# How do you find the taylor series for cos x, a=pi/3?

Dec 15, 2016

cosx = sum_(n=0)^(oo) cos(pi/3+(npi)/2)/(n!)(x-pi/3)^n

#### Explanation:

The Taylor series of $f \left(x\right)$ in $x = {x}_{0}$ is given by:

sum_(n=0)^(oo) f^((n))(x_0)/(n!)(x-x_0)^n

So we must calculate the derivatives of all orders of $\cos \left(x\right)$.

Now we note that:

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

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

and in general:

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

which can be proved by mathematical induction.

Thus the Taylor expansion is:

sum_(n=0)^(oo) cos(pi/3+(npi)/2)/(n!)(x-pi/3)^n