$\cos(x)\cos(mx)$ is a polynomial in $\sin(x)^{2}$ with degree $\le (m-1)/2$

Question

For all $m \ge 1$ odd show that:

1) $\cos(x)\cos(mx)$ is a polynomial in $\sin(x)^{2}$ with degree $\le (m+1)/2$

2) $\frac{\sin(mx)}{\sin(x)}$ is a polynomial in $\sin(x)^{2}$ with degree $\le (m-1)/2$

Hint: Use induction!

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My idea was to use some addition theorems but i dont know which. Maybe someone here has an idea how to proof it.

I proved 2) by using

• $\sin(mx+2x)=\sin(mx)\cos(2x)+\sin(2x)\cos(mx)$
• $\sin(2x)=2\sin(x)\cos(x)$
• $\cos(2x)=1-2\sin(x)^{2}$

AND 1)...

So please help me to prove 1).

Answer 1

First of all, your probably have a typo on your statement (1),

... with degree $\leq (m-1)/2$

, which should actually be

... with degree $\leq (m+1)/2$

A simple check, when $m=1$, $\cos(x)\cos(mx)$ is a first-degree polynomial of $\sin^2x$

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Now assume (1)&(2) holds for $m-2$,

$$ \cos(x)\cos(mx) = \cos(x)\left\{\cos[(m-2)x]\cos(2x) - \sin[(m-2)x]\sin(2x)\right\} $$

Let me use $P_n(x)$ denote a polynomial of $x$ with degree $\leq n$.

The first term on the RHS: $$ \cos(x)\cos[(m-2)x]\cos(2x) = P_{\frac{m-1}{2}}(\sin^2x) (1 - 2 \sin^2 x) = P_{\frac{m+1}{2}}(\sin^2x) $$

The second term on the RHS: $$ \begin{aligned} \cos(x)\sin[(m-2)x]\sin(2x) = & \cos(x)[ P_{\frac{m-3}{2}}(\sin^2x) \sin(x) ]\sin(2x)\\ = & P_{\frac{m-3}{2}}(\sin^2x) 2\sin^2(x)\cos^2(x) \\ = & P_{\frac{m+1}{2}}(\sin^2x) \end{aligned} $$

Thus $\cos(x)\cos(mx)$ is also a $ P_{\frac{m+1}{2}}(\sin^2x)$.

Answer 2

$$e^{imx}=(\cos x+i\sin x)^m=\sum_{k=0}^m\binom mk i^k\cos^{m-k}x\sin^kx. $$

Taking the real part and noting that only the terms in even $k$ remain, we see that every term of $\cos x\cos mx$ is a product of even powers of $\cos x$ and $\sin x$.