# Calculating the Accuracy of a Taylor Approximation for $\cos(x)$

I recently have been going over Power Series and Taylor Series (and their theorems) but ran into the following multiple choice problem. I am not allowed to use a calculator so I am unsure as to how we find the interval of accuracy.

On which interval does the polynomial $$y=-\cfrac{3\pi}{2}+ x - \cfrac{(2x-3\pi)^3}{48}$$ best approximates the function $$y=\cos (x)$$.

By playing with the given function I did find that it can be re-written as $$y=\bigg(x-\cfrac{3\pi}{2}\bigg) - \cfrac{8(x-\frac{3\pi}{2})^3}{48}$$ and this leads me to the fact that this is a Taylor Series for $$\sin (x)$$ centered at $$\cfrac{3\pi}{2}$$.

By graphing it I find the solution is roughly the interval [3,6], but I am not sure how to do this without calculator. Any help is appreciated. Thank you!

• You found that $y \approx \sin \left( {x - \frac{{3\pi }}{2}} \right) = \cos x$ near $x=\frac{{3\pi }}{2}$. So the interval in question should be centered at $x=\frac{{3\pi }}{2}$. What are the possible answers?
– Gary
Dec 2, 2020 at 9:51

The key point is that you properly identified that the given expression if the expansion of $$\sin(x)$$ around $$x=\frac 32\pi$$. So the next term would be $$\frac{1}{120} \left(x-\frac{3 \pi}{2}\right)^5$$ If you use the range $$\pi \leq x \leq 2\pi$$, the value of this correction is, in absolute value, $$\frac{\pi ^5}{3840}\approx 0.08$$ which, compared to $$1$$ is quite small. So, the approximation is more than decent for this interval.
What you could also check is where happen the maximum and minimum values of the approximation. they are are $$x_\pm=\frac{3 \pi}{2}\pm \sqrt 2$$ and $$\sqrt 2$$ is not "so far" from $$\frac \pi 2$$. At these points, the function value is $$\pm \frac{2 \sqrt{2}}{3}$$ which is not "very far" from $$1$$.
All of the above confirms the range $$\pi \leq x \leq 2\pi$$.