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authorsiveshs <siveshs@gmail.com>2010-07-02 03:45:38 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-02 03:45:38 +0000
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@@ -23,7 +23,7 @@ $\qquad\sin(2x) = 2\sin(x)\cos(x)$
Rearranging,
$$\begin{array}{ccl}
-\sin(2x).\cos(x) & = & (2\sin(x)\cos(x))\cos(x)\\
+\sin(2x).\cos(x) & = & [2\sin(x)\cos(x)]\cos(x)\\
& = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
& = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
& = & \cos y+i\sin y\end{array}$$