# Trig Functions...Simplifying..urgent help needed

• Nov 25th 2007, 07:32 PM
aikenfan
Trig 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 PM
Jhevon
Quote:

Originally Posted by aikenfan
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

you could note that both sides are $\cos 2x$, but that would be too easy.

instead, use the fact that $\sin^2 x + \cos^2 x = 1$. 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 PM
aikenfan
Do I do this for both problems?
• Nov 25th 2007, 07:45 PM
aikenfan
for number 5, I get
cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x
• Nov 25th 2007, 07:49 PM
Jhevon
Quote:

Originally Posted by aikenfan
for number 5, I get
cos^2x - sin^2x = sin^2x + cos^2x - 2sin^2x

no.

start with the left hand side only. forget about the right. then use the identity i showed you to transform it into the right hand side (or vice versa, start with the right side and simplify to get the left)
• Nov 25th 2007, 07:53 PM
aikenfan
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 PM
Jhevon
Quote:

Originally Posted by aikenfan
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

no, it's not correct. you still did not do what i asked.

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

consider the LHS

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

$= 1 - 2 \sin^2 x$

$= RHS$

or, we could consider the RHS:

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

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

$= LHS$

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

but you could just as easily simplify both sides to $\cos 2x$
• Nov 26th 2007, 02:03 PM
aikenfan
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 PM
Jhevon
Quote:

Originally Posted by aikenfan
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

what you did is incorrect. 1 - 2sin^2(x) is not 2cos^2(x)

here you should have changed the sin^2(x) into 1 - cos^2(x). you are aiming for something with only cos^2(x), why change the cosine to sine?
• Nov 27th 2007, 01:24 PM
aikenfan
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 AM
Jhevon
Quote:

Originally Posted by aikenfan
So do I take the 1 - sin^x - sin^2x and change it to:
1 - sin^2x - 1 - cos^2x?

no, change the sin^2 x into 1 - cos^2 x
• Nov 29th 2007, 02:34 PM
aikenfan
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 PM
Jhevon
Quote:

Originally Posted by aikenfan
Ok, I think that I have got it...i now have 1 - cos^2x - cos^x
and that simplifies to 1 - 2cos^2x ?

$\cos^2 x - \sin^2 x = \cos^2 x - \left( 1 - \cos^2 x \right)$

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

$= 2 \cos^2 x - 1$

as desired