Is this true (Trigonometry) and if so, can it be proved by induction?

Question

I'm considering functions of the form $\cos{(n\theta)}+\sin{(n\theta)}$ and have noticed that when $n=4k+1$ for positive integers $k$ the expression seems to be divisible by $\cos{\theta}+\sin{\theta}$ (also works in the trivial case when $k=0$) and when $n=4k-1$ the expression is divisible by $\cos{\theta}-\sin{\theta}$

I've tested this idea for a few integers $k$ and it seems to work, so I'm wondering if it can be proven. Induction seems like the logical choice but exactly how to proceed is, for now, beyond my grasp.

Does anyone have any ideas? (or a counter-example?)

Answer 1

Let $\varepsilon=\pm1$. First note that $\cos t+\varepsilon\sin t=\sqrt2\cos(t-\varepsilon\pi/4)$, so that your question amounts to

Prove that $\cos(\theta-\varepsilon\pi/4)$ divides $\cos((4k+1)\theta-\varepsilon\pi/4)$.

Moreover, letting $x:=\theta-\varepsilon\pi/4$, we have $\cos((4k+1)x)=\cos((4k+1)\theta-\varepsilon k\pi-\varepsilon\pi/4)=(-1)^k\cos((4k+1)\theta-\varepsilon\pi/4)$, so the task simplifies to:

Prove that $\cos x$ divides $\cos((4k+1)x)$.

Actually, we even have $$\forall n\text{ odd},\quad\cos x\mid\cos(nx)=T_n(\cos x)\quad\text{in}\quad\Bbb Z[\cos x]$$ since the $n$th Chebyshev polynomial $T_n(X)\in\Bbb Z[X]$ has the parity of $n$ (there are many ways to prove it, including induction).

Edit: alternatively,

• either notice directly (using De Moivre's formula and the binomial theorem) that $$\begin{align}\cos((2m+1)x)&=\operatorname{Re}\left((\cos x+i\sin x)^{2m+1}\right)\\&=\sum_{j=0}^m\binom{2m+1}{2j}(-1)^j\sin^{2j}x\cos^{2m+1-2j}x\\ &=\cos x\sum_{j=0}^m\binom{2m+1}{2j}(\cos^2x-1)^j\cos^{2m+1-2j}x \end{align}$$
• or use induction: $$\cos((2m+1)x)=2\cos x\cos(2mx)-\cos((2m-1)x).$$