# Typical Calculus problem on Taylor polynomials seems contrived

Here is a typical Calculus problem involving Taylor polynomials. The solution and my comment follow.

Use the linear Taylor polynomial associated with the sine function to estimate the value of $\sin(50^{\circ})$ with an error of less than $0.02$.

Solution

Lagrange's remainder term for the linear Taylor polynomial associated with the sine function at $60^{\circ}$ is \begin{equation*} R_{1}(x) = - \frac{\sin{\theta}}{2!} \, \left(x - \frac{\pi}{3}\right)^{2} . \end{equation*} for some real number $\theta$ between $\pi/3$ and $x$. $50^{\circ}$ is equivalent to $5\pi/18$ radians, and \begin{equation*} \frac{\pi}{3} - \frac{5\pi}{18} = \frac{\pi}{18} < \frac{1}{5} . \end{equation*} Since the sine function is bounded by $1$, the evaluation of the linear Taylor polynomial $T_{1}(x)$ at $5\pi/18$ is an estimate for $\sin(50^{\circ})$ with an error of less than $0.02$. \begin{equation*} T_{1}\left(\frac{5\pi}{18}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} \left(\frac{5\pi}{18} - \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{\pi}{36} . \end{equation*}

Comment

Isn't this answer disingenuous? If the author of the Calculus textbook is pretending that we do not have a calculator to evaluate $\sin(50^{\circ})$, we also should have a calculator to give estimates for $\sqrt{3}$ ... or for $\pi$.

• I don't believe the author is pretending no calculator is available. I think the point of the problem (and ones like it) is simply to practice estimating with Taylor polynomials and Lagrange error terms. One can ask similar questions when derivatives are first introduced since the first few exercises can be trivially done with any standard CAS and even some websites these days. – tilper Sep 7 '16 at 17:04
• @tilper If that were the case, I think better directions would be to say "write a linear Taylor polynomial that can be used to estimate $\sin(50^{\circ})$. – user74973 Sep 7 '16 at 17:15
• When you use calculator and get an answer, you might ask yourself two questions: how does calculator gets the number and how accurate is the number? The example you showed is really telling you the hows that, without a calculator, how do you calculate $\sin(50^{\circ})$ and how accurate is the calculated result. – user115350 Sep 7 '16 at 19:19

• Is there is a bound to the error in using Newton's method? If we are allowed to use $3.14159$ for an estimate of $\pi$, why can't we just use a table to estimate $\sin(50)^{\circ}$. I am being facetious, of course. – user74973 Sep 7 '16 at 17:04