C code that does this here
The general polynomial form is:
$$\begin{align}
f(u) &= \sum\limits_{n=0}^{N} \ a_n \ u^n \\
\\
&= a_{\small{0}} + \Bigg(a_{\small{1}} + \bigg(a_{\small{2}} + \Big(a_{\small{3}} + \,... \big(a_{\small{N-2}} + (a_{\small{N-1}} + a_{\small{N}} \,u \,)u \, \big)u \ ...\Big)u \, \bigg)u \, \Bigg)u\\
\end{align}$$
The latter form is using Horner's method, which is highly recommended, especially if you're doing this in single-precision floating point.
Then for a few specific functions:
square root:
$$ \begin{align}
f(x-1) & \approx \sqrt{x} \quad \quad 1 \le x \le 2 \quad \quad N=4\\
a_0 & = 1.0 \\
a_1 & = 0.49959804148061 \\
a_2 & = -0.12047308243453 \\
a_3 & = 0.04585425015501 \\
a_4 & = -0.01076564682800 \\
\end{align} $$
If $2 \le x \le 4$, use the above to evaluate $\sqrt{\tfrac{x}{2}}$ and multiply that result with $\sqrt{2}$ to get $\sqrt{x}$. As with $\log_2(x)$, apply power of $2$ scaling to scale the argument to the necessary range.
base 2 logarithm:
$$ \begin{align}
x\cdot f(x-1) & \approx \log_2(x) \quad \quad 1 \le x \le 2 \quad \quad N=5\\
a_0 & = 1.44254494359510 \\
a_1 & = -0.7181452567504 \\
a_2 & = 0.45754919692582 \\
a_3 & = -0.27790534462866 \\
a_4 & = 0.121797910687826 \\
a_5 & = -0.02584144982967 \\
\end{align} $$
base 2 exponential:
$$ \begin{align}
f(x) & \approx 2^x \quad \quad 0 \le x \le 1 \quad \quad N=4\\
a_0 & = 1.0 \\
a_1 & = 0.69303212081966 \\
a_2 & = 0.24137976293709 \\
a_3 & = 0.05203236900844 \\
a_4 & = 0.01355574723481 \\
\end{align} $$
sine:
$$ \begin{align}
x\cdot f(x^2) & \approx \sin\left(\tfrac{\pi}{2} x \right) \quad \quad -1 \le x \le 1 \quad \quad N=4 \\
a_0 & = 1.57079632679490 \\
a_1 & = -0.64596406188166 \\
a_2 & = 0.07969158490912 \\
a_3 & = -0.00467687997706 \\
a_4 & = 0.00015303015470 \\
\end{align} $$
cosine (use sine):
$$ \cos(\pi x) = 1 \, - \, 2 \, \sin^2 \left(\tfrac{\pi}{2} x \right) $$
tangent:
$$ \tan(x) = \frac{\sin(x)}{\cos(x)} $$
inverse tangent:
$$ \begin{align}
\frac{x}{f(x^2)} & \approx \arctan(x) \quad \quad -1 \le x \le 1 \quad \quad N=4 \\
a_0 & = 1.0 \\
a_1 & = 0.33288950512027 \\
a_2 & = -0.08467922817644 \\
a_3 & = 0.03252232640125 \\
a_4 & = -0.00749305860992 \\
\end{align} $$
$$ \arctan(x) = \tfrac{\pi}{2} - \arctan\left( \tfrac{1}{x} \right) \quad \quad 1 \le x $$
$$ \arctan(x) = -\tfrac{\pi}{2} - \arctan\left( \tfrac{1}{x} \right) \quad \quad x \le -1 $$
inverse sine:
$$ \arcsin(x) = \arctan\left( \frac{x}{\sqrt{1-x^2}} \right)$$
inverse cosine:
$$\begin{align}
\arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\
&= \frac{\pi}{2} - \arctan\left( \frac{x}{\sqrt{1-x^2}} \right)\\
\end{align}$$