Can we multiply the angle of a trig function under some constraints? Here is the basic version:  We are given the functions $\sin{(\theta)}$ and $\cos{(\theta)}$, with a seeming contradiction...  we don't know $\theta$.  The question is if we can find $\sin{(c\theta)}$ and $\cos{(c\theta)}$, with $c \ne 1$ and $c \ne 0$, by using basic arithmetic (and some trigonometry) on these functions.  The rules are that we can add or subtract any number, and multiply or divide by any number, as long as it is not a function of $\theta$, in order to get subresults.  We can also add any subresults together.  For example, we can find:
$$\sin{(\theta)}\cdot \cos{(\phi)} + \cos{(\theta)}\cdot \sin{(\phi)} = \sin{(\theta + \phi)}$$
The question is if we can do this in a constant, finite number of these arithemtic operations.  For example, it may take 100 multiplies and adds to find $\sin{(c\theta)}$, but this is acceptable.
I believe that trigonometry should be able to tell us if we can find $\sin{(c\theta)}$ and $\cos{(c\theta)}$, using these basic rules.  I'm hoping that someone knows a way to do this, but I'd also be happy if someone could prove that this is impossible.
A VARIATION
Here we are allowed a constant, finite number of trigonometric functions to begin with, and we have to find the multiples by $c$ as above.
For example, we may start with $\sin{(\theta)}\cdot\cos{(\theta)}$, $\tan{(\theta)}$, $\sin{(\theta)}$, and $\cos{(\theta)}$, and then we would have to find $\sin{(c\theta)}\cdot\cos{(c\theta)}$, $\tan{(c\theta)}$, $\sin{(c\theta)}$, and $\cos{(c\theta)}$.  The idea is that for any function of $\theta$ that we start with, we also have to find that function with the same multiple, $c$, of $\theta$.
Note that we must always include $\sin{(\theta)}$ and $\cos{(\theta)}$.
Also, we only need to do this for one value of $c$, again with $c \ne 0$ and $c \ne 1$.
 A: For an arbitrary real $c$, arithmetic operations are not enough to express $\sin c\theta$ or $\cos c\theta$ in terms of $\sin \theta$ and/or $\cos\theta$. In addition to arithmetic operations, we would also need powers; this also includes radicals (e.g. for rational $|c|<1$). Moreover, for irrational $c$ (as well as some non-integer rational $c$) even radicals might not be enough.
For instance, if $c=0.2$ we would need radicals, and even that might not be enough. One way to see this is to use generalization of de Moivre's formula: write a complex number $z$ on the unit circle $|z|=1$ in polar form:
$$
z=\cos x+i\sin x. \tag{1}
$$
We then find that (one of the values of) the 5th root of $z$ is
$$z^{1/5} = \cos \frac{x}{5} + i\sin \frac{x}{5}. \tag{2}
$$
Switching to Cartesian form and raising the latter expression to the 5th power, we have
$$
z = \cos^5{x\over5} + 5 i \cos^4{x\over5} \sin{x\over5} 
- 10 \cos^3{x\over5} \sin^2{x\over5}
$$
$$- 10 i \cos^2{x\over5} \sin^3{x\over5} 
+ 5 \cos{x\over5} \sin^4{x\over5} + i \sin^5{x\over5}.   \tag{3}
$$
Equating the real/imaginary parts of $(1)$ and $(3)$, we get 5th-degree algebraic equations for the unknown values $\cos{x\over5}$ and $\sin{x\over5}$.
Specifically, letting $y=\cos{x\over5}$ and $\cos x= C$, we get the following algebraic equation for $y$ equating the real parts of $(1)$ and $(3)$:
$$
y^5 - 10 y^3 (1-y^2) + 5 y (1-y^2)^2 = C,
$$
$$
16 y^5 - 20 y^3 + 5 y = C.
$$
We know that, by Abel-Ruffini theorem, the roots of algebraic equations of degree higher than 4th may or may not be expressible by radicals. (And, in general, neither the values of radicals nor roots of higher-degree algebraic equations would be reachable using just a finite number of arithmetic operations.)
