How do you evaluate $e^( ( 7 pi)/6 i) - e^( ( 17 pi)/8 i)$ using trigonometric functions?

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Answer 1 · 2017-07-22T15:02:51

The answer is $=-sqrt3/2-sqrt(2+sqrt2)/2-i(1/2+sqrt(2-sqrt2)/2)$

We need

$cos2theta=2cos^2theta-1$ , $=>$ , $costheta=sqrt((1+cos2theta)/2)$

$cos2theta=1-2sin^2theta$ , $=>$ , $sintheta=sqrt((1-cos2theta)/2)$

We apply Euler's relation

$e^(itheta)=costheta+isintheta$

$e^(7/6pii)=cos(7/6pi)+isin(7/6pi)$

$e^(17/8pii)=cos(17/8pi)+isin(17/8pi)$

So,

$cos(7/6pi)=cos(pi+1/6pi)=-cos(1/6pi)=-sqrt3/2$

$sin(7/6pi)=sin(pi+1/6pi)=-sin(1/6pi)=-1/2$

$cos(17/8pi)=cos(2pi+1/8pi)=cos(1/8pi)=sqrt((1+cos(pi/4))/2)=sqrt((1+sqrt2/2)/2)=sqrt(2+sqrt2)/2$

$sin(17/8pi)=sin(2pi+1/8pi)=sin(1/8pi)=sqrt((1-cos(pi/4))/2)=sqrt((1-sqrt2/2)/2)=sqrt(2-sqrt2)/2$

Therefore,

$e^(7/6pii)-e^(17/8pii)=-sqrt3/2-1/2i-(sqrt(2+sqrt2)/2+sqrt(2-sqrt2)/2i)$

$=-sqrt3/2-sqrt(2+sqrt2)/2-i(1/2+sqrt(2-sqrt2)/2)$

How do you evaluate e^( ( 7 pi)/6 i) - e^( ( 17 pi)/8 i) e^( ( 7 pi)/6 i) - e^( ( 17 pi)/8 i) using trigonometric functions?