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](https://en.wikipedia.org/wiki/Horner%27s_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:

$$\begin{align}
\arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\
           &= \frac{\pi}{2} - \arctan\left( \frac{x}{\sqrt{1-x^2}} \right)\\
\end{align}$$