2

That's a modified exercise taken from a admission test to a university.

Let $x$, $y$ and $\theta$ real numbers such that $$\left \{ \begin{array}{l} x\sin \theta + y \cos \theta = 2a \sin \theta\\ x \cos \theta - y \sin \theta = a \cos \theta\\ \end{array} \right.$$ where $a$ is a real constant. How can we determine two real constants $\mu$ ($\mu>0$) and $k$ such that $$(x+y)^\mu+(x-y)^\mu=k$$ without using brute force?

RicardoCruz
  • 3,693
  • Hi! Almost 9 years have passed! Hope I am not disturbing you. I guess you have already solved this. If not, are you interested in a solution? I can help :) – ACB Dec 16 '21 at 12:00
  • Hi! Yes, I'm interested in your approach, @Flagged. – RicardoCruz Dec 17 '21 at 01:12
  • Sure, though a minor correction. I think it should be $\cos 2\theta$ in the R.H.S. of second equation in order to get the desired answer. (This is a standard exercise in trigonometry) – ACB Dec 17 '21 at 01:32
  • Also in the meantime, I have found this question on another post. If you are satisfied with the answers there, please let me know. – ACB Dec 17 '21 at 02:02
  • Yes, @Flagged, there is really a typo, but in the first equation. The R.H.S. should be $2a\sin \theta$. Corrected. – RicardoCruz Dec 18 '21 at 13:38
  • Without further requirements on $k,\mu$, they can be anything you want. We'd have some real numbers, exponentiated, equal to another real number. What is the goal of the question? – FShrike Dec 18 '21 at 15:06
  • Related. (Same method is applied here) – ACB Dec 30 '21 at 15:38

0 Answers0