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

Printable View

- Nov 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 - Nov 25th 2007, 07:35 PMJhevon
you could note that both sides are $\displaystyle \cos 2x$, but that would be too easy.

instead, use the fact that $\displaystyle \sin^2 x + \cos^2 x = 1$. start with the left hand side of the equation and use the identity i typed abpve to replace either the sine or cosine to get the right hand side - Nov 25th 2007, 07:43 PMaikenfan
Do I do this for both problems?

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

cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x - Nov 25th 2007, 07:49 PMJhevon
- Nov 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 - Nov 25th 2007, 08:02 PMJhevon
no, it's not correct. you still did not do what i asked.

$\displaystyle \cos^2 x - \sin^2 x = 1 - 2\sin^2 x$

consider the LHS

$\displaystyle \cos^2 x - \sin^2 x = \left( 1 - \sin^2 x \right) - \sin^2 x$ ...........by the identity i showed you

$\displaystyle = 1 - 2 \sin^2 x$

$\displaystyle = RHS$

or, we could consider the RHS:

$\displaystyle 1 - 2 \sin^2 x = \sin^2 x + \cos^2 x - 2 \sin^2 x$

$\displaystyle = \cos^2 x = \sin^2 x$

$\displaystyle = LHS$

either way, we see that $\displaystyle \cos^2 x - \sin^2 x = 1 = 2 \sin^2 x$

but you could just as easily simplify both sides to $\displaystyle \cos 2x$ - Nov 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 - Nov 26th 2007, 05:37 PMJhevon
- Nov 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? - Nov 29th 2007, 08:04 AMJhevon
- Nov 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 ? - Nov 29th 2007, 09:17 PMJhevon