What are the sine and cosine of dyadic angles? What are the values of sine and cosine of dyadic angles? We know
$$\cos\pi = -1 \qquad \cos\frac{\pi}{2} = 0 \qquad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\,,$$
and we can calculate sine by appealing to symmetry. But I don't think I've seen the sine and cosine values, presented in terms of radicals, of $\pi/8$, $3\pi/8$, $\pi/16$, $\pi/2^n$, etc. Do these have nice algebraic presentations? How can we calculate these?

This answer has some approaches that were not the easiest ways to solve the problem. I'm recording them here in a separate answer so as not to lose the information (even if what this tells us is not to do it this way.)

For a complete table of sines and cosines of $\pi/16,$ we only need to compute values for $0,$ $\pi/16,$ $\pi/8,$ $3\pi/16,$ and $\pi/4.$
We already have cosines of all of these (in another answer) except $\cos(3\pi/16).$
Using the angle-sum formula for cosine,
\begin{align}
\cos \frac{3\pi}{16}
&= \cos\frac\pi8 \cos\frac\pi{16} - \sin\frac\pi8 \sin\frac\pi{16} \\
&= \frac14\sqrt{2+\sqrt2} \sqrt{2+\sqrt{2+\sqrt2}}
   - \frac14\sqrt{2-\sqrt2} \sqrt{2-\sqrt{2+\sqrt2}} \\
\end{align}
It should be possible to simplify this further (see other answer) but the path is not obvious.

As an alternative to partitioning $m\pi/2^n$ into powers of two, there are the (somewhat) well-known formulas for the sine and cosine of arbitrary multiples of angles:
\begin{align}
\sin(k\theta) &= \sum_{j=0}^n {n \choose j} \cos^j\theta \, \sin^{n-j}\theta \, \sin\frac{(k-j)\pi}{2}, \\
\cos(k\theta) &= \sum_{j=0}^n {n \choose j} \cos^j\theta \, \sin^{n-j}\theta \, \cos\frac{(k-j)\pi}{2}.
\end{align}
There is a relatively simple proof using complex numbers. That may not be a good proof to expect trig students to be able to follow, but it might give some incentive to study complex numbers later.
\begin{align}
\cos \frac{3\pi}{16}
&= \cos^3 \frac{\pi}{16} - 3 \cos \frac{\pi}{16} \sin^2 \frac{\pi}{16} \\
&= \cos \frac{\pi}{16} \left(\cos^2 \frac{\pi}{16} - 3 \sin^2 \frac{\pi}{16} \right) \\
&= \cos \frac{\pi}{16} \left(\frac14\left(2+\sqrt{2+\sqrt{2}}\right)
       - \frac34 \left(2-\sqrt{2+\sqrt{2}}\right) \right) \\
&= \cos \frac{\pi}{16} \left(-1 + \sqrt{2+\sqrt{2}}\right)\\
&= \frac12 \left(-1 + \sqrt{2+\sqrt{2}}\right) \sqrt{2+\sqrt{2+\sqrt{2}}}
\end{align}
Again it should be possible to simplify this further but the path is not obvious.
