General expression for $\sin(2^n x)$ Are there general expressions for $\sin(2^n x)$ and $\cos(2^n x)$ that only involve $\sin x$ and $\cos x$, and that moreover involve only polynomial (in $n$) number of terms? 
Edit:
$2^n$ is not polynomial in $n$. A proof that no such expression exits (perhaps using the uniquness of Chebyshev polynomials?) would be gladly accepted.
 A: If I understand the spirit of the question, no such expressions exist. First, $\cos(2^n \theta)$ is even, so a polynomial in $\sin \theta$ and $\cos \theta$ that is equal to $\cos(2^n \theta)$ contains only even powers of $\sin\theta$, and therefore may be written as a polynomial in $\cos\theta$, as marty cohen notes.
Now, $\cos(2^n \theta)$ has $2^{n+1}$ simple zeros in the interval $[0, 2\pi]$. By contrast, if $p_n$ is a polynomial with real coefficients and "polynomial-in-$n$ terms" (my interpretation of the original question), then $p_n$ has polynomial-in-$n$ roots in the interval $[-1, 1]$ by Descartes rule of signs, and consequently $p_n(\cos \theta)$ vanishes only polynomial-in-$n$ times in $[0, 2\pi]$.
A: I might be missing something here... but the identity $\cos(2x)=2\cos^2(x)-1$ can be trivially applied $n$ times to obtain an expression with polynomial length:
$$\begin{eqnarray}
\cos(2^nx) & = & 2\left(\cos(2^{n-1}x)\right)^2-1 \\
& = & 2\left(2\left(\cos(2^{n-2}x)\right)^2-1\right)^2-1 \\
& = & 2\left(2\left(2\left(\cos(2^{n-3}x)\right)^2-1\right)^2-1\right)^2-1 \\
& & \vdots \\
& = & 2\left(2\left(\ldots \left(2\left(2\cos^2 x-1\right)^2-1\right)^2\ldots\right)^2-1\right)^2-1 \\
\end{eqnarray}$$
Note that this results in just one occurrence of the $\cos x$ term and the whole expression contains $n$ twos, $n$ squarings and $n$ subtractions of $1$. Thus, length-wise it's polynomial. Of course, getting rid of the squares would blow the size of the formula up exponentially.
Similar approach can be used for $\sin(2^nx)$; it suffices to notice that $$\sin(2^nx)=2^n\sin(x)\cos(x)\cos(2x)\cos(4x)\ldots\cos(2^{n-1}x)$$ and apply the expression above to the $\cos(2^kx)$ terms (resulting in expression of length quadratic in $n$). If one is interested in evaluating the expression, it's possible to do much better (thanks to the high level of redundancy).
