# $$ \sum_{i=1}^{89}{sin(i°)}=\frac{\sqrt{90}}{2^{89}}$$
###### tags: `三角` `複數`

## 解一

## 解二
$$ \begin{align}
&sin(x)sin(60°-x)sin(60°+x)\\
=& \frac{1}{2}(cos(60°)+cos(2x-60°))sin(60°+x)\\
=& \frac{1}{2}(\frac{1}{2}sin(60°+x)+sin(60°+x)cos(2x-60°))\\
=& \frac{1}{4}(sin(60°+x)+sin(3x)+sin(120°-x))\\
=& \frac{1}{4}sin(3x)
\end{align}
$$
$$
\begin{align}
&\sum_{i=1}^{89}{sin(i°)}\\
=&\sum_{i=1}^{29}\left(\frac{1}{4}sin(3i°)\right)sin(30°)sin(60°)\\
=&\frac{\sqrt{3}}{2^{60}}\sum_{i=1}^{29}\left(sin(3i°)\right)\\
=&\frac{\sqrt{3}}{2^{60}}\sum_{i=1}^{9}\left(\frac{1}{4}sin(9i°)\right)sin(30°)sin(60°)\\
=&\frac{\sqrt{9}}{2^{80}}\sum_{i=1}^{9}\left(sin(9i°)\right)\\
=&\frac{\sqrt{9}}{2^{84}}\sum_{i=1}^{4}\left(sin(18i°)\right)sin(45°)\\
=&\frac{\sqrt{18}}{2^{87}}\left(sin(36°)+sin(72°)\right)\\
\end{align}
$$