How do you multiply $e^(( 5 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-03-13T16:30:01

The answer is $=1/2((sqrt(2+sqrt2))-i(sqrt(2-sqrt2)))$

Apply Euler's Identity

$e^(itheta)=costheta+isintheta$

$e^(5/8pii)=cos(5/8pi)+isin(5/8pi)$

$e^(3/2pii)=cos(3/2pi)+isin(3/2pi)$

$cos2theta=2cos^2theta-1=1-2sin^2theta$

$costheta=sqrt((1+cos2theta)/2)$

$sintheta=sqrt((1-sin2theta)/2)$

$cos(5/8pi)=sqrt((1+cos(10/8pi)/2))$

$cos(10/8pi)=cos(5/4pi)=cos(pi+1/4pi)$

$=cospicos(1/4pi)-sin(pi)sin(1/4pi)$

$=-1*sqrt2/2-0$

$=-sqrt2/2$

$cos(5/8pi)=sqrt((1-sqrt2/2)/2)=(sqrt(2-sqrt2))/(2)$

$sin(5/8pi)=sqrt((1-sin(10/8pi)/2)$

$sin(10/8pi)=sin(5/4pi)=2*-sqrt2/2=-sqrt2$

$sin(5/8pi)=sqrt((1+sqrt2/2)/2)=1/2sqrt(2+sqrt2)$

$e^(3/2pii)=cos(3/2pi)+isin(3/2pi)=0-i$

And finally,

$e^(5/8pii)*e^(3/2pii)=((sqrt(2-sqrt2))/(2)+i1/2sqrt(2+sqrt2))*(-i)$

$=1/2(sqrt(2+sqrt2))-i1/2(sqrt(2-sqrt2))$

How do you multiply e^(( 5 pi )/ 8 i) * e^( 3 pi/2 i ) e^(( 5 pi )/ 8 i) * e^( 3 pi/2 i ) in trigonometric form?