Question

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

Answer 1

`sin(28^@) approx 0.46977`

Explanation:

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`