Find any rational approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}}$

Question

I need to find any ratinal approximation of $a = \frac{\sin(\sqrt{2})}{\sqrt{2}} $ with accuracy $d = \frac{1}{500}$.

I generally know how to do these type of tasks (using Taylor expansion), but I got problem with this one.

I defined function $f(x) = \begin{cases} 1 \ \mbox{dla} \ x=0 \\ \frac{\sin x}{x} \ \mbox{dla} \ x \neq 0 \end{cases}$, to handle problem with dividng by zero in $f(0)$.

I calculate first and second derivative: $f'(x) = \frac{x^2 \cdot \cos x - \sin x}{x^2}$ $f''(x) = \frac{x^3 \cdot \sin x - 2x^2 \cdot \cos x + \sin x }{x^4}$

But well, what now? I have no idea how to use the Lagrange remainder theorem here.

Answer 1

$$\frac{1}{x}\sin x=\frac{1}{x}(x-\frac{x^3}{3!}+\frac{x^5}{5!}...)=1-\frac{x^2}{3!}+\frac{x^4}{5!}...$$ hence $$=$$

Answer 2

$\text{sinc}(x)$ is an entire function fulfilling $$ \text{sinc}(x) = \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n+1)!} \tag{1}$$ hence $$ \text{sinc}(\sqrt{2})=\sum_{n\geq 0}\frac{(-1)^n 2^n}{(2n+1)!}\tag{2}$$ The terms of the last series are decreasing in absolute value, hence $$ \left|\,\text{sinc}(\sqrt{2})-\sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!}\right|\leq \frac{2^3}{(2\cdot 3+1)!}=\frac{1}{630}\tag{3} $$ and $$ \sum_{n=0}^{3}\frac{(-1)^n 2^n}{(2n+1)!} = \color{red}{\frac{44}{63}}\tag{4} $$ is an approximation fulfilling the given constraints.