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

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Chapter2: Second-order Linear Odes
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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
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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