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Math Help - Please help with this...

  1. #1
    Bar0n janvdl's Avatar
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    Please help with this...

    The question says:

    Find the value of Sin(A + B) if Cos A = 2/3 and Sin B = 3/4

    We know that Sin(A + B) = SinA Cos B + CosA SinB
    So that means: Sin(A + B) = Sin A Cos B + (1/2)

    But now im stuck...
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  2. #2
    Bar0n janvdl's Avatar
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    Ah no wait. I figured it out


    But please help with this one:

    Prove that (sin2x + cos 2x)^2 = 1 + Sin 4x

    This is what i did:
    Left side = (cos^2x - 2cos.sin.2x + sin^2x)
    = (2x(sin^2 + cos^2) - 2cos.sin.2x)
    = 2x - 2cos.sin.2x
    Last edited by janvdl; May 5th 2007 at 05:41 AM.
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  3. #3
    Bar0n janvdl's Avatar
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    And prove that:
    1-sin 2x
    sinx - cosx = -----------
    sinx - cosx
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by janvdl View Post
    Ah no wait. I figured it out


    But please help with this one:

    Prove that (sin2x + cos 2x) = 1 + Sin 4x

    This is what i did:
    Left side = (cos^2x - 2cos.sin.2x + sin^2x)
    = (2x(sin^2 + cos^2) - 2cos.sin.2x)
    = 2x - 2cos.sin.2x
    This is not an identity. And how did you get that LHS?

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

    So
    sin(2x) + cos(2x) = 1 +sin(4x)

    sin(2x) + cos(2x) = 1 + 2sin(2x)cos(2x)

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

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

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

    2cos(x) - 4cos(x)(1 - 2sin(x)) = 2sin(x) <-- We must consider sin(x) = 0 as one of the solutions when we cancel it here.

    cos(x)[2 - 4 + 8sin^2(x)] = 2sin(x) <-- Square both sides

    cos^2(x)[-2 + 8sin^2(x)]^2 = 4sin^2(x)

    (1 - sin^2(x))[4 - 32sin^2(x) + 64sin^4(x)] = 4sin^2(x)

    Let's do ourselves a favor now and let y = sin^2(x)

    (1 - y)[4 - 32y + 64y^2] = 4y <-- Divide both sides by 4

    (1 - y)[1 - 8y + 32y^2] = y

    1 - 8y + 32y^2 - y + 8y^2 - 32y^3 = y

    32y^3 - 40y^2 + 10y - 1 = 0

    This does NOT have a pleasant solution. Numerically I get only one real solution: y = 0.957794.

    So
    y = sin^2(x) = 0.957794

    sin(x) = 0.97867

    x = 1.36388, 1.77771 rad
    or
    x = 0.434137*(pi), 0.565836*(pi)

    But we also have
    sin(x) = 0
    so
    x = 0, (pi) rad

    -Dan
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  5. #5
    Bar0n janvdl's Avatar
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    Er, thank you topsquark but didnt you go a little overboard?

    Im only in matric and we dont get problems with such hard-to-get solutions.
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  6. #6
    Bar0n janvdl's Avatar
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    Sorry, the left hand side should have been squared... I forgot to correct it. I'll do it now.
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  7. #7
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    Hello, janvdl!

    This is a tricky one . . .


    Prove that:
    . . . . . . . . . . . . . . . .1 - sin(2x)
    sin(x) - cos(x) . = . ------------------
    . . . . . . . . . . . . . . sin(x) - cos(x)

    I cross-multiplied ... just to verify the identity.
    . . [I know it's a illegal move in the proof of an identity.]
    But that gave me the Big Hint that we're all looking for.

    Watch this . . .

    [sin(x) - cos(x)]▓ .= .sin▓(x) - 2Ěsin(x)Ěcos(x) + cos▓(x) .= .1 - 2Ěsin(2x)


    Get it? .That numerator is a perfect square . . .

    . . . . . . . . . . . . . . .1 - sin(2x) . . . . . [sin(x) - cos(x)]▓
    The right side is: . ----------------- . = . -------------------- . = . sin(x) - cos(x)
    . . . . . . . . . . . . . sin(x) - cos(x) . . . . . sin(x) - cos(x)

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  8. #8
    Bar0n janvdl's Avatar
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    Soroban, please explain this...

    Quote Originally Posted by Soroban View Post
    Watch this . . .

    [sin(x) - cos(x)]▓ .= .sin▓(x) - 2Ěsin(x)Ěcos(x) + cos▓(x) .= .1 - 2Ěsin(2x)
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  9. #9
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    Quote Originally Posted by janvdl View Post
    Soroban, please explain this...
    Goeien avond,

    I can't see Soroban around, therefore I'll give you a hint:

    Soroban wrote:

    [sin(x) - cos(x)]▓ = sin▓(x) - 2Ěsin(x)Ěcos(x) + cos▓(x) = 1 - 2Ěsin(2x) Now change the order of the summands:

    [sin(x) - cos(x)]▓ = sin▓(x) + cos▓(x) - 2Ěsin(x)Ěcos(x)= 1 - 2Ěsin(2x)


    I assume that you know the value of sin▓(x) + cos▓(x)
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by earboth View Post
    Goeien avond,

    I can't see Soroban around, therefore I'll give you a hint:

    Soroban wrote:

    [sin(x) - cos(x)]▓ = sin▓(x) - 2Ěsin(x)Ěcos(x) + cos▓(x) = 1 - 2Ěsin(2x) Now change the order of the summands:

    [sin(x) - cos(x)]▓ = sin▓(x) + cos▓(x) - 2Ěsin(x)Ěcos(x)= 1 - 2Ěsin(2x)


    I assume that you know the value of sin▓(x) + cos▓(x)
    Ah ok, i think i understand a little better.
    Thanks Earboth and Soroban!
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  11. #11
    Bar0n janvdl's Avatar
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    By the way Earboth, i understand German. (Some of it anyway)

    But Goeien Avond looks like Dutch.

    I think German is Gutenaden.
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  12. #12
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    Hello, janvdl!

    You are doing strange things to the equation . . .


    Prove that: .(sin2x + cos 2x)▓ .= .1 + sin4x

    Square out the left side:

    . . . . sin▓(2x) + 2Ěsin(2x)Ěcos(2x) + cos▓(2x)

    . .= . sin▓(2x) + cos▓(2x) + 2Ěsin(2x)Ěcos(2x)
    . - - - \______. .______/ . . \_____ -______/
    . . . . . . . . . . \/ . . . . . . . . . . . . .\/
    . .= . . . . . . . 1 . . . . . . + . . . . sin(4x)

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