Question

If $\sum _{i=1}^n{\cos }^{-1}{\theta }_1=0$, then find the value of $\sum _{i=1}^{n}i-{\theta }_i$.

Answer 1

If $\sum _{i=1}^n{\cos }^{-1}{\alpha }_i$=0

then ${\cos }^{-1}{\alpha }_i$=0 for all $i$

So ${\alpha }_i=1$ because ${\cos }^{-1}1=0$

Now, $\sum _{i=1}^n(i-{\alpha }_i)$

$=\sum _{i=1}^{n}i-\sum _{i=1}^n{\alpha }_i=\frac{n(n+1)}{2}-n$

$=\frac{n(n-1)}{2}$