# 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.