Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 26

Estimate $\sin 1^\circ$ by using linear approximation or differentials
Using Linear Approximation
$L(x) = f(a) + f'(a)(x-a)$
Let $\displaystyle f(x) = \frac{1}{x}$, $x = a = 1000$


$
\begin{equation}
\begin{aligned}
f(a) = f(1000) &= \frac{1}{1000}\\
\\
f'(a) = f'(1000) &= \frac{d}{dx} \left( \frac{1}{x} \right)\\
\\
f'(1000) &= \frac{(x) \frac{d}{dx} (1) - (1) \frac{d}{dx} (x) }{x^2}\\
\\
f'(1000) &= \frac{-(1)(1)}{x^2}\\
\\
f'(1000) &= \frac{-1}{x^2}\\
\\
f'(1000) &= \frac{-1}{(1000)^2}
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
L(x) & = \frac{1}{1000} + \left( \frac{-1}{(1000)^2}\right) (x - 1000)\\
\\
L(x) & = \frac{1}{1000} - \frac{x+1000}{(1000)^2}\\
\\
L(x) & = \frac{1000-x+1000}{(1000)^2}\\
\\
L(x) & = \frac{2000-x}{(1000)^2}\\
\\
L(x) & = \frac{2000-1002}{(1000)^2}\\
\\
L(x) & = \frac{998}{(1000)^2}\\
\\
L(x) & = \frac{998}{1000000}\\
\\
L(x) & = 0.000998 \quad \text{or} \quad L(x) = 9.98 \times 10^{-4}
\end{aligned}
\end{equation}
$


So,
$\displaystyle \frac{1}{1002} \approx 9.98 \times 10^{-4}$