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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?

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    $\begingroup$ Think you mean $\pi/4$ instead of $\pi/2$. $\endgroup$ Commented Apr 20, 2016 at 0:05

2 Answers 2

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

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    $\begingroup$ +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
    Commented Apr 20, 2016 at 0:05
  • $\begingroup$ @MathMajor, the solution is $xy = \frac{\pi}{4} + n \pi$ $\endgroup$ Commented Apr 20, 2016 at 0:10
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$$sin(xy)=cos(xy)\iff xy= \frac{\pi}{4}+k\pi;\space k\in \mathbb Z$$

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

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  • $\begingroup$ Following the picky comment by Deepak: intermediate step(s) would be welcome in the first line. $\endgroup$
    – user65203
    Commented May 7, 2019 at 7:09
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    $\begingroup$ @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
    Commented May 7, 2019 at 8:52

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