# How do you calculate Arccos(cos 7pi/2)?

Nov 21, 2015

arcos(cos) cancel each other out so $\text{arcos} \left\{\cos \left(\frac{7}{2} \pi\right)\right\} = \frac{7}{2} \pi$

#### Explanation:

The problem with trig is that unless the range is defined there are multiple answers as the cycles just keep on going. So technically it should be written as $n \frac{7}{2} \pi$

Note that if you rotate $2 \pi$ radians you have completed 1 cycle and you are back where you started. I suppose that if you are talking about work done (physics) the number of rotations is important. However, if you are talking purely numbers then it is not so critical

$\frac{7}{2} \pi$ radians is actually defining the number of rotations. However, just being interested in the angular measure of rotational state you would have to take into account that $2 \pi \text{ radians } = {360}^{o} \equiv 1$ full cycle

$\textcolor{b r o w n}{\text{Assumption: rotation is anticlockwise:}}$

So $\frac{7}{2} \pi$ radians is $\left(\frac{7}{2} \pi \div i \mathrm{de} 2 \pi\right) = \frac{7}{4}$ cycles$= 1 \frac{3}{4}$cycles.

Assuming you start at the standard point of ${0}^{o} ,$ you end up at the same point as$\frac{3}{4}$ cycles

$\frac{3}{4} \times {360}^{o} = {270}^{o} = \frac{3}{4} \times 2 \pi = \frac{3}{2} \pi \textcolor{w h i t e}{.}$radians. This is the same as$- \frac{1}{2} \pi$ radians

$\textcolor{b l u e}{\text{In this solution I have assigned negative to be}}$
$\textcolor{b l u e}{\text{clockwise rotation and obviously positive to be anticlockwise}}$

Nov 21, 2015

$\arccos \left(\cos \left(\frac{7 \pi}{2}\right)\right) = \frac{\pi}{2}$

#### Explanation:

$\cos \left(\frac{7 \pi}{2}\right) = \cos \left(- \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2}\right) = 0$

In general $\arccos \left(x\right)$ is an angle $\theta$
$\textcolor{w h i t e}{\text{XXX}}$such that $0 \le \theta < \pi$
$\textcolor{w h i t e}{\text{XXX}}$and $\cos \left(\theta\right) = x$

$\arccos \left(\frac{7 \pi}{2}\right) = \arccos \left(0\right) = \frac{\pi}{2}$