G Guest Guest Jun 20, 2006 #1 I need some help getting through the following trig problem: Find the exact value of cos^-1(cos(4pi/3)) Thanks
I need some help getting through the following trig problem: Find the exact value of cos^-1(cos(4pi/3)) Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Jun 20, 2006 #2 If I have read the problem correctly then: \(\displaystyle \L \arccos \left( {\cos \left( {\frac{{4\pi }}{3}} \right)} \right) = \arccos \left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}\)
If I have read the problem correctly then: \(\displaystyle \L \arccos \left( {\cos \left( {\frac{{4\pi }}{3}} \right)} \right) = \arccos \left( { - \frac{1}{2}} \right) = \frac{{2\pi }}{3}\)
