In general, if we have a function `f ( x )` which is differentiable at `x = x_0 ,` then we can use the formula

`f ( x_0 + dx ) approx f( x_0 ) + f ' ( x_0 ) * dx`

for the values of `dx` small enough.

It is a clever question: What does it mean, "dx small enough"? There are several estimates using the first or higher-order derivatives of `f ( x ) .`

For your task, the function is `f ( x ) = sin ( x ),` `x_0 = pi / 6,` and `dx = 0.05 .`

This function is infinitely differentiable, and all its derivatives are bounded by `1,` so all theoretical estimates can easily be applied.

We know `f ' ( x ) = cos ( x ) ,`

so

`f ' ( x_0 ) = cos ( pi / 6 ) = sqrt ( 3 ) / 2 .`

Also we know

`f ( x_0 ) = sin ( pi / 6 ) = 1 / 2 .`

Now everything is ready and we obtain

`sin ( pi / 6 + 0.05 ) approx 1 / 2 + ( sqrt ( 3 ) / 2 ) * 0.05 approx` **0.5433**.

The actual value is about 0.5427.