• Sep 22nd 2006, 01:40 PM
TomCat
Hey guys... I logged on today only to find my other thread full of a bunch of arguing and it was also locked. I thought I'd post my Prof's reply for you guys to see what, exactly we are doing.

Here's what my prof. had to say...
Quote:

Your question is a good one. We must change the right side of the
eqution into a fraction. Yes, we need to add (or subtract)
1/sin^2(theta) - 1
So we need to have the same denominator to get
1/sin^2(theta) - sin^2(theta)/sin^2(theta)
Together we form the quotient [1 - sin^2(theta)] / sin^2(theta)
At this point you will have one quotient equaling another quotient, so
you can do the reciprocal to both sides of the equation; and then take
the square root on both sides.

-Tom

PS: Here was my original post...

http://www.mathhelpforum.com/math-help/21282-post1.html
• Sep 22nd 2006, 09:44 PM
Jameson
Yes I locked the thread because it was becoming too heated. I hope your question is answered. :)
• Sep 22nd 2006, 11:12 PM
CaptainBlack
Quote:

Originally Posted by TomCat
Hey guys... I logged on today only to find my other thread full of a bunch of arguing and it was also locked. I thought I'd post my Prof's reply for you guys to see what, exactly we are doing.

[snip]

PS: Here was my original post...

http://www.mathhelpforum.com/math-help/21282-post1.html

I have cleaned up the original thread and you should now be able to
follow the gist of what is there without the distractions.

RonL