Solving $C\cos(\sqrt\lambda\theta)+D\sin(\sqrt\lambda\theta)=C\cos(\sqrt\lambda(\theta+2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$

algebra-precalculusmultivalued-functionsperiodic functionssolution-verificationtrigonometry

I want to solve
$$C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$$
The solution must be valid for all $\theta$ in $\mathbb{R}$ and all $m$ in $\mathbb{Z}$, but $C$, $D$, and $\lambda$ are to be determined and can be in $\mathbb{C}$.

The solutions I've found by guessing are $(C\ $arbitrary$, D\ $arbitrary$, \lambda = n^2)$, where $n$ is any integer, and $(C = 0, D = 0, \lambda\ $arbitrary$)$.

Is there some algebra I can do to show that these are the only solutions, or find the rest of the solutions to this equation?

Best Answer

Let $f_{C,D,\lambda}(\theta) = C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta).$ Your want the values of the parameters $C,$ $D,$ and $\lambda$ such that

$$ f_{C,D,\lambda}(\theta) = f_{C,D,\lambda}(\theta + 2m\pi) $$

for all $\theta \in \mathbb R$ and all $m \in \mathbb Z.$ In other words, $f_{C,D,\lambda}$ must either be constant or must be non-constant with period $2\pi$ (implying a minimal period that divides $2\pi$).

You found the constant versions of $f_{C,D,\lambda},$ which require either $C=D=0$ or $\lambda = 0.$

For non-constant $f_{C,D,\lambda},$ note that $f_{C,D,\lambda}$ is sinusoidal with period $2\pi/\sqrt\lambda.$ Hence $2\pi/\sqrt\lambda$ must divide $2\pi,$ hence $\sqrt\lambda$ must be a non-zero integer. You found those solutions too.

Even if we allow $C,$ $D,$ $\lambda,$ and/or $\theta$ to be complex, $f_{C,D,\lambda}$ is still a single-valued function that is either constant or has period $2\pi/\sqrt\lambda.$ We can write $$ f_{C,D,\lambda}(\theta) = \frac12(C-iD)e^{i\sqrt\lambda \theta} + \frac12(C+iD)e^{-i\sqrt\lambda \theta}. $$ For non-constant $f_{C,D,\lambda}$, therefore, $2\pi/\sqrt\lambda$ must still divide $2\pi$ evenly, hence $\sqrt\lambda$ must still be a non-zero integer, and $\lambda$ must still be the square of a non-zero integer, which gives a set of solutions that you already found.

As far as I can see there are no other solutions.