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-   -   Proving Trig Identities (https://www.askmehelpdesk.com/showthread.php?t=143757)

  • Oct 22, 2007, 06:41 PM
    tali31
    Proving Trig Identities
    how would you prove sin4x - cos4x = sin2x-cos2x as well as sin4x + 2sin2xcos2x +cos4x = 1
    :confused:
  • Oct 22, 2007, 06:46 PM
    s_cianci
    In both cases, you'll want to factor the left side of the equation. Treat sin^4 x - cos^4 x as a difference of two squares and treat sin^4 x + 2sin^2x cos^2x + cos^4x as a perfect-square trinomial. Also the 1st. Pythagorean Identity will come into play ; sin^2x + cos^2x = 1.
  • Oct 22, 2007, 11:03 PM
    terryg752
    I think you have written the question incorrectly.

    These identities are not correct.

    I think you should have written:

    1. (Sinx)^4 - (Cos x)^4

    = [ (sinx)^2 - (cos)^2 ] [ (sinx)^2 + (Cos x)^2) ]

    = (sinx)^2 - (cos)^2

    2. sin^4 x + 2 sin^2 co^2 x + cos^4 x

    = (sin^2 x + cos^2 x)^2

    = 1^2

    = 1

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