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Math Help - Trig Inequality

  1. #1
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    Trig Inequality

    cos(4x) > cos(6x)

    I've tried to let A = 2x, and solve it using duoble angles...

    cos(2A) > cos(2A+A) ...then letting cos2AcosA - sin2AsinA replace cos(2A+A), etc. but it gets really messy and I get lost.

    Can anyone please do this step by step?

    Thanks in advance!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Sconts!

    I found an approach, but it's just as bad as your expansion . . . LOL!


    cos(4x) .> .cos(6x)
    I'll assume that the answer are between 0 and 2π.

    I used a sum-to-product identity: . cos A - cos B .= .-2新in[(A+B)/2]新in[(A-B)/2]


    We have: .cos(4x) - cos(6x) .> .0

    . . which becomes: .-2新in(5x)新in(-x) .> .0

    And we have: .2新in(5x)新in(x) .> .0


    There are two cases to consider:
    . . [1] Both factors are positive.
    . . [2] Both factors are negative.


    [1] sin(x) > 0 .and .sin(5x) > 0

    If sin(x) > 0, then: .0 < x < π

    If sin(5x) > 0, then:
    . . 0 < 5x < π . . . . . .0 < x < π/5
    . .2π < 5x < 3π . . 2π/5 < x < 3π/5
    . .4π < 5x < 5π . . 4π/5 < x < π

    Hence: .x ε (0, π/5) U (2π/5, 3π/5) U (4π/5, π)


    [2] sin(x) < 0 .and .sin(5x) < 0

    If sin(x) < 0, then: .π < x < 2π

    If sin(5x) < 0, then:
    . . 5π < 5x < 6π . . . . . .π < x < 6π/5
    . . 7π < 5x < 8π . . . 7π/5 < x < 8π/5
    . . 9π < 5x < 10π . . 9π/5 < x < 2π

    Hence: .x ε (π, 6π/5) U (7π/5, 8π/5) U (9π/5, π)


    The solution is the union of all the intervals in blue.



    But there must be a better way!
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  3. #3
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    Joined
    Feb 2007
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    Thank you very much for the help, you saved me

    I kept using half angle and cos(a+B) formulas and it got wayyy too messy, haha (I probably just didn't know how to simplify it, but okay )

    Once again, thanks
    Good method!
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