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Introduction to ITF
Question
The principal value of
S
i
n
−
1
S
i
n
2
π
3
is
Easy
A
2
π
3
B
π
3
C
−
π
3
D
−
2
π
3
Solution
S
i
n
-
1
S
i
n
2
π
3
=
S
i
n
−
1
S
i
n
π
−
π
3
=
S
i
n
−
1
S
in
π
3
=
π
3
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