Find $\alpha$, given:

  • $\sin(\alpha + \frac{\pi}{6})$

  • $\tan(\alpha)=-5$

  • And terminal side is in Quadrant II

I only got as far as figuring out that the coordinate for the point in QII is $(-1,5)$ and the length of that segment is $\sqrt{26}$. But I'm stuck in getting $\alpha$.

  • $\begingroup$ ...what value is the sine of alpha plus a sixth of pi (the first one)? $\endgroup$ – Parcly Taxel Aug 24 '16 at 14:10
  • $\begingroup$ that is the only information given. $\endgroup$ – Paul Aug 24 '16 at 14:46
  • $\begingroup$ Isn't $\alpha$ simply $\tan^{-1}(-5)$? $\endgroup$ – Parcly Taxel Aug 24 '16 at 14:47
  • $\begingroup$ That's what I was thinking. I'll ask my professor. Thanks! $\endgroup$ – Paul Aug 24 '16 at 15:07

Unfortunately, it's not quite as simple as taking the arctangent of $\tan(\alpha)$. Were your point in Quadrant I, it would be that simple, but that's not the case. On the bright side, it's not a whole lot more complicated than that.

Well, maybe a little more complicated.

The answer you're looking for is $$\alpha=\arctan(-5)+\pi$$ I could just leave it at that and say that the $+\hspace{1mm}\pi$ comes from some magical property of the arctangent function, but that's no fun. The fun part is why we need to add some integer multiple to the output of the arctangent function to get the correct answer, and why that integer changes depending on what quadrant the angle you want to find happens to be in.

Call that integer $n$. In Quadrant I, $n=0$. In Quadrants II and III, $n=1$, and in Quadrant IV, $n=2$ (or $0$). To get a feeling for why this is, let's look at the graph of $y=\arctan(\tan(x))$. I've added dashed lines at odd integer multiples of $\frac{\pi}{2}$, and the regions between them correspond to $n=-2$, $n=-1$, $n=0$, $n=1$, and $n=2$, respectively, from left to right.


Seeing as $\arctan$ and $\tan$ are inverse functions, we would expect the graph of $y=\arctan(\tan(x))$ to look like the graph of $y=x$. That's clearly not the case; instead, it looks like the graph of $y=x$ has been split up into $\frac{\pi}{2}\times\frac{\pi}{2}$ chunks that are then all lined up on the $x$-axis.

The reason why the above graph looks the way it does has to do with how we define the arctangent function. The tangent function isn't inherently one-to-one, so in order to define its inverse, we need to restrict its domain to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, which means that the range of the arctangent function is also $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. Unfortunately, angles greater in magnitude than $\frac{\pi}{2}$ do exist. So what's going to happen when you want the arctangent function to produce one of those angles? It'll produce some number in its range that is an integer multiple of $\pi$ away from the angle you're looking for.

For example, $\arctan(\tan(2\pi))$ should obviously return $2\pi$, right? Instead, the result is $0$, which, by some stroke of sheer mathematical luck, happens to be an integer multiple of $\pi$ away from $2\pi$ (joking a little here). For a more interesting example, $\arctan\left(\tan\left(\frac{7\pi}{4}\right)\right)$ returns $-\frac{\pi}{4}$. Since $\frac{7\pi}{4}$ and $-\frac{\pi}{4}$ are equivalent on the unit circle, this might be fine in some settings, but in others, all you need to do is add $2\pi$ to get back to $\frac{7\pi}{4}$.

Finally, notice that the act of adding integer multiples of $\pi$ to your answers here is like moving the line segments in the graph I included up or down by integer multiples of $\pi$ to reconstruct the graph of $y=x$.


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