How do you evaluate $e^( ( 3 pi)/8 i) - e^( ( 11 pi)/6 i)$ using trigonometric functions?

Question

1513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-23T17:17:28

$e^((3pi)/8)-e^((11pi)/16)=-2sin((pi)/32)sin((5pi)/32)-2icos((pi)/32)sin((5pi)/32)$

$e^((3pi)/8)-e^((11pi)/16)$
$=cos((3pi)/8)+isin((3pi)/8)-(cos((11pi)/16)+isin((11pi)/16))$

$=(cos((3pi)/8)-cos((11pi)/16))+i(sin((3pi)/8)-sin((11pi)/16))$

$-2sin(((3pi)/8+(11pi)/16)/2)sin(((3pi)/8-(11pi)/16)/2)+i(2cos(((3pi)/8+(11pi)/16)/2)sin(((3pi)/8+(11pi)/16)/2))$

$((3pi)/8+(11pi)/16)/2=pi/2(3/8+11/16)=pi/2((6+11)/16)=17pi/32$

$(17pi)/32=2pi-pi/32$

$((3pi)/8+(11pi)/16)/2=2pi-pi/32$

$((3pi)/8-(11pi)/16)/2=pi/2(3/8-11/16)=pi/2((6-11)/16)=-(5pi)/32$

$(-5pi)/32=2pi-5pi/32$

$((3pi)/8-(11pi)/16)/2=2pi-5pi/32$

$=-2sin(2pi-pi/32)sin(2pi-(5pi)/32)+i(2cos(2pi-pi/32)sin(2pi-(5pi)/32)$

$e^((3pi)/8)-e^((11pi)/16)=-2sin((pi)/32)sin((5pi)/32)-2icos((pi)/32)sin((5pi)/32)$

How do you evaluate e^( ( 3 pi)/8 i) - e^( ( 11 pi)/6 i) e^( ( 3 pi)/8 i) - e^( ( 11 pi)/6 i) using trigonometric functions?