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Math Help - Reversing the inequality direction, less than to greater and equal to

  1. #1
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    Thumbs up Reversing the inequality direction, less than to greater and equal to

    Hi Forum
    Using intervals and inequalities can get a little tricky, so here is my question.
    Find all of the degree-measure of all the angles x in the interval 0\leq x \leq 360 for which
    2(sin{x}+cos{x})<\sqrt{1+2sin{x}cos{x}}
    4(sin{2x}+1)<1+sin{2x}
    sin{2x}<\frac{-3}{3}
    sin{2x}+1\color{blue}{\geq} 0

    How could this happen?
    I know that the sign of the inequality change when dealing with negative numbers or reciprocals, but...
    This actually becomes greater then or equal. Can someone explain?
    Thanks!
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    Re: Reversing the inequality direction, less than to greater and equal to

    Quote Originally Posted by Zellator View Post
    4(sin{2x}+1)<1+sin{2x}
    sin{2x}<\frac{-3}{3}
    How could this happen?
    First, that whole process makes no sense to me.
    sin{2x}<\frac{-3}{3}=-1 is false for every x.

    The sine function has no values less than -1.
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    Thumbs up Re: Reversing the inequality direction, less than to greater and equal to

    Now that you mention it, Plato
    The solution says
    2(sinx+cosx)<\sqrt{1+{sinx}+{cosx}} \rightarrow sin2x+1\geq 0
    which is only true when sinx+cosx<0.
    But
    sinx+cosx=\sqrt{2} sin{(x+45^\circ)}
    so
    135^\circ<x<315^\circ

    It doesn't make sense this way, but reversing the direction we get something.
    Have you check my steps? Is there something amiss?
    Thanks Plato.
    Your help is really appreciated!
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  4. #4
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    Re: Reversing the inequality direction, less than to greater and equal to

    I'd prefer to simplify the RHS using algebra

    1 + 2\sin(x)\cos(x) = (\sin(x)+\cos(x))^2 which means the RHS becomes \sqrt{1+2\sin(x)\cos(x)} = \sqrt{(\sin(x) + \cos(x))^2}


    2(\sin(x) + \cos(x)) < |\sin(x) + \cos(x)|

    2(\sin(x)+\cos(x)) - |(\sin(x) + \cos(x))| < 0


    Hence either 3(\sin(x)+\cos(x)) < 0 \text{  or  } \sin(x)+\cos(x) < 0


    sin and cos are both positive in the first quadrant so we can discard that as a solution

    In quadrant 2 sin is positive but cos is negative so in this quadrant if \cos(x) > \sin(x) then the inequality is satisfied, 135 degrees is this quadrant so 135^o < x \leq 180^0 works.

    In quadrant 3 sin and cos are both negative so their sum will also be negative, hence all of quadrant 3 satisfies.

    In quadrant 4 sin is negative and cos positive so the opposite of quadrant 2 applies and we want the first half of the quadrant - namely 270 \leq x < 315


    Combining we get 135^o < x < 315^o.

    No doubt there is a more efficient way though
    Last edited by e^(i*pi); June 29th 2011 at 01:23 PM.
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    Thumbs up Re: Reversing the inequality direction, less than to greater and equal to

    Quote Originally Posted by e^(i*pi) View Post
    I'd prefer to simplify the RHS using algebra

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

    which means the RHS becomes \sqrt{1+2\sin(x)\cos(x)} = \sqrt{(\sin(x) + \cos(x))^2}
    135^o < x < 315^o.

    Hi e^(i*pi)!
    Thanks so much for your reply!
    There may be some othe way to solve this but, this is great, no doubt about that.

    Just one thing bothers me
    When you squared the equation you didn't forget the 2 in {\color{blue}2^2}(\sin(x)+\cos(x))^2 ?

    Later you don't use the 4 anyway, right? (I need confirmation on this one)
    Since you only considered (\sin(x)+\cos(x))<0

    I'll keep this solution in mind, thanks for your help, so far!
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    Re: Reversing the inequality direction, less than to greater and equal to

    Quote Originally Posted by Zellator View Post
    Hi e^(i*pi)!
    Just one thing bothers me
    When you squared the equation you didn't forget the 2 in {\color{blue}2^2}(\sin(x)+\cos(x))^2 ?
    I didn't square anything. I rewrote the RHS and left the LHS.

    I used algebra to show that \sqrt{1+2\sin(x)\cos(x)} = \sqrt{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|. This is because \sqrt{a^2} = |a|

    This is equal to the unchanged LHS: 2(\sin(x)+\cos(x)) = |\sin(x) + \cos(x)|. The square root cancelled of it's own accord due to the algebraic manipulation of the RHS.


    Later you don't use the 4 anyway, right? (I need confirmation on this one)
    Since you only considered (\sin(x)+\cos(x))<0

    I'll keep this solution in mind, thanks for your help, so far!
    Yes, there are two equations that would need solving but since they only differ by amplitude their intervals would be equal to each other
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    Re: Reversing the inequality direction, less than to greater and equal to

    Quote Originally Posted by e^(i*pi) View Post
    I didn't square anything. I rewrote the RHS and left the LHS.

    I used algebra to show that \sqrt{1+2\sin(x)\cos(x)} = \sqrt{\sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x)} = \sqrt{(\sin(x) + \cos(x))^2} = |\sin(x) + \cos(x)|. This is because \sqrt{a^2} = |a|

    This is equal to the unchanged LHS: 2(\sin(x)+\cos(x)) = |\sin(x) + \cos(x)|. The square root cancelled of it's own accord due to the algebraic manipulation of the RHS.




    Yes, there are two equations that would need solving but since they only differ by amplitude their intervals would be equal to each other
    I see now, thanks!
    It's a great idea to use 1 as cos^2x+sen^2x and use it for factoring.
    Thanks for explaining why you disconsidered one of the results, that's a great help indeed.
    I'll put the information to good use!
    Thanks e^(i*pi)!

    All the best!
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