# How can you find the taylor expansion of f(x) =sinx about x=pi/6?

Mar 5, 2017

sinx = sum_(n=0)^oo sin(((3n+1)pi)/6)/(n!)(x-pi/6)^n

#### Explanation:

The general expression of the Taylor series of $f \left(x\right)$ around $x = a$ is:

f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n

To find the Taylor series of $\sin x$ we must therefore evaluate the derivatives of the function for all orders. The function $\sin x$ is indeed differentiable indefinitely in $\mathbb{R}$, and to obtain a synthetic notation for its derivatives we can use the following trigonometric equality:

$\sin \left(x + \frac{\pi}{2}\right) = \cos x$

So we have:

$\frac{d}{\mathrm{dx}} \sin x = \cos x = \sin \left(x + \frac{\pi}{2}\right)$

${d}^{2} / {\mathrm{dx}}^{2} \sin x = \frac{d}{\mathrm{dx}} \sin \left(x + \frac{\pi}{2}\right) = \cos \left(x + \frac{\pi}{2}\right) = \sin \left(x + \pi\right)$

and clearly in general:

${d}^{n} / {\mathrm{dx}}^{n} \sin x = \sin \left(x + \frac{n \pi}{2}\right)$

For $x = \frac{\pi}{6}$ we have:

${\left[{d}^{n} / {\mathrm{dx}}^{n} \sin x\right]}_{x = \frac{\pi}{6}} = \sin \left(\frac{\pi}{6} + \frac{n \pi}{2}\right) = \sin \left(\frac{\left(3 n + 1\right) \pi}{6}\right)$

The Taylor series is then:

sinx = sum_(n=0)^oo sin(((3n+1)pi)/6)/(n!)(x-pi/6)^n