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anonymous

  • one year ago

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

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  1. jim_thompson5910
    • one year ago
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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

  2. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Awesome! Thank you. owbucks on the way :-)

  5. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Thanks for you time :-) @jim_thompson5910

  7. jim_thompson5910
    • one year ago
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    You're welcome. I'm glad to be of help.

  8. anonymous
    • one year ago
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    Jim is correct

  9. UsukiDoll
    • one year ago
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    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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