# Calculating exact values of “weird” functions like arcsin(sin 100)

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

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

• 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? – 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 $$2\pi$$: $$100-32\pi$$.

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

hint

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

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

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