I'm going to be simplifying two functions, in the section below, to their exact values in terms of $x$. The first one being:




To solve $\tan(\theta)$ from here, I drew a triangle of $\cos(\theta)=x$ to solve for $\tan(\theta)$.

Which is $\frac{\sqrt{1-x^2}}{x}$ for $[-1,0)\cup(0,1]$

Now consider this other example $\tan(\arccos(x)+\pi )$

There are two ways to approach this problem from my understanding. You could make $\theta=\arccos(x)+\pi$ or alternatively you could make $\theta=\arccos(x)$

This problem was what me and my roommate made up for practice but we solved it in different ways, and this is what lead us to this strange observation.

I'm going to simplify it using $\theta=\arccos(x)+\pi$





Given that $\cos(\theta)=-x$, $\tan(\theta)=\frac{\sqrt{1-x^2}}{x}$ for $[-1,0)$ while it is $\tan(\theta)=\frac{\sqrt{1-x^2}}{-x}$ for $(0,1]$

Both my roommate and I came to the conclusion that this was the wrong answer. This was wrong because in the first example we established that $\tan(\theta)$ has a solution $\frac{\sqrt{1-x^2}}{x}$, therefore $\tan(\arccos(x)+\pi)$ should contain the same solution, as we translated the tan function by $\pi$, but the function has a period of $\pi$ so it should not change anything.

Now because of this observation we decided to work backwards to check if there was a mistake.






From here we continued to try to understand the differences in reasoning and answers.

Given that $\cos(\theta)=-x$ can be simplified to $\theta=\arccos(x)+\pi$, then it can also be simplified into $\arccos(-x)=\theta$ which would suggest that

$$\tan(\arccos(-x)) = \tan(\arccos(x)+\pi)$$

but putting both functions in Wolfram Alpha did not result in the same graph. So my question is why is this case? And what errors did I make in my calculations that would result in something happening like this?

Wolfram Alpha:



Sorry if this formatting is terrible, I did this late at night. I might fix it later.

  • $\begingroup$ Since $\theta$ lies in $[\pi,2\pi]$, $\sin\theta$ is always negative. So $\tan\theta=\dfrac{\sqrt{1-x^2}}{x}$ $\endgroup$ – AnotherJohnDoe Sep 19 '18 at 4:22
  • $\begingroup$ Also, $\cos\theta=-x\nRightarrow\theta=cos^{-1}(-x)$, because $\theta\ge\pi$ $\endgroup$ – AnotherJohnDoe Sep 19 '18 at 4:25

The range of $\arccos x$ is always in $[0,\pi]$

so $\tan \arccos x = \frac {\sqrt{1-x^2}}{x}$

however if you say $\cos \theta = x$ there are two values of $\theta$ in $[0,2\pi)$ that $\theta$ could take.

and $\tan \theta = \pm \frac{\sqrt{1-x^2}}{x}$ depending on which value of $\theta$ you choose.

You need to keep track of what quadrants you might be in. This seems to be the source of your problem.

The you could us the identity $\tan x+\pi = \tan x$

$\tan(\arccos x+ \pi) = \tan (\arccos x)$

Annother route would be to apply your angle addition rules

$\tan(\arccos x+ \pi) = \frac {\tan \arccos x + \tan \pi}{1-(\tan\arccos x)(\tan \pi)}$

and of course $\tan \pi = 0$


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