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Math Help - hard!

  1. #1
    Junior Member
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    Apr 2007
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    hard!

    can u help me solve this question

    If xcos θ + y sinθ = a cos 2θ and;
    xsinθ - ycos θ = 2a sin 2θ
    prove that (x+y)2/3 + (x-y)2/3 = 2a2/3
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, calculus_jy!

    This is The Problem from Hell . . . (coming to a theater near you)


    If . \begin{array}{cc}x\cos\theta + y\sin\theta\:=\:a\cos2\theta & [1]\\ x\sin\theta - y\cos\theta\:=\:2a\sin2\theta & [2]\end{array}

    prove that: . (x+y)^{\frac{2}{3}} + (x-y)^{\frac{2}{3}} \:=\:2a^{\frac{2}{3}}
    Multiply [1] by \cos\theta:\;\;x\cos^2\!\theta + y\sin\theta\cos\theta \:=\:a\cos2\theta\cos\theta
    Multiply [2] by \sin\theta:\;\;x\sin^2\!\theta - y\sin\theta\cos\theta\:=\:2a\sin2\theta\sin\theta

    Add: . x\cos^2\!\theta + x\sin^2\!\theta \:=\:a\cos2\theta\cos\theta + 2a\sin2\theta\cos\theta

    . . . . x(\cos^2\!\theta + \sin^2\!\theta) \:=\:a(\cos^2\!\theta - \sin^2\!\theta)\cos\theta + 2a(2\sin\theta\cos\theta)\sin\theta

    . . . . . . . . . . . . . x \:=\:a\cos^3\!\theta - a\sin^2\!\theta\cos\theta + 4a\sin^2\!\theta\cos\theta

    . . . . . . . . . . . . . x \:=\:a\cos^3\!\theta + 3a\sin^2\!\theta\cos\theta


    In a similar fashion solve for y:\;\;y \:=\:-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta


    We have: . x + y \;=\;(a\cos^3\!\theta + 3a\cos\theta\sin^2\!\theta) + (-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta)
    . . . . . . . . . . . . = \;a(\cos^3\!\theta - 3\cos^2\!\theta\sin\theta + 3\cos\theta\sin^2\!\theta - \sin^3\!\theta)
    . . . . . . . . . . . . = \;a(\cos\theta - \sin\theta)^3

    We have: . x - y \;=\;(a\cos^3\!\theta + 3a\cos\theta\sin^2\!\theta) - (-3a\cos^2\!\theta\sin\theta - a\sin^3\!\theta)
    . . . . . . . . . . . . = \;a(\cos^3\!\theta + 3\cos^2\!\theta\sin\theta + 3\cos\theta\sin^2\!\theta + \sin^3\!\theta)
    . . . . . . . . . . . . = \;a(\cos\theta + \sin\theta)^3

    Then: . \begin{array}{cccc}x + y \:=\:a(\cos\theta - \sin\theta)^3 & \Rightarrow & (x + y)^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}(\cos\theta - \sin\theta)^2 & [3]\\<br />
x - y \:=\:a(\cos\theta + \sin\theta)^3 & \Rightarrow & (x - y)^{\frac{2}{3}} \:=\:a^{\frac{2}{3}}(\cos\theta + \sin\theta)^2 & [4]\end{array}

    Add [3] and [4]:
    . . (x+y)^{\frac{2}{3}} + (x - y)^{\frac{2}{3}} \;=\;a^{\frac{2}{3}}\bigg[(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2\bigg]

    . . . . . . . . . . . . . . = \;a^{\frac{2}{3}}\left[\cos^2\!\theta - 2\sin\theta\cos\theta + \sin^2\!\theta + \cos^2\!\theta + 2\sin\theta\cos\theta + \sin^2\!\theta\right]

    . . . . . . . . . . . . . . = \;a^{\frac{2}{3}}\left[2\cos^2\!\theta + 2\sin^2\!\theta\right]

    . . . . . . . . . . . . . . = \;2a^{\frac{2}{3}}\left[\cos^2\!\theta + \sin^2\!\theta\right]


    Therefore: . (x + y)^{\frac{2}{3}} + (x - y)^{\frac{2}{3}} \;=\;2a^{\frac{2}{3}} . . . . ta-DAA!!

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  3. #3
    Junior Member
    Joined
    Apr 2007
    Posts
    32

    Great thx

    i really appreciate ur help, and u said it is a problem from hell, just wondering how u manage to start solving problems lyk this, besides practicing and using ur brain, can u offer me few tips??

    Also i hav 2 other problems from hell, i will post it in the geometry section and wonder if u can help me with them, they took me ages!
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