Why can't sin(x^2) be expressed in terms of sin(x)?

[Prem1998]

I've read explanations on the internet that the product of two angles is not something fundamentally important. So, the sin of product of two angles cannot be expressed in terms of sin of individual angles. But in calculus, we often come across these functions, we deal with functions involving sin of squares of angles, roots of angles, and even logarithm of angles. So, these functions could be important.
And, if a formula doesn't exist, then is there a proof that such a formula of sin(x^2) in terms of sin(x) can't exist?

 

[robphy]

You probably need to restrict the types of functions you allow.
You probably don't want something like $\sin(x^2)=\sin( (\arcsin(\sin x))^2)$.

 

[Prem1998]

You probably need to restrict the types of functions you allow.
You probably don't want something like $\sin(x^2)=\sin( (\arcsin(\sin x))^2)$.

Obviously, I don't want that.
I want an algebraic expression involving sinx. Just like how sin(2x) is expressed in terms of sinx

 

[mfb]

sin(x) is periodic with a period of 2 pi, sin(x²) is not. You cannot use a periodic function (and nothing else with x-dependence) to make a non-periodic function.

 

[Prem1998]

sin(x) is periodic with a period of 2 pi, sin(x²) is not. You cannot use a periodic function (and nothing else with x-dependence) to make a non-periodic function.

Is there a theorem which proves this result on periodic functions?

 

[mfb]

Isn't it trivial?

$\sin(0) = 0$ and $\sin(2\pi) = 0$
No matter how you use sin(x) in a function f(sin(x)), for those two different x-values the formula will always have the same result: $f(\sin(0)))=f(\sin(2\pi))$. Which is in contradiction to what you want to get, as $\sin(0^2) = 0$ but $\sin((2\pi)^2) \neq 0$. The only way to fix this is to introduce an explicit x-dependence in f, e.g. add sin(x²)-sin(x) to sin(x), but as you can see that doesn't really lead to a formula that "depends on sin(x)".