0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Consider $f(\theta)=(\theta-\theta_1)(\theta-\theta_2)$. No need for the $\theta_1-\theta_2$ condition. $\endgroup$ – Macavity Mar 12 '16 at 9:13
1
$\begingroup$

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

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

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.