6
$\begingroup$

Given $\sin(xy)=\cos(xy)$, what is the best way to isolate $y$? Since $\sin(\frac{\pi}{2}) = \cos(\frac{\pi}{2})$ it would seem intuitive to say that $xy=\frac{\pi}{2}$ and thus that $y=\frac{\pi}{2x}$

Is this the correct approach, or am I missing something important?

$\endgroup$
  • 9
    $\begingroup$ Think you mean $\pi/4$ instead of $\pi/2$. $\endgroup$ – Edward Jiang Apr 20 '16 at 0:05
17
$\begingroup$

Hint: $$\tan {(xy)} =1$$ and $\tan ^{-1}$ both sides.

$\endgroup$
  • 11
    $\begingroup$ +1 but OP should note that $\tan^{-1}$ will not give the entire solution set, only the principle branch. To fix this note that $\tan$ is $\pi$-periodic. So if $z_0$ is a solution, so is $z_0 +n\pi$ for any $n \in \mathbb{Z}$. $\endgroup$ – MathMajor Apr 20 '16 at 0:05
  • $\begingroup$ @MathMajor, the solution is $xy = \frac{\pi}{4} + n \pi$ $\endgroup$ – user1543042 Apr 20 '16 at 0:10
11
$\begingroup$

$$sin(xy)=cos(xy)\iff xy= \frac{\pi}{4}+k\pi;\space k\in \mathbb Z$$

Thus $$\color{red}{y=\frac{(4k+1)\pi}{4x}}$$

$\endgroup$
  • $\begingroup$ Following the picky comment by Deepak: intermediate step(s) would be welcome in the first line. $\endgroup$ – Yves Daoust May 7 at 7:09
  • $\begingroup$ @YvesDaoust I've deleted my previous comments but I'll leave you with this: your jibe insinuating that I'm "picky" is unwarranted. Your original comment was inaccurate and I merely pointed it out. Why you are so unduly defensive about that, I have no clue. But if you want the last word here, feel free to seize it. $\endgroup$ – Deepak May 7 at 8:52

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.