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Trig Expansion.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} \noindent{\large Trigonimetrical Function Expansion using Complex Numbers}\\ \\ Useful in integration by expanding $\sin^n\theta$ or $\cos^n\theta$ into a series of $\sin k\theta$ or $\cos k\theta$ . \begin{align*} \text{In this context, }&z=\cos\theta+i\sin\theta\:,\quad\therefore|z|=|z^n|=|z^{-1}|=|z^{-n}|=1\text{ , where $n\in\mathbb{R+}$}\\ &z^n=\cos n\theta+i\sin n\theta\:,\qquad\:\: z^{-n}=\cos n\theta-i\sin n\theta\\ &z^n+z^{-n}=2\cos n\theta\:,\qquad\:\:\:\:\: z^n-z^{-n}=2i\sin n\theta\\ &(z+z^{-1})^n=2^n\cos^n\theta\:,\qquad (z-z^{-1})^n=2^n i^n\sin^n\theta\\ % % &\boxed{\cos^n\theta=\frac{1}{2^{n-1}}\sum_{r=0}^k C^n_r\cdot\cos(n-2r)\theta\:,\quad\text{where $n=2k+1$ .}}\\ \text{e.g. } n=3, k=1:\quad&\cos^3\theta=\frac{1}{2^{3-1}}\sum_{r=0}^1 C^3_r\cdot\cos(3-2r)\theta =\frac{1}{4}\left(\cos 3\theta+3\cos\theta\right)\:,\quad \cos 3\theta=4\cos^3\theta-3\cos\theta\\ % n=5, k=2:\quad&\cos^3\theta=\frac{1}{2^{5-1}}\sum_{r=0}^2 C^5_r\cdot\cos(5-2r)\theta =\frac{1}{16}\left(\cos 5\theta+5\cos 3\theta+10\cos\theta\right)\\ % % &\boxed{\cos^n\theta=\frac{1}{2^{n-1}}\left(\frac{\:1\:}{2}C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot\cos(n-2r)\theta\right)\:,\quad\text{where $n=2k$ .}}\\ \text{e.g. } n=2, k=1:\quad&\cos^2\theta=\frac{1}{2^{2-1}}\left(\frac{\:1\:}{2}C^2_1+\sum_{r=0}^0 C^2_r\cdot\cos(2-2r)\theta\right) =\frac{\:1\:}{2}\left(1+\cos2\theta\right)\:,\quad \cos 2\theta=2\cos^2\theta-1\\ % n=4, k=2:\quad&\cos^4\theta=\frac{1}{2^{4-1}}\left(\frac{\:1\:}{2}C^4_2+\sum_{r=0}^1 C^4_r\cdot\cos(4-2r)\theta\right) =\frac{\:1\:}{8}\left(3+\cos4\theta+4\cos2\theta\right)\\ % % &\boxed{\sin^n\theta=\frac{(-1)^k}{2^{n-1}}\sum_{r=0}^k(-1)^r\:C^n_r\cdot\sin(n-2r)\theta\:,\quad\text{where $n=2k+1$ .}}\\ \text{e.g. } n=3, k=1:\quad&\sin^3\theta=\frac{(-1)^1}{2^{3-1}}\sum_{r=0}^1(-1)^r\:C^3_r\cdot\sin(3-2r)\theta =\frac{-1}{4}\left(\sin3\theta-3\sin\theta\right)\:,\quad \sin 3\theta=3\sin\theta-4\sin^3\theta\\ % n=5, k=2:\quad&\sin^5\theta=\frac{(-1)^2}{2^{5-1}}\sum_{r=0}^2(-1)^r\:C^5_r\cdot\sin(5-2r)\theta =\frac{1}{16}\left(\sin5\theta-5\sin3\theta+10\sin\theta\right)\\ % % &\boxed{\sin^n\theta=\frac{1}{2^{n-1}}\left(\frac{\:1\:}{2}C^n_k+(-1)^k\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot\cos(n-2r)\theta\right)\:,\quad\text{where $n=2k+1$ .}}\\ \text{e.g. } n=2, k=1:\quad&\sin^2\theta=\frac{1}{2^{2-1}}\left(\frac{\:1\:}{2}C^2_1+(-1)^1\sum_{r=0}^0(-1)^r\:C^2_r\cdot\cos(2-2r)\theta\right) =\frac{\:1\:}{2}\left(1-\cos2\theta\right)\:,\quad \cos 2\theta=1-2\sin^2\theta\\ % n=4, k=2:\quad&\sin^4\theta=\frac{1}{2^{4-1}}\left(\frac{\:1\:}{2}C^4_2+(-1)^2\sum_{r=0}^1(-1)^r\:C^4_r\cdot\cos(4-2r)\theta\right) =\frac{\:1\:}{8}\left(3+\cos4\theta-4\cos2\theta\right)\\ \\ \text{Ways to remember:}\quad\text{1. }&\text{\small Coefficents of terms inside the brackets are taken from the first half of the Pascal's Triangle.}\\ &\text{(1, 1-3, 1-4, 1-5-10, 1-6-15, 1-7-21-35, 1-8-28-56, \ldots)}\\ \text{2. }&\text{\small For cos, all coefficients are positive; for sin, the first term is positive if $n$ or $n-1$ is divisible by 4, or negative}\\ &\text{\small otherwise. The signs of the rest of the terms alternate.}\\ \text{3. }&\text{\small Coefficient of $\theta$ starts from $n$ and decreases by 2 each term on. (2, 3-1, 4-2, 5-3-1, 6-4-2, 7-5-3-1, 8-6-4-2, \ldots)}\\ \text{4. }&\text{\small When $n$ is even (i.e. odd Pascal's numbers), divide the middle number by 2 and add to the sequence.}\\ &\text{(2$\div$2, 6$\div$2, 20$\div$2, 70$\div$2, \ldots)}\\ \text{5. }&\text{\small The expansion of $\sin^n \theta$ with an even $n$ contains a series of $\cos k\theta$ (instead of $\sin k\theta$ as one would expect).}\\ \end{align*} \end{document}