Question

How do you find the linearization at a=pi/6 of `f(x)=sin2x`?

Answer 1

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

Explanation:

`L(x) = f(a)+f'(a)(x-a)`

`f(pi/6) = sin(pi/3) = sqrt3/2`

`f'(x) = 2cos2x`, so `f'(pi/6) = 2cos(pi/3) = 1`

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