# How do you determine the third Taylor Polynomial at x=0 given f(x)=cos(pi-5x)?

Dec 12, 2016

${T}_{3} \left(x\right) = - 1 + \frac{25}{2} {x}^{2}$

#### Explanation:

First we note that:

$\cos \left(\pi - 5 x\right) = \cos \left(\pi\right) \cos \left(5 x\right) + \sin \left(\pi\right) \sin \left(5 x\right) = \left(- 1\right) \cos \left(5 x\right) + 0 \cdot \sin \left(5 x\right) = - \cos \left(5 x\right)$

Then we calculate the derivatives up to the third order:

$f \left(x\right) = - \cos \left(5 x\right)$

${f}^{\left(1\right)} \left(x\right) = 5 \sin \left(5 x\right)$

${f}^{\left(2\right)} \left(x\right) = 25 \cos \left(5 x\right)$

${f}^{\left(3\right)} \left(x\right) = - 125 \sin \left(5 x\right)$

Finally we calculate the Taylor polynomial of degree 3 around $x = 0$

T_3(x) = sum_(n=0)^3 (f^((n))(0))/(n!)x^n=-1+25/2x^2