Find exact value of $\tan(\frac{17\pi}{12})$ using 2 methods, one of which fails.

When I try to use an identity to break this apart and get an exact value for it, I stumbled on an interesting snag which I would like to get an explanation for.

$\tan(\frac{17\pi}{12})$

So one way to break this apart is to break off the $\frac{17\pi}{12}$ into $\frac{8\pi}{12}$ (=$\frac{2\pi}{3}$)and $\frac{9\pi}{12}$ (=$\frac{3\pi}{4}$) then use the identity $$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$

Which works just fine and evaluates to $2+\sqrt3$

Now:

My understanding is that we should be able to break that original fraction off in any way we please and arrive at the answer. So let's do so in this way:

Break off the $\frac{17\pi}{12}$ into $\frac{18\pi}{12}$ (=$\frac{3\pi}{2}$) and $\frac{-\pi}{12}$

Now Notice that $\tan(\frac{3\pi}{2})$ is undefined due to the asymptote there [same as $\tan(\frac{\pi}{2})$].

So why is it that when we break it that way it gets to an undefined state? Are there any general rules I need to follow in breaking angles down? How come the above won't work that way?

-Thanks

Solution 1:

You can't directly substitute the $\frac{3\pi}{2}$ directly as the reasons you noted. However, we can take a limit. We know that $$\tan \left(\frac{17\pi}{12}\right)=\lim_{x\to \frac{18\pi}{12}} \frac{\tan x+\tan \left(-\frac{\pi}{12}\right)}{1-\tan x\tan\left(-\frac{\pi}{12}\right)}$$ Note that as $x\to \frac{18\pi}{12}$, we have $\tan x\to \infty$, so this is equivalent to $$\tan \left(\frac{17\pi}{12}\right)=\lim_{u\to \infty} \frac{u+\tan \left(-\frac{\pi}{12}\right)}{1-u\tan\left(-\frac{\pi}{12}\right)}$$ Using horizontal asymptote theorem (or you can conclude this with algebraic or other methods), we know that this is equivalent to $$\tan \left(\frac{17\pi}{12}\right)=\frac{1}{-\tan\left(-\frac{\pi}{12}\right)}$$ $$\tan \left(\frac{17\pi}{12}\right)=\frac{1}{\tan\left(\frac{\pi}{12}\right)}$$ However, this is also not too hard to prove geometrically (draw a unit circle), and is not much useful for calculating the value of $\tan\left(\frac{17\pi}{12}\right)$. The reason why we split $\frac{17\pi}{12}$ into $\frac{2\pi}{3}$ and $\frac{3\pi}{4}$ is because we already know the values of $\tan\left(\frac{2\pi}{3}\right)$ and $\tan\left(\frac{3\pi}{4}\right)$. On the other hand, using your method I'm assuming you don't already know the value of $\tan\left(\frac{\pi}{12}\right)$, so it's not as useful.