Find interval containing $\tan 0.7$ using Taylor expansion and Lagrange

Using a second degree Taylor series about $x = 0$ (I'm aware this is called Mclaurin too) and the Lagrange remainder I am supposed to find an interval where I am certain that $\tan 0.7$ will be. I follow the same procedure as my book for a similiar problem, starting with the second degree polynomial:

\begin{equation*} \begin{aligned} & f(x) = \tan x & f(0) = 0 \\ & f'(x) = 1 + \tan^2 x & f'(0) = 1 \\ & f''(x) = 2\tan x(1 + \tan^2 x) & f''(0) = 0 \\ \end{aligned} \end{equation*}

Using this I get

\begin{equation*} \begin{aligned} p_2(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2} = x \\\\ \tan 0.7 = f(0.7) \approx p_2(0.7) = 0.7 \end{aligned} \end{equation*}

Now I find the error function

\begin{equation*} \begin{aligned} & f^{(3)}(x) = 2(3\tan^4 x + 4\tan^2 x + 1) \\\\ & E_2(x) = \frac{f^{(3)}(s)}{6}x^{3} \end{aligned} \end{equation*}

Now, for $0 \lt s \lt 0.7$, I have the inequality

$$\left|f^{(3)}(s)\right| \leq f^{(3)}(0) = 2$$

which gives

$$\left|E_2(0.7)\right| \leq \frac{2}{6}0.7^3 = \frac{343}{3000} \approx 0.114\overline{3}$$

At this point, my book had an interval containing the approximated value. I however, do not. I have this interval:

$$(p_2(0.7) - E_2(0.7), p_2(0.7) + E_2(0.7)) \approx (0.585, 0.814)$$

But my calculator gives $\tan 0.7 \approx 0.842$. How can I get an interval containing the value that I want? I guess I could choose another value for in the inequality for $f^{(3)}(s)$ instead of $0$ so the numerator will be larger and thus the interval larger. But how am I certain it will contain $\tan 0.7$

Double check whether $|f^{(3)}(s)|<|f^{(3)}(0)|$ is actually true on that interval. It should be positive and growing above 0, so $|f^{(3)}(s)|>|f^{(3)}(0)|$. Compare it maybe to $|f^{(3)}(s)|<|f^{(3)}(\pi/4)|$ since $\tan(\pi/4)=1$ is a known value which you can then use to get a better error value for your interval around 0,7