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authorsiveshs <siveshs@gmail.com>2010-07-02 18:45:36 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-02 18:45:36 +0000
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@@ -31,11 +31,11 @@ $$\begin{array}{ccl}
Based on the double angle formula,
-$\qquad\sin(3x) = 3\sin(x) - 4\sin^3(x)$
+$$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
Rearranging,
-$\qquad\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$
+$$\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$$
Substituting back in the former equation, we get