Exponential Sine Identity in Geometric Algebra Given
$$e^{\textbf{i}\theta}=cos(\theta) + \textbf{i}sin(\theta) \ \ \ \ (1)$$
and
$$e^{-\textbf{i}\theta}=cos(\theta) - \textbf{i}sin(\theta) \ \ \ \ (2)$$
To find $sin(\theta)$ first you should perform:
$$\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=\textbf{i}sin(\theta) \ \ \ \ (3)$$
The seemingly obvious thing to do would then be to just "divide" the $\bf{i}$ over to get $\frac{1}{\textbf{i}}\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2\textbf{i}}$ (which is the answer I'm working toward). But that introduces a scalar (1) dividing a bivector ($\textbf{i}$). How can you just divide a scalar by a bivector?
UPDATE
Despite the help given here in the comments, I've managed to find an answer. It is a simple answer:
Following $(3)$, the next step would be to product both sides of the equation by $-\textbf{i}$ on the left giving
$$-\textbf{i}\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2}=sin(\theta) \ \ \ \ (4)$$
Then all there is to note is that $\frac{1}{\textbf{i}}$ is simply an abbreviation for $-\textbf{i}(=\textbf{i}^{-1})$. With that in mind, $(4)$ becomes
$$\frac{e^{\textbf{i}\theta}-e^{-\textbf{i}\theta}}{2\textbf{i}}=sin(\theta) \ \ \ \ (5)$$
 A: $\mathbf{B}^{-1}$ for bivector $\mathbf{B}$, is just a bivector, and differs from $\mathbf{B}$ by a scalar factor.  When $\mathbf{i}$ is a unit bivector satisfying $\mathbf{i}^2 = -1$, you can see by inspection that $\mathbf{i}^{-1} = -\mathbf{i}$.
To gain a bit more insight into why this works, consider the fact that you can always factor a bivector $ \mathbf{i} $ into a product of orthogonal vectors $ \mathbf{u} \mathbf{v} $, (at least in 2-3 dimensions), after which you can invert the bivector as follows
$$   \mathbf{i}^{-1} = \frac{1}{\mathbf{v}} \frac{1}{\mathbf{u}} = \frac{\mathbf{v} \mathbf{u}}{\mathbf{u}^2 \mathbf{v}^2}.$$
The simplest examples are those where the bivector is some scalar multiple of two clearly orthonormal factors, such as 
$$(\mathbf{e}_1 \mathbf{e}_2)^{-1} = \frac{1}{{\mathbf{e}_2}} \frac{1}{{\mathbf{e}_1}} = \mathbf{e}_2 \mathbf{e}_1,$$
A less trivial example is $ \mathbf{i} = \frac{1}{{\sqrt{3}}}\left( { \mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_2 \mathbf{e}_3 + \mathbf{e}_3 \mathbf{e}_1 } \right) $, where possible factorizations include
$$\begin{aligned}   \mathbf{i}    &= \frac{1}{{\sqrt{3}}}    \left( { \mathbf{e}_1 + \mathbf{e}_2 - 2 \mathbf{e}_3 } \right) \frac{ \mathbf{e}_2 - \mathbf{e}_1 }{2}  \\ &= \frac{1}{{\sqrt{3}}}    \frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2} \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right),\end{aligned}$$
but once you find such factors, you can easily compute $\mathbf{i}^{-1} = -\mathbf{i}$ using these factors, for example:
$$\begin{aligned}\mathbf{i}^{-1} &= 2 \sqrt{3} \frac{ 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 }{6} \frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2 } \\ &= \frac{1}{{2 \sqrt{3}}} \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right) \left( { \mathbf{e}_3 - \mathbf{e}_2 } \right) \\ &= \frac{1}{{\sqrt{3}}} \left( { \mathbf{e}_1 \mathbf{e}_3 + \mathbf{e}_2 \mathbf{e}_1 + \mathbf{e}_3 \mathbf{e}_2 } \right) \\ &= -\mathbf{i}.\end{aligned}$$
