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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Question
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 ()](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0bb1775-0071-46f8-a936-f230fc3332cf%2Fb407f075-b99c-4bde-b37f-a3cbb2b1f05c%2Fnltl8m_processed.jpeg&w=3840&q=75)
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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