$\sin(x^2)$ in terms of $\sin(x)$ and $\cos(x)$ One of my students asked me "Can you write $\sin(x^2)$ in terms of $\sin(x)$"? I said I'd think about it. Having thought about it for a while, I now know that I definitely don't know the answer!
Lets relax the question to "...$\sin(x^2)$ in terms of $\sin(x)$ and $\cos(x)$".
For $x$ an integer we have $\sin(x^2)=Im[(\cos x+i\sin x)^x]$, so we can do it for fixed integers. But that isn't awfully satisfying.
Another chain of thought is that $\sin(x^2)$ isn't periodic, and so we cannot simply do something with Fourier series. However, my knowledge of Fourier series has been lost to time, and I do not know if

not periodic $\Rightarrow$ cannot be written as a sum of trig functions

is true. Sounds nice thought, doesn't it?
 A: A fourier series is usually an infinite series, and it is possible to write a whole deal of functions, even for example $f(x)=x^2$ (where no trigonometric function appears), as an infinite sum of trigonometric funcions (as long as you limit yourself to an interval).
However, I don't think your student really had Fourier series in mind. I imagine his question was more like "can  you simplify $\sin x^2$ in a similar way to $\sin(2x)$", and in that case, you are on the right track: the problem is that $\sin(x^2)$ is not periodic.
It is very simple to see that any finite sum, product or fraction of $\sin$ and $\cos$ is also periodic, because any function $f(x)=F(\sin x, \cos x)$ is also periodic:
$$f(x+2\pi)=F(\sin (x+2\pi), \cos(x+2\pi)) = F(\sin(x),\cos(x)) = f(x)$$ 
A: Suppose $F:\mathbb R^2 \to \mathbb R$ is such a function, so that $\sin(x^2)=F(\sin(x),\cos(x))$ for all $x \in \mathbb R$. Then we have
$$\sin((x+2\pi)^2) = F(\sin(x+2\pi),\cos(x+2\pi)) = F(\sin(x),\cos(x)) = \sin(x^2)$$
for all $x \in \mathbb R$. But this is clearly false (just take $x=0$, for instance).
A: Another idea could be thinking of $x^2$ as $x+x+...+x$ - ($x$ times).
Then $\sin(x^2) = \sin(x+(x+...+x)) = sin(x)cos(x+(x+...+x)) + cos(x)sin(x+(x...+x))$ and you can continue expanding the formula. I'm not sure how much could be simplified, so you might end with an ugly monster, although all in terms of $\sin(x)$ and $\cos(x)$
