Question 9: Using the fact that i²= -1 and the identities: and cos(x)=(-1)" x2n (2n)!' n=0 x2n+1 sin(x)=(-1)" (2n+1)!' n=0 show that z = reio =rcos + ir sin 0.
Question 9: Using the fact that i²= -1 and the identities: and cos(x)=(-1)" x2n (2n)!' n=0 x2n+1 sin(x)=(-1)" (2n+1)!' n=0 show that z = reio =rcos + ir sin 0.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section: Chapter Questions
Problem 12RE
Related questions
Question
![Question 9: Using the fact that 2-1 and the identities:
cos(x) = (-1)"
n=0
x2n
(2n)!'
and
sin(x):
=
Σ(-1)" -
n=0
x2n+1
(2n+1)!'
show that
z=re₁ = r cos + ir sin 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6ab55cb-bf08-43d5-b773-a7fb5882b2b9%2F13d358bf-533f-402a-9c04-4b4be3df98f4%2Ftz21zh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 9: Using the fact that 2-1 and the identities:
cos(x) = (-1)"
n=0
x2n
(2n)!'
and
sin(x):
=
Σ(-1)" -
n=0
x2n+1
(2n+1)!'
show that
z=re₁ = r cos + ir sin 0.
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