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Trig Multi-angle.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Trigonimetrical Multi-angle Expansion using Complex Numbers} \begin{align*} \text{In this context, }&z=\cos\theta+i\sin\theta\:,\quad\therefore z^n=\cos n\theta+i\sin n\theta\text{ , where $n\in\mathbb{R+}$}\\ z^n&=\left(\cos\theta+i\sin\theta\right)^n=\sum_{r=0}^n C^n_r\cos^{n-r}\theta\cdot i^r\sin^r\theta\\ % \text{Ways to remember }&\text{the following formulae:}\\ \text{1. }&\text{\small For cos, coefficents are the \it odd \rm terms of the Pascal's Triangle; for sin, they are the \it even \rm ones.}\\ &\text{(cos: 1-1, 1-3, 1-6-1, 1-10-5, 1-15-15-1, 1-21-35-7, \ldots; sin: 2, 3-1, 4-4, 5-10-1, 6-20-6, 7-35-21-1, \ldots)}\\ \text{2. }&\text{\small The odd terms are positive; the even terms are negative. ($+ - + - \ldots$)}\\ \text{3. }&\text{\small For cos, first term is $\cos^n\theta$ (or $\cos^n\theta\:\sin^0\theta$); for sin, first term is $\cos^{n-1}\theta\:\sin\theta$.}\\ \text{4. }&\text{\small After the first term, the index of cos decreases by 2 and that of sin increases by 2 each term on.}\\ \text{5. }&\text{\small For tan, it is a ratio of $\tfrac{\sin n\theta}{\cos n\theta}$ with a factor of $\cos^n\theta$ taken out from each term to make it $\tan^k\theta$.}\\ % % % Odd % % \text{When $n$ is odd:}\quad\\ &\boxed{\cos n\theta=\sum_{s=0}^k(-1)^s C^n_{2s}\cos^{n-2s}\theta\cdot\sin^{2s}\theta\:,\quad\text{where $n=2k+1$ .}}\\ % \text{e.g. } n=3, k=1:\quad&\cos 3\theta=\sum_{s=0}^1(-1)^s C^3_{2s}\cos^{3-2s}\theta\cdot\sin^{2s}\theta =\cos^3\theta-3\cos\theta\sin^2\theta\\ &=\cos^3\theta-3\cos\theta(1-\cos^2\theta) =4\cos^3\theta-3\cos\theta\\ % n=5, k=2:\quad&\cos 5\theta=\sum_{s=0}^2(-1)^s C^5_{2s}\cos^{5-2s}\theta\cdot\sin^{2s}\theta =\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta\\ &=\cos^5\theta-10\cos^3\theta(1-\cos^2\theta)+5\cos\theta(1-\cos^2\theta)^2 =16\cos^5\theta-20\cos^3\theta+5\cos\theta\\ % \\ &\boxed{\sin n\theta=\sum_{s=0}^k(-1)^s C^n_{2s+1}\cos^{n-(2s+1)}\theta\cdot\sin^{2s+1}\theta\:,\quad\text{where $n=2k+1$ .}}\\ % \text{e.g. } n=3, k=1:\quad&\sin 3\theta=\sum_{s=0}^1(-1)^s C^3_{2s+1}\cos^{3-(2s+1)}\theta\cdot\sin^{2s+1}\theta =3\cos^2\theta\sin\theta-\sin^3\theta\\ &=3(1-\sin^2\theta)\sin\theta-\sin^3\theta =3\sin\theta-4\sin^3\theta\\ % n=5, k=2:\quad&\sin 5\theta=\sum_{s=0}^2(-1)^s C^5_{2s+1}\cos^{5-(2s+1)}\theta\cdot\sin^{2s+1}\theta =5\cos^4\theta\cdot\sin\theta-10\cos^2\theta\cdot\sin^3\theta+\sin^5\theta\\ &=5(1-\sin^2\theta)^2\sin\theta-10(1-\sin^2\theta)\sin^3\theta+\sin^5\theta =16\sin^5\theta-20\sin^3\theta+5\sin\theta\\ \\ &\boxed{\tan n\theta=\frac {\sum_{s=0}^k(-1)^s C^n_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^k(-1)^s C^n_{2s}\tan^{2s}\theta} \:,\quad\text{where $n=2k+1$ .}}\\ % \text{e.g. } n=3, k=1:\quad&\tan 3\theta =\frac {\sum_{s=0}^1(-1)^s C^3_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^1(-1)^s C^3_{2s}\tan^{2s}\theta} =\frac {3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}\\ % n=5, k=2:\quad&\tan 5\theta =\frac {\sum_{s=0}^2(-1)^s C^5_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^2(-1)^s C^5_{2s}\tan^{2s}\theta} =\frac {5\tan\theta-10\tan^3\theta+\tan^5\theta} {1-10\tan^2\theta+5\tan^4\theta}\\ \end{align*} \begin{align*} % % % Even % % \text{When $n$ is even:}\quad\\ &\boxed{\cos n\theta=\sum_{s=0}^k(-1)^s C^n_{2s}\cos^{n-2s}\theta\cdot\sin^{2s}\theta\:,\quad\text{where $n=2k$ .}}\\ % \text{e.g. } n=2, k=1:\quad&\cos 2\theta=\sum_{s=0}^1(-1)^s C^2_{2s}\cos^{2-2s}\theta\cdot\sin^{2s}\theta =\cos^2\theta-\sin^2\theta\\ % n=4, k=2:\quad&\cos 4\theta=\sum_{s=0}^2(-1)^s C^4_{2s}\cos^{4-2s}\theta\cdot\sin^{2s}\theta =\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ \\ &\boxed{\sin n\theta=\sum_{s=0}^{k-1}(-1)^s C^n_{2s+1}\cos^{n-(2s+1)}\theta\cdot\sin^{2s+1}\theta\:,\quad\text{where $n=2k$ .}}\\ % \text{e.g. } n=2, k=1:\quad&\sin 2\theta=\sum_{s=0}^0(-1)^s C^2_{2s+1}\cos^{2-(2s+1)}\theta\cdot\sin^{2s+1}\theta =2\cos\theta\sin\theta\\ % n=4, k=2:\quad&\sin 4\theta=\sum_{s=0}^1(-1)^s C^4_{2s+1}\cos^{4-(2s+1)}\theta\cdot\sin^{2s+1}\theta =4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\\ &=4\cos\theta\sin\theta\:(\cos^2\theta-\sin^2\theta)\qquad (=2\cdot\sin 2\theta\cdot\cos 2\theta)\\ \\ \therefore\quad&\boxed{\tan n\theta=\frac {\sum_{s=0}^{k-1}(-1)^s C^n_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^k(-1)^s C^n_{2s}\tan^{2s}\theta} \:,\quad\text{where $n=2k+1$ .}}\\ % \text{e.g. } n=2, k=1:\quad&\tan 2\theta =\frac {\sum_{s=0}^0(-1)^s C^2_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^1(-1)^s C^2_{2s}\tan^{2s}\theta} =\frac {2\tan\theta} {1-\tan^2\theta}\\ % n=4, k=2:\quad&\tan 4\theta =\frac {\sum_{s=0}^1(-1)^s C^4_{2s+1}\tan^{2s+1}\theta} {\sum_{s=0}^2(-1)^s C^4_{2s}\tan^{2s}\theta} =\frac {4\tan\theta-4\tan^3\theta} {1-6\tan^2\theta+\tan^4\theta}\\ \\ \text{An Interesting Example:}\quad&\text{Equation }x^4+4x^3-6x^2-4x+1=0\text{ can be expressed as:}\\ &x^4-6x^2+1=4x-4x^3\:,\qquad\text{i.e. }1=\frac{4x-4x^3}{1-6x^2+x^4}\\ &\text{Let }x=\tan\theta,\quad\text{then }RHS=\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}=\tan 4\theta\\ &\therefore\quad\tan 4\theta=1\:,\qquad 4\theta=k\pi+\tfrac{\pi}{4}\quad\text{where }k=0,1,2,3\\ &\theta=\left(4k+1\right)\tfrac{\pi}{16}=\tfrac{\pi}{16},\:\tfrac{5\pi}{16},\:\tfrac{9\pi}{16},\:\tfrac{13\pi}{16}\\ \\ &x=\tan^{-1}\left(\tfrac{\pi}{16}\right),\:\tan^{-1}\left(\tfrac{5\pi}{16}\right),\:\tan^{-1}\left(\tfrac{9\pi}{16}\right),\:\tan^{-1}\left(\tfrac{13\pi}{16}\right)\\ \end{align*} \end{document}