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authorsiveshs <siveshs@gmail.com>2010-07-02 04:00:43 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-02 04:00:43 +0000
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@@ -38,7 +38,7 @@ 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}] $
+$\sin(2x).\cos(x) = 2 \sin(x) - 2 [\frac{3\sin(x)-\sin(3x)}{4}]$
##What is the Fourier series actually?</b>