How do you evaluate # e^( (3 pi)/2 i) - e^( ( 11 pi)/12 i)# using trigonometric functions?

Question

1247 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-04T11:34:09

#e^((3pi)/2i)- e^((11pi)/12i) ~~ 0.966 - 1.259 i#

We know #e^(itheta) = cos theta +i sin theta#

#(3pi)/2 =(3*180)/2= 270^0 , (11 pi)/12= (11*180)/12 = 165^0#

#cos 270 =0 ; sin(270) = -1 ; cos 165 ~~ -0.966 ; #

#sin 165 ~~ 0.259 #

#:. e^((3pi)/2i) = cos ((3pi)/2)+ i sin ((3pi)/2) = 0 - i*1=0-i #

#:. e^((11pi)/12i) = cos ((11pi)/12)+ i sin ((11pi)/12) ~~ -0.966 +0.259 i#

# :. e^((3pi)/2i) - e^((11pi)/12i) ~~ (0- i) - (- 0.966 + 0.259 i)# or

# ~~ (0.966) + (-1-0.259)i ~~ 0.966 - 1.259 i#