Problems

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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 \(\frac{\sqrt{7}}{2}\)

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Solution:

1. \begin{align*} \frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\ &= \tan \tfrac{\alpha}{2} \end{align*}
2. \begin{align*} \int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\ &= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\ &= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C \end{align*}

\begin{align*} \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\ &= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\ &= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\ &= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\ &= \frac{\pi-\alpha}{2} \end{align*}

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Solution: First notice that \(y = \cos 2 \theta = 1 - 2\sin^2 \theta = 1- 2x^2\), therefore \(P\) is lies on that parabola.

The arc length is \begin{align*} && s &= \int_{-\pi/2}^{\pi/2} \sqrt{\left ( \frac{\d x}{\d \theta} \right)^2+\left ( \frac{\d y}{\d \theta} \right)^2} \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \sqrt{\cos^2 \theta+16 \sin^2 \theta \cos^2 \theta } \d \theta\\ && &= \int_{-\pi/2}^{\pi/2} \cos \theta\sqrt{1+16 \sin^2 \theta} \d \theta\\ u = \sin \theta, \d u = \cos \theta \d \theta && &= \int_{u=-1}^{u=1} \sqrt{1+16 u^2} \d u\\ 4u = \sinh v, 4\d u = \cosh v: && &= \int_{v=-\sinh^{-1} 4}^{v=\sinh^{-1} 4} \sqrt{1+\sinh^2 v} \tfrac14\cosh v \d v\\ && &= \frac14 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} \cosh^2 v \d v\\ && &= \frac18 \int_{-\sinh^{-1} 4}^{\sinh^{-1} 4} (1 + \cosh 2v) \d v\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \frac12\sinh 2v \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \frac18\left [ \sinh v \sqrt{1 + \sinh^2 v} \right]_{-\sinh^{-1} 4}^{\sinh^{-1} 4}\\ && &= \frac14 \sinh^{-1} 4 + \left (\frac18 \cdot 4 \sqrt{17} \right) - \left (\frac18 \cdot (-4) \sqrt{17} \right)\\ && &= \frac14 \sinh^{-1} 4 + \sqrt{17}\\ \end{align*} The volume of revolution is \begin{align*} && V &=\pi \int_{-1}^1 (2-2x^2)^2 \d x \\ &&&= \pi \left [4x-\frac83x^3+\frac45x^5 \right]_{-1}^1 \\ &&&= \pi \left ( 8-\frac{16}3+\frac85 \right) \\ &&&= \frac{64}{15}\pi \end{align*}

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Solution: \begin{align*} &&\sin \alpha &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \\ &&&= 4 \sin \frac{\alpha}{4} \cos \frac{\alpha}{4} \cos \frac{\alpha}{2} \\ \Rightarrow && \frac{\sin \alpha}{4 \sin \frac{\alpha}{4}} &= \cos \frac{\alpha}{2} \cos \frac{\alpha}{4} \end{align*} We proceed by induction on \(n\). Clearly this is true for \(n = 1\) (as we just established). Assume it is true for \(n=k\). Then: \begin{align*} && \frac{\sin \alpha}{2^n \sin \frac{\alpha}{2^n}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} \cos \frac{\alpha}{2^{n+1}}} &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right) \\ \Rightarrow && \frac{\sin \alpha}{2^{n+1} \sin \frac{\alpha}{2^{n+1}} } &= \cos\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{4}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)\cos \left ( \frac{\alpha}{2^{n+1}} \right) \\ \end{align*} Therefore it is true for \(n=k+1\) Therefore since it is true for \(n=1\) and if it is true for \(n=k\) it is also true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) \begin{align*} \lim_{n \to \infty} \frac{\sin \alpha}{\cos\left(\frac{\alpha}{2}\right)\cdots\cos\left(\frac{\alpha}{2^{n}}\right)} &= \lim_{n \to \infty} 2^n \sin \frac{\alpha}{2^n} \\ &= \lim_{n \to \infty} \alpha \frac{\sin \frac{\alpha}{2^n}}{\frac{\alpha}{2^n}} \\ &= \alpha \lim_{t \to 0} \frac{\sin t}{t} \\ &= \alpha \end{align*} When \(\alpha = \frac{\pi}{2}\) notice that \(\sin \alpha =1\), \(\cos \frac{\alpha}{2} = \sqrt{\frac12}\) and \(2\cos^2 \frac{\alpha}{2^{n+1}}-1 = \cos \frac{\alpha}{2} \Rightarrow \cos \frac{\alpha}{2^{n+1}} = \sqrt{\frac12 + \cos \frac{\alpha}{2^n}}\) exactly the series we see.

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Solution:

We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

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Solution: Suppose \(z^7 = 1\), then we can write \(z = \cos \theta + i \sin \theta\) and we must have that: \begin{align*} 0 &= \textrm{Im}((\cos \theta + i \sin \theta)^7) \\ &= \binom{7}{6}\cos^6 \theta \sin \theta - \binom{7}{4} \cos^4 \theta \sin^3 \theta + \binom{7}{2} \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= 7 \cos^6 \sin \theta - 35 \cos^4 \theta \sin ^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= -\cos^7 \theta \l \tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7 \tan \theta\r \\ &= \cos^7 \theta \cdot t (t^7-21t^4+35t^2-7) \end{align*} Where \(t = \tan \theta\). So if \(z\) is a root of \(z^7 = 1\) and \(\cos \theta \neq 0, \tan \theta \neq 0\) then \(t\) is a root of the equation. Thererefore the roots are: \(\tan \frac{2\pi k}{7}\) where \(k = 1, 2, \ldots 6\). Noting that \(\tan \frac{\pi}7 = -\tan \frac{6\pi}{7}, \tan \frac{3\pi}{7} = -\tan \frac{4 \pi}{7}, \tan \frac{5\pi}{7} = -\tan \frac{2 \pi}{7}\) we can conclude that: \begin{align*} && 7 &= \prod_{k=1}^k \tan \frac{k \pi}{6} \\ &&&= \l \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \r^2 \\ \Rightarrow&& \pm \sqrt{7} &= \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \end{align*} However, we know that \(\tan \frac{2\pi}{7}\) is positive, \(\tan \frac{4\pi}{7},\tan \frac{6\pi}{7}\) are negative, therefore the result must be positive, ie \(+\sqrt{7}\) Using a similar method, we notice that: \begin{align*} 0 &= \textrm{Im} \l (\cos \theta + i \sin \theta)^n \r \\ &= \cos^n \theta \cdot t (t^{n-1} + \cdots - \binom{n}{n-1}) \end{align*} Therefore \(\prod_{k=0}^{n-1} \tan \frac{k \pi}{n} = n\) and since \(\tan \frac{(2k+1) \pi}{n} = \tan \frac{(n-2k-1)\pi}{n}\) is a map of all the odd numbers to the even numbers (and vice versa) when \(n\) is odd. We also know that the terms less where \(\tan \theta\) has \(\theta < \frac{\pi}{2}\) are positive, and the others even, we can determine the signs: \begin{align*} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} \tan \frac{6 \pi}{9} \tan \frac{8 \pi}{9} & = 3 \\ \tan \frac{2 \pi}{11} \tan \frac{4 \pi}{11} \tan \frac{6 \pi}{11} \tan \frac{8 \pi}{11} \tan \frac{10 \pi}{11} &= -\sqrt{11} \end{align*}