Question

How do you find the linearization at a=pi/6 of `f(x)=sinx`?

Answer 1

`L(x) = 1/2+sqrt3/2(x-pi/6)`

Explanation:

`f(x) = sinx`, so `f'(x) = cosx`

At `a = pi/6`, we have `y = f(pi/6) = 1/2` and `f'(pi/6) = sqrt3/2`.

The linearization is the tangent line. So

`L(x) = f(a)+f'(a)(x-a) = 1/2+sqrt3/2(x-pi/6)`