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Problem
Problem
If
∑
i
=
1
2
n
sin
−
1
x
i
=
n
π
\sum_{i=1}^{2n} \sin^{-1}x_i = n\pi
∑
i
=
1
2
n
sin
−
1
x
i
=
nπ
, then find the value
of
∑
i
=
1
2
n
x
i
.
\text{of } \sum_{i=1}^{2n} x_i.
of
i
=
1
∑
2
n
x
i
.
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