Using Taylor series approximation fınd the value of Sin(45), given values: Sin(40) = 0.6428 and Sin(60) = 0.8660 O Sin(45)= 0.6703 O Sin(45)= 0.6986 O Sin(45)= 0.6446 O Sin(45)= 0.8191
Using Taylor series approximation fınd the value of Sin(45), given values: Sin(40) = 0.6428 and Sin(60) = 0.8660 O Sin(45)= 0.6703 O Sin(45)= 0.6986 O Sin(45)= 0.6446 O Sin(45)= 0.8191
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Using Taylor series approximation find the value of Sin(45), given values:
Sin(40) = 0.6428 and Sin(60) = 0.8660
Sin(45)= 0.6703
Sin(45)= 0.6986
Sin(45)= 0.6446
Sin(45)= 0.8191](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F26fe8f9d-a829-4ffe-94fa-74d9058056a0%2F56c126a2-ab94-4b38-afb3-3f9d2b8e8fa4%2Fknr5ni_processed.png&w=3840&q=75)
Transcribed Image Text:Using Taylor series approximation find the value of Sin(45), given values:
Sin(40) = 0.6428 and Sin(60) = 0.8660
Sin(45)= 0.6703
Sin(45)= 0.6986
Sin(45)= 0.6446
Sin(45)= 0.8191
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