# How do you evaluate cos^(-1) [cos((-5pi)/3)]?

Jul 21, 2017

$\frac{- 5 \pi}{3}$

#### Explanation:

${\cos}^{-} 1 \left[\cos \left(\frac{- 5 \pi}{3}\right)\right]$

Let ${\cos}^{-} 1 \left[\cos \left(\frac{- 5 \pi}{3}\right)\right] = \theta$ , then

$\cos \theta = \cos \left(\frac{- 5 \pi}{3}\right)$

$\therefore \theta = \frac{- 5 \pi}{3}$

$\therefore {\cos}^{-} 1 \left[\cos \left(\frac{- 5 \pi}{3}\right)\right] = \frac{- 5 \pi}{3}$ [Ans]

Jul 22, 2017

${\cos}^{-} 1 \left(\cos \left(\frac{- 5 \pi}{3}\right)\right) = \frac{\pi}{3}$

#### Explanation:

Here's the trick to this problem: ${\cos}^{-} 1 \left(\theta\right)$ is only defined on the interval $0 \le \theta \le \pi$.

So, what this problem is asking us to do is find the angle between $0$ and $\pi$ whose cosine is the same as the cosine of $\frac{- 5 \pi}{3}$.

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Adding $2 \pi$ to an angle doesn't change any of its trig functions, so we can say that:

$\cos \left(\frac{- 5 \pi}{3}\right) = \cos \left(\frac{- 5 \pi}{3} + 2 \pi\right)$

$\cos \left(\frac{- 5 \pi}{3}\right) = \cos \left(\frac{\pi}{3}\right)$

And since $\frac{\pi}{3}$ is between $0$ and $\pi$, we can say that:

${\cos}^{-} 1 \left(\cos \left(\frac{- 5 \pi}{3}\right)\right) = \frac{\pi}{3}$