The answer is error lower than $5\cdot10^{-6}$.
\begin{align} f(0)& = \cos(x)\\ f'(0)& = -\sin(x)\\ f''(0)& = -\cos(x)\\ f'''(0)& = \sin(x)\\ f^4(0)& = \cos(x) \end{align}
We also have that $$P_2(x) = 1 - \frac{1}{2}x^2 = f(0) + f'(0)x + \frac{f''(0)x^2}{2}$$ and $$P_3(x) = 1 - \frac{1}{2}x^2 = f(0) + f'(0)x + \frac{f''(0)x^2}{2} + \frac{f'''(0)x^3}{6}$$
Using the Lagrange Remainder: for $n=3$,
$$| R_3(x)| \leq \frac{\cos^4(c)}{4!}x^4 \leq \frac{(0.1)^4}{24} = 0.00000041 < 0.000005$$
I'm not understanding why the value of $n$ is equal to 3.
If the result is equal to $1-x^2/2$ in both $P_2(x)$ and $P_3(x)$, why do I have to take $n$ as 3? Why don't use it as 2?