# Taylor series without calculus

Is calculus really necessary for obtaining a Taylor series? Years ago, while perusing an old Schaum’s outline, I found a very curious formula for the area of a circular segment:  $A = \displaystyle \frac{R^2}{2} (\theta - \sin \theta)$

The thought occurred to me that $\theta$ was the first term in the Taylor series expansion of $\sin \theta$ about $\theta = 0$, and perhaps there was a way to use this picture to generate the remaining terms of the Taylor series.

This insight led to a paper which was published in College Mathematics Journal: cmj38-1-058-059. To my surprise and delight, this paper was later selected for inclusion in The Calculus Collection: A Resource for AP and Beyond, which is a collection of articles from the publications of Mathematical Association of America specifically targeted toward teachers of AP Calculus.

Although not included in the article, it can be proven that this iterative method does indeed yield the successive Taylor polynomials of $\sin \theta$, adding one extra term with each successive step.

I carefully scaffolded these steps into a project that I twice assigned to my TAMS precalculus students. Both semesters, my students got it… and they were impressed to know the formula that their calculators use to compute $\sin \theta$. So I think this project is entirely within the grasp of precocious precalculus students. I personally don’t know of a straightforward way of obtaining the expansion of $\cos \theta$ without calculus. However, once the expansion of $\sin \theta$ is known, the expansion of $\cos \theta$ can be surmised without calculus. To do this, we note that $\cos \theta = 1 - 2 \sin^2 \left( \displaystyle \frac{\theta}{2} \right) = 1 - 2 \left( \displaystyle \frac{\theta}{2} - \frac{(\theta/2)^3}{3!} + \frac{(\theta/2)^5}{5!} \dots \right)^2$

Truncating the series after $n$ terms and squaring — and being very careful with the necessary simplifications — yield the first $n$ terms in the Taylor series of $\cos \theta$.

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