How do you use a linear approximation to estimate sin(28) degrees?

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Answer 1 · 2016-08-18T10:30:29

#sin(28^@) approx 0.46977#

We know that #sin(30^@)=0.5# then

using the Taylor expansion we have

#f(x) = f(x_0)+f'(x_0)(x-x_0)+O_2(x-x_0)#

Supposing that #O_2(x-x_0) approx 0#

#sin(28^@) approx sin(30^@)+cos(30^@)(28/180 pi-pi/6) = 1/2 - pi/(60 sqrt[3])#
or

#sin(28^@) approx 0.46977#