How do you evaluate $e^( ( 13 pi)/12 i) - e^( ( 3 pi)/2 i)$ using trigonometric functions?

Question

1506 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-14T16:24:57

$-(1/2)sqrt(2+sqrt 3)+i(1-(1/2)sqrt(2-sqrt 3))$

Use $e^(pii)=-1 and e^(pi/2i)=i$ .

$e^(13/12pi) i-e^(3/2pii)$

$=e^(pii)(e^(pi/6i)-e^(pi/2i))$

$=-(cos(pi/6)+i sin(pi/6)-i)$

$=-(sqrt((1+cos(pi/3))/2)+i(sqrt((1-cos(pi/3))/2)-i)$

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

$-(1/2)sqrt(2+sqrt 3)+i(1-(1/2)sqrt(2-sqrt 3))$

How do you evaluate e^( ( 13 pi)/12 i) - e^( ( 3 pi)/2 i)e^( ( 13 pi)/12 i) - e^( ( 3 pi)/2 i) using trigonometric functions?