0
$\begingroup$

I'm looking for a solution to $b\sin({\theta})\cos({\phi})+a\cos({\theta})\sin({\phi})=0$ for the variable $\phi$.

In the equation both $a$ and $b$ are real numbers; in particular, I have $\frac{a}{b}>1$.

I'm looking for a solution for $\phi$ in the interval $\phi \in [0,\pi]$.

$\theta$ is also a real parameter which varies in the interval $\theta \in [0,\pi]$ (the same of $\phi$).

I'm trying to solve this equation with mathematica with the command "Reduce", I attach a part of the code:

 k[\[Theta]_, \[Phi]_] := -b Cos[\[Phi]] Sin[\[Theta]] - 
  a Cos[\[Theta]] Sin[\[Phi]]

Assuming[a \[Element] Reals && a > 0 && b \[Element] Reals && b > 0, 
Reduce[k[\[Theta], \[Phi]] == 0, \[Phi], 
 Reals]]

But of course I get all the periodic solutions as results and they're hard to interpret. Is there a way to restrict the solutions only in the intervals I'm interested in? Or is there a quick solution to this problem?

$\endgroup$
4
$\begingroup$

We have $a\sin\theta\cos\phi=-b\cos\theta\sin\phi$

$\iff\dfrac{\sin\phi}{\cos\phi}=-\dfrac ab\dfrac{\sin\theta}{\cos\theta}\iff \tan\phi= -\dfrac ab\tan\theta$

$\phi=n\pi+\arctan\left(-\dfrac ab\tan\theta\right)$ where $n$ is any integer

$\endgroup$

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.