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anonymous

  • one year ago

Trig / Pre Cal / identities These things are killing me. Please help guide me. Thank you. \[ (\sin^4 x-\cos^4 x) = 2\sin^2 x -1 \] \[ (\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 \] Am I on the right path?

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  1. Nnesha
    • one year ago
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    apply special identity \[\huge\rm sin^2 x + \cos^2 =1 \] solve for sin^2

  2. phi
    • one year ago
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    yes that is a good start

  3. anonymous
    • one year ago
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    so we have \[ (1)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1 \]

  4. phi
    • one year ago
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    I would next replace the cos^2

  5. Nnesha
    • one year ago
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    sin^2 x + cos ^2 = 1 solve this identity for cos^2

  6. anonymous
    • one year ago
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    \[ (1)(\sin^2 x + 1-\sin^2 x )= 2\sin^2 x -1 \]

  7. Nnesha
    • one year ago
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    \[ \rm (\sin^2 x-(\color{reD}{{1-sin^2x}}))\] write that in the parentheses and distribute by negative one

  8. phi
    • one year ago
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    close, but you lost a minus sign -(1 - sin^2)

  9. anonymous
    • one year ago
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    \[ (1)(\sin^2 x - (1-\sin^2 x))= 2\sin^2 x -1 \] Now what?

  10. phi
    • one year ago
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    -(a +b) means -1*(a+b) = -1*a + -1*b

  11. anonymous
    • one year ago
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    Ah so \[ (1)(\sin^2 x - 1+\sin^2 x )= 2\sin^2 x -1 \] I see it now.

  12. anonymous
    • one year ago
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    Thank you both!!!

  13. Nnesha
    • one year ago
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    :-)

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