# How do you use Taylor series for sin(x) at a = pi/3?

Sep 15, 2015

Just plug values for $x$ into it that are close to $\frac{\pi}{3}$ (like $x = 1$) to get good estimates for $\sin \left(x\right)$ near such values of $x$. Truncating the series at higher powers gives better approximations for a given $x$.

#### Explanation:

Since $f \left(x\right) = \sin \left(x\right)$ implies that $f ' \left(x\right) = \cos \left(x\right)$, $f ' ' \left(x\right) = - \sin \left(x\right)$, $f ' ' ' \left(x\right) = - \cos \left(x\right)$, $f ' ' ' ' \left(x\right) = \sin \left(x\right)$, etc..., it follows that if $a = \frac{\pi}{3}$, then $f \left(a\right) = \frac{\sqrt{3}}{2}$, $f ' \left(a\right) = \frac{1}{2}$, $f ' ' \left(a\right) = - \frac{\sqrt{3}}{2}$, $f ' ' ' \left(a\right) = - \frac{1}{2}$, $f ' ' ' ' \left(a\right) = \frac{\sqrt{3}}{2}$, etc...

The Taylor series for $\sin \left(x\right)$ at $a = \frac{\pi}{3}$ is therefore:

f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+(f''''(a))/(4!)(x-a)^4+cdots

$= \frac{\sqrt{3}}{2} + \frac{1}{2} \left(x - \frac{\pi}{3}\right) - \frac{\sqrt{3}}{4} {\left(x - \frac{\pi}{3}\right)}^{2} - \frac{1}{12} {\left(x - \frac{\pi}{3}\right)}^{3} + \frac{\sqrt{3}}{48} {\left(x - \frac{\pi}{3}\right)}^{4} + \cdots$.

If you, for example, substitute $x = 1$ (radian) into this expansion and stop at the 4th powered term, you'll get:

$\sin \left(1\right) \approx \frac{\sqrt{3}}{2} + \frac{1}{2} \left(1 - \frac{\pi}{3}\right) - \frac{\sqrt{3}}{4} {\left(1 - \frac{\pi}{3}\right)}^{2} - \frac{1}{12} {\left(1 - \frac{\pi}{3}\right)}^{3} + \frac{\sqrt{3}}{48} {\left(1 - \frac{\pi}{3}\right)}^{4} \approx 0.84147$

Using a calculator (in radian mode) gives the same approximation to 5 decimal places (actually, if you carry it out further, it's accurate to 10 decimal places).

To know how accurate you are guaranteed to be requires more theory (see especially Theorem 4 at the following link: http://www.stewartcalculus.com/data/CALCULUS%20Early%20Transcendentals/upfiles/Formulas4RemainderTaylorSeries5ET.pdf )