How do you evaluate # e^( (3 pi)/2 i) - e^( ( 5 pi)/6 i)# using trigonometric functions?

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Answer 1 · 2018-03-04T18:58:24

#e^((3pi)/2 i) - e^((5pi)/6 i) = (sqrt3-3i)/2#

Recall the identity :

#e^(color(red)idelta) = cos delta + color(red)isindelta#

Given the values of #delta#, we have :

#e^((3pi)/2i) = cos((3pi)/2)+isin((3pi)/2)#

#e^((5pi)/6i)= cos((5pi)/6)+isin((5pi)/6)#

These are fairly small values of #delta#, so you can figure them out using the unit circle and properties of the trigonometric functions.

#cos((3pi)/2) = 0#
#sin((3pi)/2) = -1#

For #delta = (5pi)/6#, we have to use the equalities #sin(color(red)pi-delta) = sin delta# and #cos(color(red)pi-delta) = color(red)-cos delta#:

#sin((5pi)/6) = sin(color(red)pi-(5pi)/6) = sin (pi/6) = 1/2#

#cos((5pi)/6) = color(red)-cos(pi/6) = - sqrt3/2#

Plugging in these values for #e^((3pi)/2i) - e^((5pi)/6i)#, we get :

#e^((3pi)/2i) - e^((5pi)/6i) = 0 + i(-1) - (-sqrt3/2)-i(1/2)#

#color(red)(e^((3pi)/2i) - e^((5pi)/6i) = (sqrt3-3i)/2)#.