Prove that: $\begin{cases} f(\theta_1)\cos \theta_1=f(\theta_2)\cos \theta_2 \\ f(\theta_1)\sin \theta_1=f(\theta_2)\sin \theta_2 \end{cases}$

Question

Let $f(\theta)$ be a continue function for $\theta\in[\theta_1,\theta_2]$. Prove that: \begin{cases} f(\theta_1)\cos \theta_1=f(\theta_2)\cos \theta_2 \\ f(\theta_1)\sin \theta_1=f(\theta_2)\sin \theta_2 \end{cases} if and only if $f(\theta_1)=f(\theta_2)$ and $\theta_2-\theta_1=2n\pi$ for some integer $n$.

It is easily to verify that $\{(\theta_1,\theta_2):\, f(\theta_1)=f(\theta_2), \, \theta_2-\theta_1=2n\pi, \, n\in\mathbb Z\}$ satisfies the system but... is it the only solution of the system?

Answer 1

$f(\theta_1)^2=f(\theta_1)^2\sin^2(\theta_1)+f(\theta_1)^2\cos^2(\theta_1)= f(\theta_2)^2\sin^2(\theta_2)+f(\theta_2)^2\cos^2(\theta_2)=f(\theta_2)^2$

then $f(\theta_1)=\pm f(\theta_2)$.

-If $f(\theta_1)= f(\theta_2)$ it is necessary that $\theta_1$ and $\theta_2$ differ for a multiple of $2\pi$.

-If $f(\theta_1)=- f(\theta_2)$ you have to find $\theta_1$ and $\theta_2$ such that \begin{align} \sin(\theta_1)&=-sin(\theta_2) \\ \cos(\theta_1) &=-\cos(\theta_2) \end{align} and it is true for $\theta_2=\theta_1+\pi+2k\pi,k\in \mathbb{Z}$ (note that this was true if $f(\theta_1)\neq 0\neq f(\theta_2)$).

There are then three sets of solutions $$f(\theta_1)=f(\theta_2) \qquad \text{and} \qquad \theta_2=\theta_1+2k\pi$$

$$f(\theta_1)=-f(\theta_2) \qquad \text{and} \qquad \theta_2=\theta_1+\pi+2k\pi$$

$$f(\theta_1)=f(\theta_2)=0 \qquad \text{and} \qquad \forall \theta_1,\theta_2 \in \mathbb{R} $$

Answer 2

Let's suppose that $f$ is a solution of the equation. Using that $\sin^2 \theta + \cos^2 \theta = 1$ we have:

$$ f(\theta_1)^2 = f(\theta_1)^2\sin^2 \theta_1 + f(\theta_1)^2 \cos^2 \theta_1 = f(\theta_2)^2\sin^2 \theta_2 + f(\theta_2)^2 \cos^2 \theta_2 = f(\theta_2)^2 $$

Hence, $\vert f(\theta_1) \vert = \vert f(\theta_2) \vert$. This fact reduces the possible solutions to the following cases:

• $f(\theta_1) = f(\theta_2) \ne 0$. Consequently, dividing in the equations we have $\cos \theta_1 = \cos \theta_2$ and $\sin \theta_1 = \sin \theta_2$. Thus, $\theta_1 - \theta_2 = 2\pi n$ with $n \in \mathbb{Z}$. Clearly, these are solutions.

• $f(\theta_1) = f(\theta_2) = 0$. Then, no condition is needed in $\theta_1$ and $\theta_2$ to obtain a solution.

• $f(\theta_1) = - f(\theta_2) \ne 0$. Dividing again we have $\cos \theta_1 = - \cos \theta_2$ and $\sin \theta_1 = - \sin \theta_2$. Hence, $\theta_1 - \theta_2 = (2n+1)\pi$ with $n \in \mathbb{Z}$. Again, these are solutions too.

Note that we haven't used the continuity of $f$. If you want to obtain only the first type of solutions, then you could add as hypothesis that $f(\theta) \ne 0 \ \forall \theta \in [\theta_1, \theta_2]$. The second case is not possible anymore. Furthermore, the third case tells us that $f(\theta_1)f(\theta_2) < 0$ and using Bolzano's theorem we would have a zero in $\left(\theta_1, \theta_2\right)$, what is not possible.