From 9f6c9f4e37ea38fe043e55adeb4880fead4448af Mon Sep 17 00:00:00 2001
From: siveshs <siveshs@gmail.com>
Date: Fri, 2 Jul 2010 03:45:38 +0000
Subject: still testing

---
 Fourier Series.page | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/Fourier Series.page b/Fourier Series.page
index 4da54cf..61253ee 100644
--- a/Fourier Series.page	
+++ b/Fourier Series.page	
@@ -23,7 +23,7 @@ $\qquad\sin(2x) = 2\sin(x)\cos(x)$
   
 Rearranging,
 $$\begin{array}{ccl} 
-\sin(2x).\cos(x) & = & (2\sin(x)\cos(x))\cos(x)\\
+\sin(2x).\cos(x) & = & [2\sin(x)\cos(x)]\cos(x)\\
  & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
  & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
  & = & \cos y+i\sin y\end{array}$$
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