Tweet
Login
Mathematics Crystal
You may switch between
tex
and
pdf
by changing the end of the URL.
Home
About Us
Materials
Site Map
Questions and Answers
Skills
Topic Notes
HSC
Integration
Others
Tangent
UBC
UNSW
Calculus Advanced
Challenges
Complex Numbers
Conics
Differentiation
Integration
Linear Algebra
Mathematical Induction
Motion
Others
Polynomial Functions
Probability
Sequences and Series
Trigonometry
/
Topics /
Complex Numbers /
Trig Expansion WO.tex
--Quick Links--
The Number Empire
Wolfram Mathematica online integrator
FooPlot
Calc Matthen
Walter Zorn
Quick Math
Lists of integrals
List of integrals of trigonometric functions
PDF
\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Trigonimetrical Function Expansion using Complex Numbers \--- The Working Out} \begin{align*} % % cos Expansions % \cos\text{ Expansions:}\quad&2^n\cos^n\theta=(z+z^{-1})^n=\sum_{r=0}^n C^n_r\cdot (z)^{n-r}\cdot (z^{-1})^r=\sum_{r=0}^n C^n_r\cdot z^{n-2r}\\ % % cos odd % \text{When }&n=2k+1\text{ where $k$ is an integer (i.e. $n$ is odd),}\\ 2^n\cos^n\theta&=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=k+1}^n C^n_r\cdot z^{n-2r}\\ &=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=0}^{n-(k+1)} C^n_{r+(k+1)}\cdot z^{n-2[r+(k+1)]}\\ &=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=0}^k C^n_{(k-r)+(k+1)}\cdot z^{n-2[(k-r)+(k+1)]}\\ &=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=0}^k C^n_{(2k+1)-r}\cdot z^{n-2[(2k+1)-r]}\\ &=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=0}^k C^n_{n-r}\cdot z^{n-2[n-r]}\\ &=\sum_{r=0}^k C^n_r\cdot z^{n-2r}+\sum_{r=0}^k C^n_r\cdot z^{-(n-2r)}\\ &=\sum_{r=0}^k C^n_r\cdot\left(z^{(n-2r)}+z^{-(n-2r)}\right)\\ &=\sum_{r=0}^k C^n_r\cdot 2\cos(n-2r)\theta\\ \therefore 2^n\cos^n\theta&=\sum_{r=0}^k C^n_r\cdot 2\cos(n-2r)\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^5\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)\\ % n=7, k=3:\quad&\cos^7\theta=\frac{1}{2^{7-1}}\sum_{r=0}^3 C^7_r\cdot\cos(7-2r)\theta =\frac{1}{64}\left(\cos 7\theta+7\cos 5\theta+21\cdot\cos 3\theta+35\cos\theta\right)\\ \end{align*} \begin{align*} % % cos even % \text{When }&n=2k\text{ where $k$ is an integer (i.e. $n$ is even),}\\ 2^n\cos^n\theta&=\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+C^n_k\cdot z^{n-2k}+\sum_{r=k+1}^n C^n_r\cdot z^{n-2r}\\ &=\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+C^n_k\cdot z^0+\sum_{r=0}^{n-(k+1)} C^n_{r+(k+1)}\cdot z^{n-2[r+(k+1)]}\\ &=\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+C^n_k+\sum_{r=0}^{k-1} C^n_{(k-1-r)+(k+1)}\cdot z^{n-2[(k-1-r)+(k+1)]}\\ &=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1} C^n_{2k-r}\cdot z^{n-2[2k-r]}\\ &=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1} C^n_{n-r}\cdot z^{n-2[n-r]}\\ &=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1} C^n_r\cdot z^{-(n-2r)}\\ &=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot\left(z^{(n-2r)}+z^{-(n-2r)}\right)\\ &=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot 2\cos(n-2r)\theta\\ \therefore 2^n\cos^n\theta&=C^n_k+\sum_{r=0}^{k-1} C^n_r\cdot 2\cos(n-2r)\theta\\ \\ &\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+\cos 2\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+\cos 4\theta+4\cos 2\theta\right)\\ % n=6, k=3:\quad&\cos^6\theta=\frac{1}{2^{6-1}}\left(\frac{\:1\:}{2}C^6_3+\sum_{r=0}^2 C^6_r\cdot\cos(6-2r)\theta\right) =\frac{1}{32}\left(10+\cos 6\theta+6\cos 4\theta+15\cos 2\theta\right)\\ \end{align*} \begin{align*} % % sin Expansions % \sin\text{ Expansions:}\quad&2^n i^n\sin^n\theta=(z-z^{-1})^n=\sum_{r=0}^n(-1)^r\:C^n_r\cdot (z)^{n-r}\cdot (z^{-1})^r=\sum_{r=0}^n(-1)^r\:C^n_r\cdot z^{n-2r}\\ % % sin odd % \text{When }&n=2k+1\text{ where $k$ is an integer (i.e. $n$ is odd),}\\ 2^n i^n\sin^n\theta&=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=k+1}^n(-1)^r\:C^n_r\cdot z^{n-2r}\\ 2^n\:i^{2k+1}\sin^n\theta&=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^{n-(k+1)}(-1)^{r+(k+1)}\:C^n_{r+(k+1)}\cdot z^{n-2[r+(k+1)]}\\ 2^n(i^2)^k\:i\sin^n\theta&=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^k(-1)^{(k-r)+(k+1)}\:C^n_{(k-r)+(k+1)}\cdot z^{n-2[(k-r)+(k+1)]}\\ 2^n(-1)^k\:i\sin^n\theta&=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^k(-1)^{(2k+1)-r}\:C^n_{(2k+1)-r}\cdot z^{n-2[(2k+1)-r]}\\ &=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^k(-1)^{n-r}\:C^n_{n-r}\cdot z^{n-2[n-r]}\\ &=\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^k(-1)^{n-r}\:C^n_r\cdot z^{-(n-2r)}\\ &=\sum_{r=0}^k(-1)^r\:C^n_r\cdot\left(z^{(n-2r)}+(-1)^{n-2r}\:z^{-(n-2r)}\right)\\ &=\sum_{r=0}^k(-1)^r\:C^n_r\cdot\left(z^{(n-2r)}-z^{-(n-2r)}\right)\\ &=\sum_{r=0}^k(-1)^r\:C^n_r\cdot 2i\sin(n-2r)\theta\\ \therefore 2^n\sin^n\theta&=(-1)^k\sum_{r=0}^k(-1)^r\:C^n_r\cdot 2\sin(n-2r)\theta\\ \\ &\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(-\sin 3\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(\sin 5\theta-5\sin 3\theta+10\sin\theta\right)\\ % n=7, k=3:\quad&\sin^7\theta=\frac{(-1)^3}{2^6}\sum_{r=0}^3(-1)^r\:C^7_r\cdot\sin(7-2r)\theta =\frac{\:1\:}{64}\left(-\sin 7\theta+7\sin 5\theta-21\sin 3\theta+35\sin\theta\right)\\ \end{align*} \begin{align*} % % sin even % \text{When }&n=2k\text{ where $k$ is an integer (i.e. $n$ is even),}\\ 2^n i^n\sin^n\theta&=\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot z^{n-2r}+(-1)^k\:C^n_k\cdot z^{n-2k}+\sum_{r=k+1}^n(-1)^r\:C^n_r\cdot z^{n-2r}\\ 2^n\:i^{2k}\sin^n\theta&=\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot z^{n-2r}+(-1)^k\:C^n_k\cdot z^0+\sum_{r=0}^{n-(k+1)}(-1)^{r+(k+1)}\:C^n_{r+(k+1)}\cdot z^{n-2[r+(k+1)]}\\ 2^n(i^2)^k\sin^n\theta&=\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot z^{n-2r}+(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^{(k-1-r)+(k+1)}\:C^n_{(k-1-r)+(k+1)}\cdot z^{n-2[(k-1-r)+(k+1)]}\\ 2^n(-1)^k\sin^n\theta&=(-1)^{k-1}\:C^n_{k-1}+\sum_{r=0}^k(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1}(-1)^{2k-r}\:C^n_{2k-r}\cdot z^{n-2[2k-r]}\\ &=(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1}(-1)^{n-r}\:C^n_{n-r}\cdot z^{n-2[n-r]}\\ &=(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot z^{n-2r}+\sum_{r=0}^{k-1}(-1)^{n-r}\:C^n_r\cdot z^{-(n-2r)}\\ &=(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot\left(z^{(n-2r)}+(-1)^{n-2r}\:z^{-(n-2r)}\right)\\ &=(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot\left(z^{(n-2r)}+z^{-(n-2r)}\right)\\ &=(-1)^k\:C^n_k+\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot 2\cos(n-2r)\theta\\ \therefore 2^n\sin^n\theta&=C^n_k+(-1)^k\sum_{r=0}^{k-1}(-1)^r\:C^n_r\cdot 2\cos(n-2r)\theta\\ \\ &\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+\cos 4\theta-4\cos 2\theta\right)\\ % n=6, k=3:\quad&\sin^6\theta=\frac{1}{2^{6-1}}\left(\frac{\:1\:}{2}C^6_3+(-1)^3\sum_{r=0}^2(-1)^r\:C^6_r\cdot\cos(6-2r)\theta\right) =\frac{1}{32}\left(10-\cos 6\theta+6\cos 4\theta-15\cos 2\theta\right)\\ \end{align*} \end{document}