How do I simplify these trig functions so that the right and left sides are the same?

http://i103.photobucket.com/albums/m...91919/11-1.jpg

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- November 25th 2007, 07:32 PMaikenfanTrig Functions...Simplifying..urgent help needed
How do I simplify these trig functions so that the right and left sides are the same?

http://i103.photobucket.com/albums/m...91919/11-1.jpg - November 25th 2007, 07:35 PMJhevon
- November 25th 2007, 07:43 PMaikenfan
Do I do this for both problems?

- November 25th 2007, 07:45 PMaikenfan
for number 5, I get

cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x - November 25th 2007, 07:49 PMJhevon
- November 25th 2007, 07:53 PMaikenfan
I did it this way, I don't know if this is correct, but it made sense and worked out...is this what you meant?

http://i103.photobucket.com/albums/m...1919/11001.jpg - November 25th 2007, 08:02 PMJhevon
- November 26th 2007, 02:03 PMaikenfan
I think that I have got it....but on the second problem, I am missing -1 in my final solution....

http://i103.photobucket.com/albums/m...1919/11002.jpg - November 26th 2007, 05:37 PMJhevon
- November 27th 2007, 01:24 PMaikenfan
So do I take the 1 - sin^x - sin^2x and change it to:

1 - sin^2x - 1 - cos^2x? - November 29th 2007, 08:04 AMJhevon
- November 29th 2007, 02:34 PMaikenfan
Ok, I think that I have got it...i now have 1 - cos^2x - cos^x

and that simplifies to 1 - 2cos^2x ? - November 29th 2007, 09:17 PMJhevon