How can you use trigonometric functions to simplify $9 e^( ( 17 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2018-07-26T16:50:18

The answer is $=9/4((-sqrt6+sqrt2)-i(sqrt2+sqrt6))$

The Euler's Identity is

$e^(itheta)=costheta+isintheta$

Therefore,

$z=9e^(17/12pii)$

$=9cos(17/12pi)+isin(17/12pi)$

$cos(17/12pi)=cos(7/6pi+1/4pi)$

$=cos(7/6pi)cos(1/4pi)-sin(7/6pi)sin(1/4pi)$

$=-sqrt3/2*sqrt2/2+1/2*sqrt2/2$

$=1/4(-sqrt6+sqrt2)$

$sin(17/12pi)=sin(7/6pi+1/4pi)$

$=sin(7/6pi)cos(1/4pi)+cos(7/6pi)sin(1/4pi)$

$=-1/2*sqrt2/2-sqrt3/2*sqrt2/2$

$=-1/4(sqrt2+sqrt6)$

Finally,

$z=9/4((-sqrt6+sqrt2)-i(sqrt2+sqrt6))$

How can you use trigonometric functions to simplify 9 e^( ( 17 pi)/12 i ) 9 e^( ( 17 pi)/12 i ) into a non-exponential complex number?