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How does one compute: $$\tan^{-1}\left [ \frac{\cos\left ( 2t \right )}{\sin\left ( 2t \right )} \right ]$$

I came across this in a text and I cannot figure out how the author arrive at the solution $2t + \frac{\pi}{2}$

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closed as off-topic by Namaste, Claude Leibovici, Leucippus, kingW3, user91500 Jul 15 '17 at 15:45

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  • $\begingroup$ Which text did you find this in? Have you tried? If yes, please show some of the workings you came up with. $\endgroup$ – Namaste Jul 15 '17 at 13:50
  • $\begingroup$ A text in chaos theory. Admittedly, simple trig stuffs from high school can be hard to recall. $\endgroup$ – Mathematicing Jul 15 '17 at 13:54
  • $\begingroup$ Mathematicing You might want to revisit the question and its answers, one of which is incorrect. $\endgroup$ – Namaste Jul 15 '17 at 18:30
  • $\begingroup$ @amWhy Haven't the below poster rectified the mistake? $\endgroup$ – Mathematicing Jul 16 '17 at 2:34
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Attention: This answer was corrected, thanks for Rory Daulton's heads up!

We know that:

$$\frac{\cos(2t)}{\sin(2t)}=\cot(2t)=\tan\left(\frac{\pi}{2}-2t\right)=-\tan\left(2t-\frac{\pi}{2}\right)$$

And also that:

$$\tan^{-1}(\tan(x))=x$$

So, the correct answer should be:

$$\frac{\pi}{2}-2t$$

Sorry for my previous mistaken anwser! For further info, refer to Rory's answer

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    $\begingroup$ That equality from cotangent to tangent is not true, almost-always. It becomes true if you put a minus sign in front of either the cotangent or the tangent. The OP's stated solution is not correct. Try graphing the original and the final expressions to see this. $\endgroup$ – Rory Daulton Jul 15 '17 at 13:59
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    $\begingroup$ I agree, @Rory. This is a very careless answer, and also, it is incorrect. $\endgroup$ – Namaste Jul 15 '17 at 18:20
  • $\begingroup$ @RoryDaulton Thank you very much for the heads up! I'm going to add a correction to it right now! I had the $\cot(x)=\tan\left(\frac{\pi}{2}-x\right)$ on my mind, but I was completely induced by the OP's suggested answer. $\endgroup$ – bertozzijr Jul 15 '17 at 20:21
  • $\begingroup$ actually, the correct answer is $\dfrac{\pi}{2} - 2t \pmod \pi$ $\endgroup$ – steven gregory Jul 16 '17 at 4:20
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The author's solution is incorrect. Perhaps you typed the question or the answer wrongly here?

$$\begin{align} \tan^{-1}\left[\frac{\cos(2t)}{\sin(2t)}\right] & = \tan^{-1}\left[\cot(2t)\right] \\ & = \tan^{-1}\left[\tan\left(\frac{\pi}2 -2t\right)\right] \\ & = \frac{\pi}2 -2t \\ \end{align}$$

That last equation is true only if $-\frac{\pi}2 \le \frac{\pi}2 -2t \le \frac{\pi}2$ --i.e. for $0\le t\le \frac{\pi}2$. That limitation may be true in the context of the problem in your book. But your given "solution" is wrong.

Here is a graph showing the equality and where it is true--the tick marks on the x-axis are at multiples of pi over 2.

enter image description here

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