Find all values for cos(i)

$$\cos(i)=\frac{e^{i^2}+e^{-i^2}}{2}=\frac{e^{-1}+e}{2}=\frac{1+e^2}{2e}$$

Let $k \in \mathbb{Z}$. We have the relation$$\cos(\theta+2k\pi)=\frac{1}{2}\left[e^{i(\theta+2k\pi)}+e^{-i(\theta+2k\pi)} \right]$$ for all $k$. Now let $\theta=i$. Our expression becomes $$\frac{1}{2}\left[e^{i(i+2k\pi)}+e^{-i(i+2k\pi)} \right]$$ $$=\frac{1}{2}\left[e^{2k\pi i-1}+e^{-2k\pi i+1} \right]$$ $$=\frac{1}{2}\left[\frac{e^{2k\pi i}}{e}+e\cdot e^{-2k\pi i} \right]$$ Now use Euler's formula on each complex exponent inside the brackets. We will get $$\frac{1}{2}\left[\frac{e^{2k\pi i}}{e}+e\cdot e^{-2k\pi i} \right]=\frac{1}{2}\left[\frac{\cos(2k\pi)+i\sin(2k\pi)}{e}+e\cdot (\cos(-2k\pi)+i\sin(-2k\pi) )\right]$$ We know $\sin(\pm2k\pi)=0$ for all $k$ so this expression disappears from inside the large brackets. Similarly, we know $\cos(\pm2k\pi)=1$ for all $k$. Plugging this back in should leave us with a final result of $$\cos(i)=\frac{1}{2}\left[\frac{1+0}{e}+e\cdot (1+0 )\right] = \frac{1}{2} \left[\frac{1}{e}+e \right]=\frac{1+e^2}{2e}$$ Note: It is important that we let $k$ be arbitrary. This establishes there is precisely one value for $\cos(i)$ which was part of your question.