calculus
posted by jimmy .
I need help on finding the local linear approximation of tan 62 degree.
i got 1.77859292096 can someone check if i got it right?
tan(60+2) = (tan 60+ tan 2) / (1 tan60 tan 2)
But tan2 appx = sin 2deg = sin2PI/180= 2PI/180
tan 62=(tan 60+2PI/180) / (1 tan60 2PI/180)
check that. It does not equal your answer.
The actual value of tan 62 is 1.880726465.
I don't know what you mean by the "local linear" approximation. If you were doing a linear Taylor series approximation around 60 degrees, you would get
tan 62 = tan 60 + [d/dx(tan x)@ 60 deg]* 2 deg*(pi/180)(deg/radian)= sqrt 3 + (8 pi/180 = 1.8716
The pi/180 factor converts degrees to radians for use in the linear approximation.
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