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Math Help - trignometry confusion

  1. #1
    Member grgrsanjay's Avatar
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    trignometry confusion

    1]the equation 1-\cos^ x + \sinx\cosx = k then there exists real solution when k belongs to

    A) 0<k<\frac{1+\sqrt2}{2}
    B) 2-\sqrt3 < k 2+\sqrt3
    C) 0<k<2-\sqrt3
    D) \frac{1-\sqrt2}{2} < k < \frac{1+\sqrt2}{2}

    2]let a,b belong to the set [0, \frac{\pi}{2}] if
    sin^6 a+3\sin^2 a\cos^2 a+\cos^6 b=1

    then \frac{255a+b}{a+b} =?
    Last edited by grgrsanjay; October 17th 2010 at 05:48 PM.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Ok. Is that it?

    1-\cos^2 (x) + \sin (x) \cos (x) = k

    Well...

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

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

    \dfrac12 \cos(2x) + \dfrac12 + \dfrac12\sin(2x) = k

    \dfrac12 (1 + \cos(2x) + \sin(2x)) = k

    Now,

    \cos(2x) + \sin(2x) = R\sin(2x + \alpha)

    R = \sqrt{1^2 + 1^2} = \sqrt2

    \alpha = \tan^{-1}\left(\dfrac11\right) = \dfrac{\pi}{4}

    \cos(2x) + \sin(2x) = \sqrt2 \sin(2x + \frac{\pi}{4})

    Hence;

    \dfrac12 (1 + \sqrt2\sin(2x + \frac{\pi}{4})) = k

    \sin(2x + \frac{\pi}{4}) = \dfrac{2k-1}{\sqrt2}

    So,

    -1 < \dfrac{2k-1}{\sqrt2} < 1

    Can you complete it now?

    And your second question is... difficult to understand.
    Last edited by Unknown008; October 17th 2010 at 06:37 AM. Reason: LaTeX typo
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  3. #3
    Member grgrsanjay's Avatar
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    in the second question i substituted \frac{\pi}{2}for a and b and got answer as 128


    need to prove this is true for all values
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  4. #4
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    You are right.
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