How did it became -2pi/27 from the equation (-2pi/3+2pi(0))/9? Please explain.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.5: Product-to-sum And Sum-to-product Formulas
Problem 38E
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How did it became -2pi/27 from the equation (-2pi/3+2pi(0))/9? Please explain.
The polar form of –256 – 256/3i is 512 (cos (-) + i sin (-)) (for steps, see
polar form calculator).
According to the De Moivre's Formula, all n-th roots of a complex number
r (cos (0) + i sin (0))
0..n – 1.
zk), k =
are given by ri (cos (0+2rk) + i sin (0+2nk
n
We have that r =
512, 0 :
and n =
3
9.
k = 0: 512 ( cos
+ i sin
- +2-7-0
9.
9.
2 (cos (-) i sin (-)) = 2 cos () – 2i sin ()
Transcribed Image Text:The polar form of –256 – 256/3i is 512 (cos (-) + i sin (-)) (for steps, see polar form calculator). According to the De Moivre's Formula, all n-th roots of a complex number r (cos (0) + i sin (0)) 0..n – 1. zk), k = are given by ri (cos (0+2rk) + i sin (0+2nk n We have that r = 512, 0 : and n = 3 9. k = 0: 512 ( cos + i sin - +2-7-0 9. 9. 2 (cos (-) i sin (-)) = 2 cos () – 2i sin ()
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