0
$\begingroup$

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

Answers:

  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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $2\pi$: $100-32\pi$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

hint

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.