# Question #34ccf

May 30, 2017

$\theta = \frac{\pi}{3}$ if $\cos \left({\cos}^{-} 1 \left(u\right)\right) = u$

#### Explanation:

First notice that we can deduct some things from the equation:

$\cos \left({\cos}^{-} 1 \left(- \frac{\pi}{3}\right)\right)$

We can't have ${\cos}^{-} 1 \left(- \frac{\pi}{3}\right)$ , it is outside the domain of the function $f \left(x\right) = {\cos}^{-} 1 \left(x\right)$.
Howeve, because ${\cos}^{-} 1 \left(\cos \left(x\right)\right) = x$ , the only way I see to move on, is to apply the same concept.

That means the equation from above tells us, that at some point, some angle was converted into $- \frac{\pi}{3}$ radiens.

Imagine $- \frac{\pi}{3}$ radiens on the unit circle. This angle is not within $0 \le \theta \le \pi$ . In fact this translates to the upper half of the unit circle (see illustrations). You can see that:
$C o s \left(- \frac{\pi}{3}\right) = \cos \left(2 \pi - \frac{\pi}{3}\right)$

Now calculate $\cos \left(2 \pi - \frac{\pi}{3}\right) = 0.5$
Using your prefered graph tool, you can also see that $\cos \left(x\right) = 0.5$ has many more solutions. One particular solution exist in between $\cos \left(- \frac{\pi}{3}\right)$ and $\cos \left(2 \pi - \frac{\pi}{3}\right)$, that is $\cos \left(\frac{\pi}{3}\right)$.
This means:
$\cos \left(- \frac{\pi}{3}\right) = \cos \left(2 \pi - \frac{\pi}{3}\right) = \cos \left(\frac{\pi}{3}\right)$

As you can see:
only one solution satisfies, $0 \le \theta \le \pi$ , that is $\theta = \frac{\pi}{3}$

Imagine $\theta = \frac{\pi}{3} = {\cos}^{-} 1 \left(\cos \left(\frac{\pi}{3}\right)\right) = {\cos}^{-} 1 \left(\cos \left(- \frac{\pi}{3}\right)\right)$ , because $\cos \left(\frac{\pi}{3}\right) = \cos \left(- \frac{\pi}{3}\right)$

and so if $\cos \left({\cos}^{-} 1 \left(u\right)\right) = u$ then we can write:
$u = - \frac{\pi}{3}$

${\cos}^{-} 1 \left(\cos \left(- \frac{\pi}{3}\right)\right) = {\cos}^{-} 1 \left(\cos \left(u\right)\right) = u = \cos \left({\cos}^{-} 1 \left(u\right)\right) = \cos \left({\cos}^{-} 1 \left(- \frac{\pi}{3}\right)\right)$

To conclude, when $\theta = \frac{\pi}{3}$ , it satisfies $0 \le \theta \le \pi$ , and that can be rewritten to form the equation $\cos \left({\cos}^{-} 1 \left(- \frac{\pi}{3}\right)\right) = \frac{\pi}{3}$ if $\cos \left({\cos}^{-} 1 \left(u\right)\right) = u$

Idon't know... It makes sense for me to do it this way, but I would like someone to double check it for fundamental flaws :)