MHB Inverse trigonometric functions

[Guest2]

What's $1. ~ \displaystyle \arccos(\cos\frac{4\pi}{3})?$ Is this correct?

The range is $[0, \pi]$ so I need to write $\cos\frac{4\pi}{3}$ as $\cos{t}$ where $t$ is in $[0, \pi]$

$\cos(\frac{4\pi}{3}) = \cos(2\pi-\frac{3\pi}{3}) = \cos(\frac{2\pi}{3}) $ so the answer is $\frac{2\pi}{3}$

 

[Greg]

There's an error in your last line.

Using symmetry,

$\cos\dfrac{4\pi}{3}=\cos\left(\pi-\dfrac{\pi}{3}\right)=\cos\dfrac{2\pi}{3}$

 

[Guest2]

There's an error in your last line.

Using symmetry,

$\cos\dfrac{4\pi}{3}=\cos\left(\pi-\dfrac{\pi}{3}\right)=\cos\dfrac{2\pi}{3}$

Thanks. My reasoning was that cosine has a period $2\pi$. Where have I messed up?

 

[Prove It]

What's $1. ~ \displaystyle \arccos(\cos\frac{4\pi}{3})?$ Is this correct?

The range is $[0, \pi]$ so I need to write $\cos\frac{4\pi}{3}$ as $\cos{t}$ where $t$ is in $[0, \pi]$

$\cos(\frac{4\pi}{3}) = \cos(2\pi-\frac{3\pi}{3}) = \cos(\frac{2\pi}{3}) $ so the answer is $\frac{2\pi}{3}$

$\displaystyle \begin{align*} \arccos{ \left[ \cos{ \left( \frac{4\,\pi}{3} \right) } \right] } &= \arccos{ \left[ \cos{ \left( \pi + \frac{\pi}{3} \right) } \right] } \\ &= \arccos{ \left[ -\cos{ \left( \frac{\pi}{3} \right) } \right] } \\ &= \arccos{ \left( -\frac{1}{2} \right) } \\ &= \pi - \arccos{ \left( \frac{1}{2} \right) } \\ &= \pi - \frac{\pi}{3} \\ &= \frac{2\,\pi}{3} \end{align*}$

 

[Greg]

My reasoning was that cosine has a period $2\pi$. Where have I messed up?

You typed a '3' where you probably intended a '4'.