Answer
$\sin (0.1) \approx 0.1$
Work Step by Step
The local linear approximation of $f(x)$ at $x_0$ can be expressed as: $f(x) \approx f(x_0)+f^{\prime}(x_0)(x-x_0)$
Next, we have: $f(x)=\sin x$ at $x_0=0.1$
The local linear approximation of $f(x)$ at $x_0$ can be expressed as: $\sin (0.1) \approx \sin (0)+\cos (0) (0.1 -0) =0.1$
Therefore, $\sin (0.1) \approx 0.1$
