How can $\sin(2nx)$ be expressed as a polynomial? Let $P_n (u)$ be a polynomial in $u$ of degree $n$. Then, prove that for every positive integer $n$, $\sin(2nx)$ is expressible as
$(\cos x)(P_{2n-1} (\sin x))$ for some $P_{2n-1}$.
I tried by starting as
$\sin(2nx)=2\sin(nx)\cos(nx),$ and ta-da, it satisfies for $n=1,2,3$. But I can't figure it in general.
P.S. Complex exponent is not allowed.
 A: Hint:
$$ \sin(2x)=(\cos x)(2 \sin x)$$
$$ \sin(4x) = 2 \sin(2x) \cos (2x) = 2(\cos x)(2 \sin x)(1- 2 \sin^2x) = (\cos x)(4\sin x - 8  \sin^3x)$$
$$ \sin(6x) =  \sin(4x + 2x)= \sin(4x)\cos(2x) +\sin(2x)\cos(4x)$$
$$= \sin(4x) (1-2\sin^2 x) + \sin(2x) ( 1- \sin^2(2x))$$
$$ = \sin(4x) (1- 2\sin^2 x) + \sin(2x) - \sin^3(2x) $$
We then use our decompositions for  $\sin(2x)$ and $\sin(4x)$.
$$ = (\cos x) P_3(\sin x) (1 - 2\sin^2 x) + (\cos x) P_1(\sin x) - \cos^3 x P_1(\sin x)^3  $$
$$ = (\cos x) P_3(\sin x) (1 - 2\sin^2 x)) + (\cos x) P_1(\sin x) -  (\cos x)\left(1- \frac{1}{4}P_1(\sin x)^2\right) P_1(\sin x)^3 $$
Note: Last step above uses
$$ \cos^{2m+1} y = \cos y (1-\sin^2y)^m = \cos y \left(1-\frac{1}{4}P_1(\sin y)^2\right)^m $$
Next level, $8x$, gets treated as $8x =  4x +4x$ (to avoid having to deal with $\cos(6x)$):
$$ \sin(8x) =  \sin(4x + 4x)= 2\sin(4x)\cos(4x) = 2\sin(4x)(1- 2\sin^2(2x)) $$
$$ = 2(\cos x) P_3(\sin x) - 2(\cos x) P_3(\sin x) 8 \cos^2 x P_1(\sin x)^2 $$
$$= 2(\cos x) P_3(\sin x) - 2(\cos x) P_3(\sin x) 8 (1 -\sin^2 x) P_1(\sin x)^2  $$
This has all ingredients of a more formal induction based proof of the decomposition (existence of $P_{2n-1}$ for all $n\geq 1$).
A: As a hint, if complex exponent is allowed: we are to factor $\dfrac{e^{ix}+e^{-ix}}{2}$ out of $\dfrac{e^{2nix}-e^{-2nix}}{2i}$.
Equivalently, we have to factor $e^{2ix}+1$ out of $f(x)=e^{4nix}-1$.
So let's denote $w=e^{2ix}$ and then $f(x)=w^{2n}-1$, so we factor $w^2-1$ out of $w^{2n}-1$ first and factor $w+1$ out of $w^2-1$ then.
As $\frac{w^{2n}-1}{w+1}$ will be a polynomial of $w$, it will be a polynomial of $(w-1)$ too, so we're mostly done.
From the above argument we can derive an explicit formula for $P(w-1)$ and then we will be able to prove the desired explicit equation say by induction, not involving the complex exponent into the solution.
A: Using the Euler
$$e^{inx}=\cos nx +i\sin nx$$
$$(\cos x+i\sin x )^n= \cos nx+i\sin nx$$ so $\sin nx$ equals the imaginary part of the left hand side i.e when $i$ have odd power
If $n$ is odd $n-1$ is even
$$i\sin nx= n i\cos^{n-1}x \sin x + p(n,3)i^3\cos^{n-3}x\sin^3x+\cdots +i^n\sin^nx$$
$$i\sin nx= ni(1-\sin^2x)^{\frac{n-1}{2}}\sin x+-ip(n,3)(1-\sin^2x)^{\frac{n-3}{2}}\sin^3x+\cdots i^n\sin^nx$$ where $p(n,r)=\frac{n!}{r!(n-r)!}$ and if $n$ is even you can do similar argument and you will get $\cos x$. Note you said in the comments that’s true for arbitrary polynomial and that’s wrong take for example $P(u)=u^2-1$ then $P(\sin x)=\sin^2x-1=-\cos^2x$ thus $\cos x(P(\sin x))= -\cos ^3 x \ne \sin nx$ for any $n\in \mathbb{N}$ to see this plug $\pi$ the rhs is 0 but the left is 1
