If $\sum_{i=1}^{n} \cos ^{-1}\left(\alpha_{i}\right)=0,$ then find $\sum_{i=1}^{n} \alpha_{i}$
[list=1]
[*] $\frac{n}{2} $
[*] $n $
[*] $n\pi $
[*] $\frac{n\pi}{2} $
[/list]
1. We start with the given equation:
   \[
   \sum_{i=1}^{n} \cos^{-1}(\alpha_{i}) = 0
   \]
2. Recall that the range of the function $\cos^{-1}(x)$ is $[0, \pi]$. This means that for any $x$ in the domain of $\cos^{-1}$, $\cos^{-1}(x) \geq 0$.
3. Since the sum of non-negative terms is zero, each term must individually be zero. Therefore, for each $i$:
   \[
   \cos^{-1}(\alpha_{i}) = 0
   \]
4. The equation $\cos^{-1}(\alpha_{i}) = 0$ implies:
   \[
   \alpha_{i} = \cos(0) = 1
   \]
5. Therefore, each $\alpha_{i} = 1$. Summing these values, we get:
   \[
   \sum_{i=1}^{n} \alpha_{i} = \sum_{i=1}^{n} 1 = n
   \]

The final answer is $\boxed{n}$