Let sin-1i=x
then sinx=i
If x belongs to complex set then I think it is defined.
We know
sinx=(eix-e-ix)/2i= -i(eix-e-ix)/2
Suppose (eix-e-ix)/2= -1 and put eix=a
Then we get a2+2a-1=0
or a= -1(+or-)(
2)Consider a=((
2)-1)
e
ix=((
2)-1)
Taking log to the base e on both sides
we get x=[ln((
2)-1)]/i
If you are satisfied then I'm waiting for a salute!!!
Cheers!!!
Moderation Team
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