$\frac{ \sin\theta }{ \theta } = \frac{2165}{2166}$ Find the approximate value of $\theta$

Question

$\dfrac{ \sin\theta }{ \theta }$ = $\dfrac{2165}{2166}$
Find the approximate value of $\theta$

What is the method to solve this question. (I have tried solving it by using Taylor series expansion, but didn't got answer.

Answer 1

$\sin x/x \approx 1-x^2/6$ so $x^2/6 \approx 1/2166$ so $x^2 \approx 6/2166 = 1/361$ so $x \approx 1/19$.

Answer 2

The solution of $$\dfrac{ \sin(\theta) }{ \theta }=\dfrac{2165}{2166}=1-\dfrac{1}{2166}$$ must be very close to $\theta=0$. So, using Taylor expansion is a good idea; with the standard series of $\sin(\theta)$ we then have $$\dfrac{ \sin(\theta) }{ \theta }=1-\frac{\theta^2}{6}+\frac{\theta^4}{120}+O\left(\theta^6\right)$$ So, ignoring the higher order terms, you can solve either $$-\frac{\theta^2}{6}=-\dfrac{1}{2166}\implies \theta= ???$$ or, if you want more accuracy, solve $$-\frac{\theta^2}{6}+\frac{\theta^4}{120}=-\dfrac{1}{2166}\implies \theta= ???$$ This last is just a quadratic in $\theta^2$.

Edit

Sooner or later, you will learn that, better than with Taylor series, we can approximate functions using Padé approximants. For example, the simplest one would be $$\dfrac{ \sin(\theta) }{ \theta }=\frac{60-7 \theta^2}{60+3 \theta^2}$$ and then, you just need to solve for $\theta^2$ $$\frac{60-7 \theta^2}{60+3 \theta^2}=\dfrac{2165}{2166}$$ which is simpler than the quadratic and at least as accurate.