$\displaystyle f(x) = sec(x)$ $\displaystyle \rightarrow$ $\displaystyle a=0$, $\displaystyle n=2$, $\displaystyle -0.3 \leq x \leq 0.3$

Here is my Taylor polynomial approximation:

$\displaystyle T_2(x) = 1 + \frac{1}{2}x^2$

So, then I found my (n+1) derivative:

$\displaystyle f^3(x) = secxtanx(tan^2x + 5sec^2x)$

So,

$\displaystyle

\left|R_n\right| \leq \frac{M}{(n+1)!} \left|x-a\right|^{n+1}$

CAn someone show me how to do this part?

I know that I need to find a suitable M. How do I do that with these trig fuctions? Can I just choose zero for my x, since it is within my interval?

Thanks!!