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

Question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T21:36:35

$- 0.188 + 0.355i$

According to Euler's formula,

$e^(ix) = cosx + isinx$ .

If we substitute the two values, beginning with $(11pi)/6$ ,

$cos((11pi)/6) + isin((11pi)/6) = cos330 + isin330$
$= - 0.991 - 0.132i$

$x = pi/8$
$cos(pi/8) + isin(pi/8) = cos22.5 + isin22.5$
$= - 0.873 - 0.487i$

Putting the two together,

$e^((11pi)/6i) - e^((pi/8)i) = - 0.991 + 0.873 - 0.132i + 0.487i$
$= - 0.118 + 0.355i$

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