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 ) ,`
`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.