# How do you find the exact value of  arccos(cos (7pi)/6)?

Oct 28, 2015

Find the exact value of $\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right)$

Ans:$\left(\frac{5 \pi}{6}\right)$

#### Explanation:

$\cos \left(\frac{7 \pi}{6}\right) = \cos \left(\frac{\pi}{6} + \pi\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$
$\cos x = - \frac{\sqrt{3}}{2}$ --> arc $x = \left(\frac{5 \pi}{6}\right)$

Jul 23, 2016

$\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \frac{5 \pi}{6}$

#### Explanation:

Typically, the arccosine function works as such:

$\arccos \left(\cos \left(x\right)\right) = x$

So here, you would think that:

$\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \frac{7 \pi}{6}$

However, this is not true!

The range of the arccosine function, that is, the values that the arccosine function can spit out, is restricted from $\left[0 , \pi\right]$.

Since $\frac{7 \pi}{6} > \pi$, the arccosine function cannot spit this out as an answer.

The best way to think about this is that $\frac{7 \pi}{6}$ is an angle in the third quadrant with a reference angle of $\frac{\pi}{6}$.

Since cosine is negative in the third quadrant, we will need an angle with a reference angle of $\frac{\pi}{6}$ that is also negative, i.e., in the second or third quadrants, that fits in the range of $\left[0 , \pi\right]$.

Since to fit in that range the angle must be in the first or second quadrant, and since cosine is negative, we want the angle from the second quadrant.

The angle with a reference angle of $\frac{\pi}{6}$ in the second quadrant is $\frac{5 \pi}{6}$.

Thus:

$\arccos \left(\cos \left(\frac{7 \pi}{6}\right)\right) = \frac{5 \pi}{6}$