Using Chebisev polynomials to express sin(nx) & cos(nx) as polinomials of sin(x) and cos(x)

Question

$Sin(nx)$ and $cos(nx)$ can be expressed as polynomials of sin(x) and cos(x). I am interested in the way of this expression and a proof (preferably at secondary-school level) as well.

Answer 1

Here's a proof that $\sin(2x)$ can't be written as a polynomial in $\sin x$.

Recall that $\sin(2x) = 2\cos(x)\sin(x)$. Now assume we have a polynomial $p \in \mathbb R[x]$ such that $\sin(2x) = p(\sin(x)).$ Then we also have $$ 2\cos(x)\sin(x) = p(\sin(x)). $$ Since the l.h.s. is $0$ at $x = 0$, we must have $0 = p(\sin(0)) = p(0)$. This means that $p$ doesn't have a constant term, so we can write $p(x) = xq(x)$ with another polynomial $q \in \mathbb R[x].$ Thus we have $$ 2\cos(x)\sin(x) = \sin(x)q(\sin(x)) $$ for any $x \in \mathbb R$. So we have $$ 2\cos(x) = q(\sin(x)) \quad \forall x \in \mathbb R\setminus\pi\mathbb Z $$ since $\pi\mathbb Z$ is the zero set of $\sin(x)$. Since both the l.h.s. and the r.h.s of the preceding equation are continuous and $\pi\mathbb Z$ is discrete in $\mathbb R$, we get by continuity $$ 2\cos(x) = q(\sin(x)) \quad \forall x \in \mathbb R. $$ Now we square this equation and get $$ 4(1-\sin^2(x)) = q(\sin(x))^2 \quad \forall x \in \mathbb R. $$ Put $r(x) = 4(1-x^2) \in \mathbb R[x]$. With this we can write $$ r(\sin(x)) = q(\sin(x))^2 \quad \forall x \in \mathbb R. $$ Since $\sin(x)$ ranges over $[-1,1]$ as $x$ ranges over $\mathbb R$, this is equivalent to $$ r(x) = q(x)^2 \quad \forall x \in [-1,1]. $$ Now we use the fact that if two polynomials coincide on a non-empty open interval (and $[-1,1]$ contains $(-1,1)$), then these polynomials must be equal. So we get $$ r(x) = q(x)^2 \quad in\ \mathbb R[x]. $$ But $$ r(x) = 4(1-x^2) = 4(1-x)(1+x)\quad in\ \mathbb R[x] $$ is not a square.

So we have reached a contradiction, and the polynomials $q$ and $p$ don't exist.

Answer 2

Here you are.

Let me state the formulas again. For any $x \in \mathbb R$, we have the following. $$ e^{ix} = \cos x + i\sin x $$ $$ (e^{ix})^n = (\cos x + i\sin x)^n = \sum_{j=0}^n{n\choose j}(\cos x)^{n-j}(i\sin x)^j $$ $$ (e^{ix})^n = e^{inx} = \cos(nx) + i\sin(nx) $$ Note that the second formula is a little bit different than in my comment.

Now we equate real and imaginary parts in the second and third formula for $(e^{ix})^n$.

From the real parts we get $$ \cos(nx) = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} {n\choose 2k} (\cos x)^{n-2k}(i\sin x)^{2k} = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} {n\choose 2k} (\cos x)^{n-2k}(-1)^k(1-\cos^2x)^k. $$ Here, the r.h.s. is a polynomial in $\cos x$.

From the imaginary parts we get $$ \begin{align*} i \sin(nx) & = \sum_{k = 0}^{\lfloor \frac{n-1}{2} \rfloor} {n\choose 2k+1} (\cos x)^{n-2k-1}(i\sin x)^{2k+1} \\ & = i\sin x\sum_{k = 0}^{\lfloor \frac{n-1}{2} \rfloor} {n\choose 2k+1} (\cos x)^{n-2k-1}(-1)^k(1-\cos^2x)^k \end{align*} $$ Here, the sum on the r.h.s. is a polynomial in $\cos x$. And of course we can cancel the common factor $i$.