Trig/ Pre Cal/ identities 2 owlbucks Is this correct? The problem \[ \cos^4 x -\sin^4 x = 2\cos^2 x - 1 \] \[ (\cos^2 x +\sin^2 x)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1\] \[ (1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1 \] \[ 2\cos^2 x - 1 = 2\cos^2 x - 1 \]

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Trig/ Pre Cal/ identities 2 owlbucks Is this correct? The problem \[ \cos^4 x -\sin^4 x = 2\cos^2 x - 1 \] \[ (\cos^2 x +\sin^2 x)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -\sin^2 x) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -(1-cos^2 x)) = 2\cos^2 x - 1 \] \[ (1)(\cos^2 x -(-1+cos^2 x)) = 2\cos^2 x - 1\] \[ (1)(\cos^2 x -1+cos^2 x) = 2\cos^2 x - 1 \] \[ 2\cos^2 x - 1 = 2\cos^2 x - 1 \]

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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
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?
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

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Awesome! Thank you. owbucks on the way :-)
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
Thanks for you time :-) @jim_thompson5910
You're welcome. I'm glad to be of help.
Jim is correct
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 \[(1)\cos^2 x -1+\cos^2 x = 2\cos^2 x - 1\] \[(1)\ 2cos^2 x -1= 2\cos^2 x - 1\]

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