## anonymous one year ago Trig/ Pre Cal/ identities 2 owlbucks Is this correct? The problem $\cos^4 x -\sin^4 x = 2\cos^2 x - 1$ $(\cos^2 x +\sin^2 x)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1$ $(1)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1$ $(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1$ $(1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1$ $(1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1$ $2\cos^2 x - 1 = 2\cos^2 x - 1$

1. jim_thompson5910

You made an error when you got to $(1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1$ but you have the next step of $(1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1$ correct

2. anonymous

I took the middle - operator and distributed it across (1 - cos^2 x), which gave me $(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1$ $(1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1$ <--- so I should leave this part out?

3. jim_thompson5910

yeah you can go from $(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1$ to $(1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1$ so leave out that step

4. anonymous

Awesome! Thank you. owbucks on the way :-)

5. jim_thompson5910

You can say $(1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1$ $(1)(\cos^2 x -(1)-(-cos^2 x)) = 2\cos^2 x - 1$ but it's not 100% necessary

6. anonymous

Thanks for you time :-) @jim_thompson5910

7. jim_thompson5910

You're welcome. I'm glad to be of help.

8. anonymous

Jim is correct

9. UsukiDoll

I saw what happened... the distribution part was too much... everything was right until $(1)(\cos^2 x -(1-\cos^2 x)) = 2\cos^2 x - 1$ distribute the negative $(1)\cos^2 x -1+\cos^2 x = 2\cos^2 x - 1$ $(1)\ 2cos^2 x -1= 2\cos^2 x - 1$