If is in degree then find the value of .
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S i n ( 1 ) = 2 1 i e − 1 8 0 i π − 2 1 i e 1 8 0 i π
⇒ 2 i 1 ( − ( e − 2 i π ) 9 0 1 + ( e 2 i π ) 9 0 1 )
⇒ 2 i 1 ( − ( − i ) 9 0 1 + ( i ) 9 0 1 )
⇒ 2 i 1 ( − ( 0 . 9 9 9 8 4 . . . − 0 . 0 1 7 4 5 . . . i ) + ( 0 . 9 9 9 8 4 . . . − 0 . 0 1 7 4 5 . . . i ) ) ≈ 0 . 0 1 7 4 5 . . .
Problem is that when taking the n th root of a complex number, there are n possible answers, not just one.