Find the exact radian value of $\displaystyle Arccos(\frac{-\sqrt{3}}{2})$

My answers: $\displaystyle \theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k

\pi$

is this right?

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- Mar 26th 2006, 01:57 PMc_323_hNeed double checking.
Find the exact radian value of $\displaystyle Arccos(\frac{-\sqrt{3}}{2})$

My answers: $\displaystyle \theta = \frac{5\pi}{6} + 2k\pi, \frac{7\pi}{6} + 2k

\pi$

is this right? - Mar 26th 2006, 02:02 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

$\displaystyle -\frac{\sqrt{3}}{2}$ to get that you need,

$\displaystyle x=\frac{5\pi}{6}$ thus, all the solutions are,

$\displaystyle x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}$ - Mar 26th 2006, 02:15 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Mar 26th 2006, 03:40 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

- Mar 26th 2006, 03:48 PMtopsquarkQuote:

Originally Posted by**ThePerfectHacker**

Your original answer is correct, c_323_h.

-Dan - Mar 26th 2006, 04:20 PMThePerfectHacker
Here are the two non-coterminal pairs.

$\displaystyle (\pi-\pi/6)+2\pi k$

$\displaystyle (\pi+\pi/6)+2\pi k$

Which simplifies to,

$\displaystyle -\pi/6+\pi(2k+1)$

$\displaystyle \pi/6+\pi(2k+1)$

Thus, all solutions are,

$\displaystyle \pm\frac{\pi}{6}+\pi k$ where $\displaystyle k\in\{...-5,-3,-1,1,3,5,...\}$ - Mar 26th 2006, 04:43 PMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Mar 26th 2006, 05:58 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

That means $\displaystyle k$ must be one of the "elements" of that set. - Mar 26th 2006, 06:05 PMc_323_h
wait a second. the range of arccos is $\displaystyle [0,\pi]$ since it is the inverse of cos. This means that arccos is only negative in II. so the only answer would be $\displaystyle \frac{5\pi}{6}$. is my reasoning correct? i was just thinking about the graph of arccos.

- Mar 27th 2006, 03:57 AMtopsquarkQuote:

Originally Posted by**c_323_h**

-Dan