How can I mentally calculate $\cos(x), x∈(0.7, 1.2)$ I'm trying to learn how to calculate trig functions in my head. I'm planning on learning $\cos(x), x∈[0,π/2]$ and then using symmetry to calculate the others.
I think the quadratic Maclaurin series at $0$ and the linear at $π/2$ could be calculated in a matter of seconds with some practice. However, I'm struggling to find something that works to 2 D.P. on the interval $(0.7, 1.2)$.
My best idea so far is to use $\color{green}{ 1.3-x/1.3}$, but that is neither fast nor accurate to 2 D.P.
Graph of $\color{red}{\cos(x)},\ \ \color{blue}{1-x^2/2},\ \ \color{green}{1.3-x/1.3},\ \ \color{blue}{π/2-x}$:

Error:

How can I quickly approximate $\cos(x)$ for $x∈(0.7, 1.2)$? Or is there a better way to do this?

Here's a quick approximation. The second-order Taylor series around $x=\pi/3$ is
$$T_2(x) = \frac12 - \frac{\sqrt3}2\left(x-\frac\pi3\right)-\frac14\left(x-\frac\pi3\right)^2.$$
Now, $\frac{\sqrt3}2 = 0.866... \approx \frac{13}{15}$, and $\frac\pi3 \approx 1.05$, so we have
$$T_2(x) \approx \frac12 - \frac{13}{15}(x-1.05) - \frac14(x-1.05)^2.$$
You'll see that the error is within $0.01$ in the desired range. Alternatively using $\frac{\sqrt3}2 \approx 0.85$ (which keeps all the constants in multiples of $0.05$) works also.
