This is pretty much the last thing I need to know for now.

Tasks (calculate):

  1. $\arccos{(\cos{12})}$

  2. $\arctan{(\tan{\sqrt{5}})}$

  3. $\arcsin{(\sin{100})}$


  1. $4\pi-12$
  2. $\sqrt{5}-\pi$
  3. $100-32\pi$

All the answers are a little bit weird too "complicated" for me to deduce proper way on how to approach these tasks.

If only I had $\pi$ values in there, for example: $\arccos{(\cos{(12\pi)})}$, then I would know how to apply "the shift" rule in there.

Thanks for taking your time in educating total noob. :D

  • $\begingroup$ How would you apply "the shift" if the argument were a multiple of $\pi$? What prevents you from doing the same thing when it isn't? $\endgroup$ – Chris Culter Jan 23 at 20:16

The first thing you would want to do is see what signs $\cos(12)$, $\tan\left(\sqrt{5}\right)$, and $\sin(100)$ have. From there, using the range of inverse functions, you find the answer.

For the first example, $\dfrac{7\pi}{2} < 12 < 4\pi$, so the angle lies in quadrant $4$. Cosine is positive there, and since the range of $\arccos(x)$ is $y \in(0, \pi)$, the angle in quadrant $1$ is desired: $4\pi-12$.

For the second example, $\dfrac{\pi}{2} < \sqrt{5} < \pi$, so the angle lies in quadrant $2$. However, recall that the range of $\arctan(x)$ is $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so the quadrant $4$ angle is desired. Also recall that $\tan(x)$ is periodic by $\pi$ radians, so if $\sqrt{5}$ is in quadrant $2$, then $\sqrt{5}+\pi$ is in quadrant $4$. Note that this angle is also represented by $\sqrt{5}-\pi$, as they are $2\pi$ radians apart.

For the third example, I would start by finding the greatest integer $n$ such that $100-n\pi > 0$. From here, $n < \dfrac{100}{\pi} \approx 31.831$. The greatest possible integer $n$ becomes $31$. This means there have been $15$ full revolutions. Simplify by subtracting $30\pi$ ($2\pi$ for each revolution) from the angle: $100-30\pi \approx 5.7522$. You can conclude that this angle lies in quadrant $4$ as $\dfrac{3\pi}{2} < 100-30\pi < 2\pi$. Fortunately, the range of $\arcsin(x)$ is $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $100-30\pi$ immediately becomes the answer. Once again, you may also use the negative value for the quadrant $4$ angle by subtracting $2\pi$: $100-32\pi$.

  • $\begingroup$ About the first example, can I denote it as $12 - 4\pi$ instead of $4\pi - 12$? $\endgroup$ – weno Jan 23 at 21:15
  • $\begingroup$ That would be the quadrant $4$ angle, which isn’t included in the range of $\arccos(x)$. $\endgroup$ – KM101 Jan 23 at 21:18


If $X\in [0,\pi]$, then

$$\arccos(\cos(X))=X=$$ $$\arccos(\cos(X+2k\pi))=$$ $$\arccos(\cos(-X+2k\pi))$$

observe that $$4\pi-12\in[0,\pi]$$


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