- 2\sin^2 \theta) - 1 \right) \\ &&&= \cos \theta \left (-8\sin^3 \theta +4\sin \theta - 1 \right) \\ &&&= \cos \theta (1 - 2 \sin \theta)(4 \sin^2 \theta+2\sin \theta -1)\\ \cos \theta = 0: && \theta &= 90^\circ \\ \sin \theta = \frac12: && \theta &= 30^{\circ} \\ && \theta &= \sin^{-1} \left ( \frac{-1\pm \sqrt5}{4} \right) \end{align*} Therefore \(\sin 18^{\circ} = \frac{\pm \sqrt{5}-1}{4}\), but since \(\sin 18^{\circ} > 0\) it must be the positive version.
2. \(\,\) \begin{align*} && 4 \sin^2 x + 1 &= 4 \sin^2 2 x \\ &&&= 16 \sin^2 x \cos^2 x \\ &&&= 16 \sin^2 x (1- \sin^2 x) \\ \Rightarrow && 0 &= 16y^2 -12y+1 \\ \Rightarrow && \sin^2 x &= \frac{3\pm \sqrt5}{8} \\ &&&= \left ( \frac{1 \pm \sqrt5}{4} \right)^2 \\ \Rightarrow && \sin x &= \pm \frac{1 \pm \sqrt{5}}{4} \end{align*}
3. \(\,\) \begin{align*} && \sin^2 x + \frac1{2^2} &= \sin^2 2x \end{align*} So if we can have \(\sin 5x = \pm \frac12\) and \(\sin 3x = \pm \frac{1 \pm \sqrt5}{4}\) then we are good, ie \begin{align*} && 5x &= 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}, 390^{\circ}, \cdots \\ \Rightarrow && x &= 6^{\circ}, 30^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ} \\ \Rightarrow && 3x &= \boxed{18^{\circ}}, \cancel{90^{\circ}}, \boxed{126^{\circ}}, \boxed{198^{\circ}}, \boxed{78^{\circ}} \end{align*} So our solutions are \(x = 6^{\circ}, 42^{\circ}, 66^{\circ}, 78^{\circ}\) although it's interesting to note that \(x = 45^{\circ}\) is another solution

Solution: \begin{align*} && \sum_{k=0}^n \sin k\theta &= \textrm{Im} \left ( \sum_{k=0}^n e^{i k \theta}\right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i \theta}-1} \right)\\ &&&= \textrm{Im} \left ( \frac{e^{i(n+\tfrac12)\theta}-e^{-i\theta/2}}{e^{i \theta/2}-e^{-i \theta/2}} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/2i}{(e^{i \theta/2}-e^{-i \theta/2})/2i} \right)\\ &&&= \textrm{Im} \left ( \frac{(e^{i(n+\tfrac12)\theta}-e^{-i\theta/2})/i}{2\sin \tfrac12 \theta} \right)\\ &&&= \frac{\cos \tfrac12 \theta - \cos(n+ \tfrac12)\theta}{2\sin \tfrac12 \theta} \end{align*}

1. When \(n\) is large we have \begin{align*} &&\sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) &= \frac{\cos \frac{\pi}{2n} - \cos \frac{(2n + 1)\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&= \frac{\cos \frac{\pi}{2n} +\cos \frac{\pi}{2n}}{2 \sin \frac{\pi}{2n}} \\ &&&\approx \frac{1- \frac{\pi^2}{4n^2}}{\frac{\pi}{2n}} \\ &&&= \frac{2n}{\pi} - \frac{\pi}{2n} \\ &&&\approx \frac{2n}{\pi} \end{align*}
2. \(\,\) \begin{align*} && \sum_{k=0}^n \sin k\theta &= \frac { \cos \tfrac12\theta - \cos (n+ \tfrac12) \theta} {2\sin \tfrac12\theta} \\ \frac{\d }{\d \theta}: && \sum_{k=0}^n k\cos k\theta &= \frac { (-\tfrac12\sin\tfrac12\theta +(n+\tfrac12) \sin (n+ \tfrac12) \theta)2 \sin \tfrac12 \theta - (\cos \tfrac12\theta - \cos (n+ \tfrac12) \theta) \cos \tfrac12 \theta} {4\sin^2 \tfrac12\theta} \\ &&&= \frac{-\sin^2 \tfrac12 \theta+(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta-\cos^2\tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{(2n+1)\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos(n+\frac12)\theta \cos\tfrac12 \theta-1}{4 \sin^2 \tfrac12 \theta} \\ &&&= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \\ \Rightarrow && \sum_{k=0}^n k\left ( 1-2\sin^2 \left ( \frac{k\theta}{2} \right) \right) &= \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{4 \sin^2 \tfrac12 \theta} \\ \Rightarrow && \sum_{k=0}^n k\sin^2 \left ( \frac{k\theta}{2} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\theta \sin \tfrac12 \theta+\cos n \theta-1}{8 \sin^2 \tfrac12 \theta} \\ \theta = \frac{\pi}{n}: && \sum_{k=0}^n k\sin^2 \left ( \frac{k\pi}{2n} \right) &= \frac{n(n+1)}{4} - \frac{2n\sin(n+\tfrac12)\frac{\pi}{n}\sin \tfrac12 \frac{\pi}{n}+\cos n \frac{\pi}{n}-1}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n(n+1)}{4} - \frac{-2n \sin^2 \frac{\pi}{2n}-2}{8 \sin^2 \tfrac12 \frac{\pi}{n}} \\ &&&= \frac{n^2+n}{4} + \frac{n}{4} + \frac{1}{4\sin^2\frac{\pi}{2n}} \\ &&&\approx \frac{n^2}4 + \frac{n}{2}+ \frac{n^2}{\pi^2} \\ &&&= \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 + \frac{n}{4} \\ &&&\approx \left ( \frac14 + \frac{1}{\pi^2} \right)n^2 \end{align*}