This is probably just me misunderstanding trig properties. I remember from my trig days that $\tan(\arctan(x)) = x$ But I can't remember if that holds the other way. Can someone help me out? Is this valid:
$$\arctan(\tan(\theta)) = \theta$$
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1Just saying, the arctan of an angle is meaningless. – zzz Jan 27 '17 at 16:04
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2$\theta$ is a variable not exclusively reserved for angles. – Umberto P. Jan 27 '17 at 16:05
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@Zxu You're right that was just laziness on my part. I've edited, to make it clear that I'm not talking about an angle. – Jonathan Mee Jan 27 '17 at 16:16
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simple solution... draw a graph of the tangent function. Then draw the arctan function on both an x-y graph and a y-x graph. – John Joy Jan 29 '17 at 10:25
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@JohnJoy Yeah Rohan demonstrates that in his answer: http://math.stackexchange.com/a/2116642/194115http://math.stackexchange.com/a/2116642/194115 – Jonathan Mee Jan 30 '17 at 03:04
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It is true that $\tan\arctan x=x$, for every $x$. The converse is not true and it cannot be, because the tangent is not an injective function.
Recall that $\arctan x$ returns a number (an angle if you prefer) in the interval $(-\pi/2,\pi/2)$. So we have the equality $$ \arctan(\tan\theta)=\theta $$ if and only if $\theta\in(-\pi/2,\pi/2)$.
A formula can be given for any $\theta$: you just need to “reduce” the angle to the right interval by subtracting an integral multiple of $\pi$ so that $$ -\frac{\pi}{2}<\theta-k\pi<\frac{\pi}{2} $$ which is equivalent to $$ -\frac{1}{2}<\frac{\theta}{\pi}-k<\frac{1}{2} $$ or $$ 0<\frac{\theta}{\pi}+\frac{1}{2}-k<1 $$ or $$ k<\frac{\theta}{\pi}+\frac{1}{2}<k+1 $$ which means $$ k=\left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor $$ Thus $$ \arctan(\tan\theta)=\theta-\pi\left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor $$ (of course when $\tan\theta$ is defined to begin with).
For instance, if $\theta=15\pi/4$, we have $$ \left\lfloor\frac{\theta}{\pi}+\frac{1}{2}\right\rfloor= \left\lfloor\frac{15}{4}+\frac{1}{2}\right\rfloor=4 $$ and $$ \arctan\tan\frac{15\pi}{4}=\frac{15\pi}{4}-4\pi=-\frac{\pi}{4} $$
egreg
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I think that I understand everything here, except for why you're taking the floor. What's that about? – Jonathan Mee Jan 27 '17 at 18:10
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@JonathanMee If $k<x<k+1$, then $k=\lfloor x\rfloor$, we need to find $k$. – egreg Jan 27 '17 at 18:29
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@egreg: if theta = 2.5*pi, then arctan(tan(2.5pi))=pi/2. The equality provided yields -pi/2. Check use of ceiling vs. floor function. – 926reals Jun 21 '18 at 01:22
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@egreg: Thank you for answering. I see tan(5pi/2)=1/0 -> undefined. In approximation we have: limit as x->0 of (1/x) = +infinity; likewise, limit as x->-0 (1/x) = -infinity. We also have that Arctan(+infinity) = pi/2 and Arctan(-infinity)= -pi/2. In order to consolidate, I grapple the ceiling and floor functions. What I come across is that the following equation provides the correct solution: Arctan(tan(x))=x-pi*ceiling(x/pi-1/2). – 926reals Jun 22 '18 at 02:11
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It only equals $\theta $ if $$-\frac {\pi}{2}<\theta < \frac {\pi}{2} $$ as the range of $\arctan$ is only from $-\frac {\pi}{2} $ to $\frac {\pi}{2} $. If $\theta $ is outside this interval, then you would need to add or subtract $\pi $ from $\theta $ until you get to the angle in this interval that has the same value of $\tan$.
For instance, $\arctan (\tan \frac {\pi}{6}) = \frac {\pi}{6} $, but $\arctan (\tan \frac {3\pi }{4}) = -\frac {\pi}{4} $.
This can also been seen here:
Hope it helps.
