# How do you simplify the expression arccos(cos ((7pi)/6))?

Sep 11, 2016

$\frac{7}{6} \pi$

#### Explanation:

Let y = cos x.

The period of cos x is $2 \pi$.

For $x \in \left[0 , 2 \pi\right]$, there are two points

$P \left(\frac{5}{6} \pi , - \frac{\sqrt{3}}{2}\right) \mathmr{and} Q \left(\frac{7}{6} \pi , - \frac{\sqrt{3}}{2}\right)$ that have the same

$y = - \frac{\sqrt{3}}{2}$.

I make 'as is where is' inversion.

Referred to P.

$- \frac{\sqrt{3}}{2} = \cos \left(\frac{5}{6} \pi\right)$. So,

$a r c \cos \left(- \frac{\sqrt{3}}{2}\right) = a r c \cos \cos \left(\frac{5}{6} \pi\right) = \frac{5}{6} \pi$.

Referred to Q,

$- \frac{\sqrt{3}}{2} = \cos \left(\frac{7}{6} \pi\right)$. So,

$a r c \cos \left(- \frac{\sqrt{3}}{2}\right) = a r c \cos \cos \left(\frac{7}{6} \pi\right) = \frac{7}{6} \pi$

Of course, the convention that $\frac{5}{6} \pi$ is the principal value# has to

be broken here. for the sake of local 1-1 mapping in the neighbor

hood of Q. We cannot move to P saying that it is convention. It is

my opinion that any convention is not the rule, for all

contexts/situations/applications

It is befitting to adopt ${f}^{- 1} f \left(x\right) = x$, for locally bijective locations

(x, y), with 1-1 correspondence at the point of continuity...
.