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Mathematics
Problem
∑
n
=
1
5
sin
−
1
(
sin
(
2
n
−
1
)
)
\sum_{n=1}^{5} \sin^{-1}(\sin(2n-1))
∑
n
=
1
5
sin
−
1
(
sin
(
2
n
−
1
))
is a. 1 b. 2 c. 3 d. 4
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