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Math Help - Need help verifying identities

  1. #1
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    Need help verifying identities

    I was given a homework sheet to verify several identities, but I could not figure out these two problems. Can someone enlighten me.


    1. cos 4x= 1-8sin²x cos²x

    2. (sin³x - cos³x)/(sinx - cosx) = 1+sinxcosx
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  2. #2
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    1. cos 4x= 1-8sin²x cos²x

    Right Side = 1 - 2 (4sin²x cos²x) = 1 - 2 (2sin x . cos x)²

    = 1 - 2 (sin 2x)² = (1)- 2 sin² 2x = (sin² 2x + cos² 2x) -2sin² 2x

    = cos² 2x - sin² 2x = cos 4x

    2. (sin³x - cos³x)/(sinx - cosx) = 1+sinxcosx

    Left side = \frac {\sin^3 x -\cos^3 x}{\sin x - \cos x}

    = \frac {(\sin x -\cos x)(\sin^2 x +\sin x . \cos x + \cos^2 x)}{\sin x - \cos x}

    because a^3-b^3 = (a-b)(a^2+ab+b^2)

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

    = 1+ \sin x . \cos x
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  3. #3
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    \cos(4x) =

    \cos[2(2x)] =

    1 - 2\sin^2(2x) =

    1 - 2(2\sin{x}\cos{x})^2 =

    finish it.


    hint for the second one ...

     a^3 - b^3 = (a-b)(a^2+ab+b^2)
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  4. #4
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    ahh, I see now. Thank you guys.
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  5. #5
    Junior Member toraj58's Avatar
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    from left:

    \cos4x = \cos2(2x)
    = \cos^22x - \sin^22x
     = (\cos^2x-sin^2x)^2 - (2\sin x\cos x)^2
     = (\cos^4x+\sin^4x)-2\cos^2x\sin^2x-4\sin^2x\cos^2x

    since we have: (\cos^4x+\sin^4x) = 1-2\sin^2x\cos^2x
    but why:
    simply we can approve it like this:

    (\sin^2x+\cos^2x)^2 = \sin^4x+\cos^4x+2\sin^2x\cos^2x
    and we know that: (\sin^2x+\cos^2x) = 1

    we replace it and:

     = (1-2\sin^2x\cos^2x)-2\cos^2x\sin^2x-4\sin^2x\cos^2x

    finally we simplifying it:

     = 1-8\sin^2x\cos^2x
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