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authorsiveshs <siveshs@gmail.com>2010-07-02 13:21:36 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-02 13:21:36 +0000
commit7f63506972de44746eb875bdb6e90c7648232be6 (patch)
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@@ -40,7 +40,7 @@ $\qquad\sin^3(x) = \frac{3\sin(x)-\sin(3x)}{4}$
Substituting back in the former equation, we get
$$
-\begin{array}
+\begin{array} {ccl}
\sin(2x) & = & 2\sin(x) - 2 [\frac{3\sin(x)-\sin(3x)}{4}]\\
& = & \frac{1}{2}\sin(x) + \frac{1}{2}\sin(3x)\\
\end{array}