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

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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#