How do you evaluate e^( (3 pi)/2 i) - e^( ( 11 pi)/12 i)e3π2i−e11π12i using trigonometric functions?

1 Answer
Binayaka C.
Jun 4, 2017

e^((3pi)/2i)- e^((11pi)/12i) ~~ 0.966 - 1.259 ie3π2i−e11π12i≈0.966−1.259i

Explanation:

We know e^(itheta) = cos theta +i sin thetaeiθ=cosθ+isinθ

(3pi)/2 =(3*180)/2= 270^0 , (11 pi)/12= (11*180)/12 = 165^03π2=3⋅1802=2700,11π12=11⋅18012=1650

cos 270 =0 ; sin(270) = -1 ; cos 165 ~~ -0.966 ; cos270=0;sin(270)=−1;cos165≈−0.966;

sin 165 ~~ 0.259 sin165≈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

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