What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$ The Question 
$\large\cos^{-1}(\cos 10)$

My Approach
$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$ 
$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$. 
So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$
But this answer is wrong and the correct answer is $\large 4\pi - 10$.
They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?
Kindly help.
 A: If $x$ is in the third quadrant, its symmetric w.r.t. the $x$-axis is in the second quadrant, and has the same cosine. This symmetric point has curvilinear abscissa $6\pi - x$, as you've found. To determine the value which is represented by the same point on the trigonometric circle, let's just make some elementary manipulations on inequalities.
If $x=10$ˆ, we have $3\pi < x <\dfrac{7\pi}2$, so
\begin{align}
&&\frac{5\pi}2 &<6\pi -x< 3\pi  \\[1ex]
&\llap{\text{whence}}\qquad& \frac \pi 2 &<(6\pi-x)-2\pi=4\pi -x <\pi&&\qquad
\end{align}
The solution follows.
A: I suppose that $10^c$ means 10 radians. 
$\pi^c=180^o\implies 1^c \sim57^o\implies 10^c\sim 570^o$
$\cos 10 =\cos (10-2\pi)=-\cos(10-3\pi)=\cos (\pi-(10-3\pi))$
$\arccos(\cos 10)=4\pi-10$ 
Your answer turns out to be wrong as $\arccos(\cos x)=x$ when $x\in [0,\pi]$. Note that $3\pi-10\notin [0,\pi]$.
A: Note
$$
\cos^{-1}(\cos(x)) =\begin{cases}
2\pi n +x\in (0,\frac\pi2)\>\>\>x \text{ in 1st quadrant}\\
2\pi n +x\in (\frac\pi2,\pi) \>\>\>x \text{ in 2nd quadrant}\\
2\pi n -x\in (\frac\pi2,\pi) \>\>\>x \text{ in 3rd quadrant}\\
  2\pi n -x\in (0,\frac\pi2)\>\>\>x \text{ in 4th quadrant}\\
\end{cases}$$
In particular, if  $x$ is in the 3rd quadrant
$$\cos^{-1}(\cos(x)) = 
\cos^{-1}(-|\cos(x)|) 
=  \pi -\cos^{-1}(|\cos(x)|)\\
= \pi -\cos^{-1}(\cos(\pi+ x-2\pi n))
=2\pi n -x\in (\frac\pi2,\pi)$$
Thus, for $x=10$ in the 3rd quadrant, let $n=2$ to obtain
$$\cos^{-1} (\cos(10))= 4\pi -10\in (\frac\pi2,\pi)
$$
A: First, note that $\arccos(\cos(x)) =x$ only if $x \in [0,\pi]$. Note that arccos(x) is inverse of cos(x).
Thus, we need to get $10$ into that range. We can subtract periods to get $10-4\pi$, but this is a negative number $\approx -2$, not in the range, which was your mistake. So, we can use the identity that $\cos(x) = \cos(-x)$, so $\cos(10-4\pi) = \cos(4\pi-10)$. $4\pi - 10$ is in range, so we are good there. Now, we need to calculate $\arccos(\cos(4\pi-10))$. However, since $4\pi - 10 \approx 2.5$, which is in the range $[0,\pi]$, we know that $\arccos(\cos(4\pi-10))$ is just $4\pi-10$.
A: The angle $\theta$ that you're looking for corresponds to the point (on the unit circle) directly above the black dot in your second drawing, which corresponds to the angle $10-2\pi$. The average of these two angles is $\pi$, so from
$${1\over2}(\theta+(10-2\pi))=\pi$$
we find $\theta=4\pi-10$.
