How can you use trigonometric functions to simplify $9 e^( ( 3 pi)/8 i )$ into a non-exponential complex number?

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Answer 1 · 2016-01-23T16:26:38

$9 [ cos((3pi)/8) + isin((3pi)/8 ) ]$

Euler's theorem states : $e^(itheta) = costheta + isintheta$

in this question $theta =(3pi)/8$

$rArr = 9[ cos((3pi)/8) + isin((3pi)/8) ]$

How can you use trigonometric functions to simplify 9 e^( ( 3 pi)/8 i ) 9 e^( ( 3 pi)/8 i ) into a non-exponential complex number?