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The small-angle approximation for cosine is: $$ \cos (x) = 1 - \frac{x^2}{2} $$ Question: How can I find a range of values of $x$ for which this approximation gives correct results rounded to 2 decimal places?

Thought: The error term of this $2^{nd}$-order Taylor approximation is $$ E(x)=\frac{sin(\eta)}{6}x^3, $$ where $\eta$ is between $x$ and $0$. Thus, $$ |E(x)|<10^{-2} \to |\sin(\eta)x^3|<6\times10^{-2}. $$ This is just my thought, but I am not sure this is the correct approach.

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  • $\begingroup$ Right. Since $|\eta|<|x|$ you will have $|\sin\eta|\le|\eta|\le |x|$ and you derive a sufficient condition on $x$ from here. $\endgroup$ – GReyes Feb 10 at 23:31
  • $\begingroup$ Observe also that you can think of your approximation as a third order approximation (the series contains only even powers) and use the corresponding error term, which is smaller and gives you a better range for the possible values of $x$. $\endgroup$ – GReyes Feb 10 at 23:34
  • $\begingroup$ @GReyes Thank you. I got the same result as that in Ross Millikan's answer. $\endgroup$ – A Slow Learner Feb 10 at 23:44
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The alternating series theorem says the truncation error is smaller than the first neglected term and of the same sign. The first term you neglect is $\frac {x^4}{4!}$ so we want $$\frac {x^4}{4!} \lt 0.01\\x^4 \lt 0.24\\|x|\lt 0.24^{1/4}\approx 0.700$$ When you demand correct rounding to a number of places it is hard to say what the allowable error is. If you are very close to a breakpoint you may have to be very accurate. I used $0.01$ as the allowable error, you can use whatever value you want.

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  • $\begingroup$ I never learned about this theorem before. However, I have been able to derive the same result based on @GReyes' comments in the OP. Could you give me a link to the theorem that you used? Thank you. $\endgroup$ – A Slow Learner Feb 10 at 23:42
  • $\begingroup$ It is powerful. Wikipedia has an entry. If you have an alternating series with the terms decreasing in magnitude, the theorem applies. $\endgroup$ – Ross Millikan Feb 10 at 23:49
  • $\begingroup$ @Just to clarify the answer, the answer should be $x \in (-0.7,0.7)$, right? Thank you. $\endgroup$ – A Slow Learner Feb 12 at 2:47
  • $\begingroup$ Yes, I will put absolute value bars on the $x$ in the last line. Thanks $\endgroup$ – Ross Millikan Feb 12 at 2:56

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