Here's the question you clicked on:
hanbru
inverse cosine of square root of 3 over 2? cos-1(square root of 3/2)
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cos(pi/6)=square of 3/2
\[\text{ArcCos}\left[\frac{\sqrt{3}}{2}\right]=\frac{\pi }{6}\] Am not sure the square root of what was.
oops i missed my word root above
was the argument of the cosine (square root 3)/2 or square root (3/2)
i assumed it to be over just the numerator
Thanks for responding.
lol everyone needs love
rob look at the post before this? do you agree with my attachment?
not sure what thread you are referring to. do you have the web address?
ok you see all the questions right? right underneath this thread by hanbru there is galaxy (if sinx=1/3 and secy=5/4, where x and y lie btw 0 and pi/2, blah blah
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4ddf14c3c9a28b0b52ad63c2