still testing

1 files changed, 15 insertions, 4 deletions

diff --git a/Fourier Series.page b/Fourier Series.page
index 751b1c3..984f3e6 100644
--- a/Fourier Series.page
+++ b/Fourier Series.page
@@ -24,11 +24,22 @@ $\qquad\sin(2x) = 2\sin(x)\cos(x)$
 Rearranging,
 $$\begin{array}{ccl} 
 \sin(2x).\cos(x) & = & [2\sin(x)\cos(x)].\cos(x)\\
- & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
- & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
- & = & \cos y+i\sin y\end{array}$$
+ & = & 2 \sin(x) [ 1 - \sin^2(x)]\\
+ & = & 2\sin(x) - 2\sin^3(x)\\
+\end{array}$$
+  
+
+Based on the double angle formula,  
+  
+$\qquad\sin(3x) = 3\sin(x) - 4\sin^3(x)$  
+  
+Rearranging,  
+  
+$\qquad\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$  
+
+Substituting back in the former equation, we get  
+$\sin(2x).\cos(x) = 2 \sin(x) - 2 [\frac{3\sin(x)-\sin(3x)}{4}] $  
   
-$\qquad\sin^2(x) = \frac{1-\cos(2x)}{2}$ 
 ##What is the Fourier series actually?</b>
 
 ##Why is Fourier series useful? </b>