how to solve arctan[tan(7pi /4)]

Question

zepdrix · Answer

$$\Large \arctan\left[\tan(7\pi/4)\right]$$So work from the inside out.
Do you know what tan(7pi/4) gives us? :)

ybarrap · Answer

Recall that arctan(tan(x)) = x, because arctan is the inverse function for tan.

ybarrap · Answer

Does this make sense? What is "x" in your equation?

zepdrix · Answer

@ybarrap 
Gotta be careful with that method since tangent is an even function! :D
Luckily we were given an angle in the 4th quadrant so it will work out just fine this time.

zepdrix · Answer

Just as an example:
$$\Large \arctan\left[\tan(5\pi/4)\right] \ne 5\pi/4$$

zepdrix · Answer

This idea applies to odd functions.
Tangent is an odd function XD lol sorry typo.

ybarrap · Answer

Good point.

ybarrap · Answer

What is going on here is as follows.

tan(3pi/4) is in the 2nd quadrant, so it equals $ -\sqrt{2}/2$.
Since arctan(-x)=-arctan(x), we have arctan($ -\sqrt{2}/2)=$-arctan($\sqrt{2}/2)$
But arctan($\sqrt{2}/2)$=$\pi/4$, so arctan($ -\sqrt{2}/2)=$ $-\pi/4$, taking into account the fact that arctan(x) is odd.

So the rule for arctan(x)   is:
arctan(tan(-x))=arctan(-tan(x))=-arctan(tan(x)), taking into account the oddness of the tan and arctan functions.
For even functions, you can just do $ f^{-1}(f(x))=x$ and not worry about the sign.

ybarrap · Answer

You have $arctan(tan(7\pi/4))$. Since $7\pi/4$ is in the same as $-\pi/4$

$arctan(tan(7\pi/4))\=arctan(tan(-\pi/4))\=arctan(-tan(\pi/4))\=-arctan(tan(\pi/4))\=-\pi/4\=7pi/4$

anonymous · Answer

omg so much information. I am looking at 6 different charts trying to get this to make perfect sense. I understand where you get these numbers from when I look at the charts. is this mostly memorizing these charts? 

It is slowly making sense. Thank you for your help!! :)

ybarrap · Answer

There is logic to it all that helps to remember things.
You're welcome.