To find the linearization of the function `f(x) = sin(x)` at `x = π/6`, follow these steps:
1. Evaluate `f(x)` at `x = a` to find `f(a)`. Since `f(x) = sin(x)`, we substitute `x = π/6` to get `f(a) = sin(π/6) = 1/2`.
2. Compute the derivative of `f(x)` which is `f'(x) = cos(x)`.
3. Evaluate `f'(x)` at `x = a` to find `f'(a)`. We substitute `x = π/6` into `f'(x)` to get `f'(a) = cos(π/6) = sqrt(3)/2`.
4. Now that we have `f(a)` and `f'(a)`, we can define the linearization function `L(x)`. For any value of `x`, `L(x)` is defined as `L(x) = f(a) + f'(a)(x - a)`.
5. Replace `f(a)` and `f'(a)` with their corresponding values, we get `L(x) = 1/2 + sqrt(3)/2 * (x - π/6)`. This is the equation of the tangent line to the curve `y = sin(x)` at the point `(π/6, 1/2)`, which is also called the linear approximation or the linearization of the function at `x = π/6`.
So, the linearization `L(x)` of the function `f(x)` at `a = π/6` is `L(x) = 1/2 + sqrt(3)/2 * (x - π/6)`.