How to solve $|1-\cos(x)| < \frac{1}{2}10^{-8}$ for x? Can you show me the steps? The answer should be that $-10^{-4}\leq x \leq 10^{-4}$ but I am not sure why.Or $-\arccos(\frac{199999999}{200000000}) \leq x \leq \arccos(\frac{199999999}{200000000})$. I get that $\arccos(\frac{200000001}{200000000}) \leq x \leq \arccos(\frac{199999999}{200000000})$ which is not the same.
Here is what I did (few steps):
\begin{array}{l}
-\left(\frac{1}{2} \cdot 10^{-8}\right) \leq 1-\cos (x) \leq \frac{1}{2} \cdot 10^{-8} \\
-\left(\frac{1}{2} \cdot 10^{-8}\right)-1 \leq-\cos (x) \leq \frac{1}{2} \cdot 10^{-8}-1 \\
\left(\frac{1}{2} \cdot 10^{-8}\right)+1 \leq \cos (x) \leq \frac{-1}{2} \cdot 10^{-8}+1
\end{array}
And yes, I am aware that the solution applies for every cycle of cos – that is, $2\pi$
Best Answer
Using $$ 1-\cos x=2\sin^2(\frac x2) $$ one changes the inequality into $$ \sin^2(\frac x2)<\frac14 10^{-8} $$ or $$ -\frac12 10^{-4}<\sin(\frac x2)<\frac12 10^{-4} $$ So $$ -2\arcsin(\frac12 10^{-4})<x<2\arcsin(\frac12 10^{-4})$$