I'm a non-mathematician trying to learn geometric algebra by working my way (slowly) through "Imaginary Numbers are not Real" by Gull, Lasenby and Doran [Found. Phys. 23(9), 1175-1201 (1993)].
So far so good and I can prove Euler's formula with GA in two dimensions - but I'm struggling to prove the following identity (Section 3, eq.13) in the three dimensional case:
$$ e^{-i\,(\tfrac{a}{2})}=\cos(\tfrac{\lvert a\rvert}{2})-i \tfrac{a}{\lvert a \rvert}\sin(\tfrac{\lvert a\rvert}{2}) \phantom{tab} $$
(where $a$ is a vector and $i$ is the pseudoscalar).
My current attempt is to define an orthonormal basis $\{\sigma_1,\sigma_2,\sigma_3\}$, put $i=\sigma_1\sigma_2\sigma_3$ and let $a=(e_1\sigma_1+e_2\sigma_2+e_3\sigma_3)$; from which I obtain:
$$ \begin{align} e^{-i\,(\tfrac{a}{2})}&=e^{-\sigma_2\sigma_3\tfrac{e_1}{2}}e^{-\sigma_3\sigma_1\tfrac{e_2}{2}}e^{-\sigma_1\sigma_2\tfrac{e_3}{2}}\\ &= (\cos\tfrac{e_1}{2}-\sigma_2\sigma_3\sin\tfrac{e_1}{2})\, (\cos\tfrac{e_2}{2}-\sigma_3\sigma_1\sin\tfrac{e_2}{2})\, (\cos\tfrac{e_3}{2}-\sigma_1\sigma_2\sin\tfrac{e_3}{2}) \end{align} $$ ... but trying to expand that I'm just getting bogged down in endless trig identities. Even looking at just the scalar part I'm not seeing anything much like $\cos(\tfrac{\lvert a \rvert}{2})$.
Is there a simpler approach that I'm missing?
I'd start by assuming the Taylor series representation of the exponential $\exp(x) = \sum_{k = 0}^\infty \frac{x^k}{k!}.$ Since $a$ and $i$ commute
$\begin{aligned}\exp(-i a/2) &= \sum_{k = 0}^\infty \frac{i^k (-a/2)^k}{k!} \\ &= \sum_{\mbox{$k$ even}} \frac{i^k (-a/2)^k}{k!}+ \sum_{\mbox{$k$ odd}} \frac{i^k (-a/2)^k}{k!} \\ &= \sum_{r = 0}^\infty \frac{i^{2r} (-a/2)^{2r}}{(2r)!}+ \sum_{r = 0}^\infty \frac{i^{2r + 1} (-a/2)^{2r + 1}}{(2r + 1)!} \\ &= \sum_{r = 0}^\infty \frac{(-1)^{r}(-a/2)^{2r}}{(2r)!}+ i \sum_{r = 0}^\infty \frac{(-1)^{r} (-a/2)^{2r + 1}}{(2r + 1)!} \\ \end{aligned}$
The even sum has the structure of a cosine, and is in fact a scalar, since $a^{2k} = \left\lvert {a} \right\rvert^{2k}$.
The odd sum has the structure of a sine, but it is a vector. We may express this sum as a vector times scalar sum, by noting that $a^{2k+1} = \left\lvert {a} \right\rvert^{2k} a = \left\lvert {a} \right\rvert^{2k+1} a / \left\lvert {a} \right\rvert.$
This leaves us with
$\begin{aligned}\exp(-i a/2) &= \sum_{r = 0}^\infty \frac{(-1)^{r}(-\left\lvert {a} \right\rvert/2)^{2r}}{(2r)!}+ i \frac{a}{\left\lvert {a} \right\rvert}\sum_{r = 0}^\infty \frac{(-1)^{r} (-\left\lvert {a} \right\rvert/2)^{2r + 1}}{(2r + 1)!} \\ &= \cos\left( {-1\left\lvert {a} \right\rvert/2} \right) + i \frac{a}{\left\lvert {a} \right\rvert} \sin\left( { -\left\lvert {a} \right\rvert/2 } \right) \\ &= \cos\left( {\left\lvert {a} \right\rvert/2} \right) - i \frac{a}{\left\lvert {a} \right\rvert} \sin\left( { \left\lvert {a} \right\rvert/2 } \right) \\ \end{aligned}$