# Need double checking.

• Mar 26th 2006, 01:57 PM
c_323_h
Need 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 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Find the exact radian value of $\displaystyle cos^-1(\frac-{\sqrt3}{2})$

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

is the right?

Notice that,
$\displaystyle -\frac{\sqrt{3}}{2}$ 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 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
Notice that,
$\displaystyle -\frac{\sqrt{3}}{2}$ to get that you need,
$\displaystyle x=\frac{5\pi}{6}$ thus, all the solutions are,
$\displaystyle x=\frac{5\pi}{6}+\pi k,k\in\mathbb{Z}$

why pi and not 2pi?
• Mar 26th 2006, 03:40 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
why pi and not 2pi?

• Mar 26th 2006, 03:48 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker

ThePerfectHacker, $\displaystyle cos(5 \pi/6 + \pi) = + \sqrt3/2$ so your form for the solution is not correct. :eek:

-Dan
• Mar 26th 2006, 04:20 PM
ThePerfectHacker
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 PM
c_323_h
Quote:

Originally Posted by ThePerfectHacker
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 $\displaystyle k\in\{...-5,-3,-1,1,3,5,...\}$

what does $\displaystyle \in$ mean in layman's terms?
• Mar 26th 2006, 05:58 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
what does $\displaystyle \in$ mean in layman's terms?

It means "element of"

That means $\displaystyle k$ must be one of the "elements" of that set.
• Mar 26th 2006, 06:05 PM
c_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 AM
topsquark
Quote:

Originally Posted by c_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.

Technically you are correct here. However, in all cases that I have seen, when you are asked to solve this kind of equation the solution set is assumed to be over the whole real line, not just the range given by arccos.

-Dan