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Math Help - Solving for x

  1. #1
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    Solving for x

    First off, I am new here and don't know how to use math symbols such as pi, so I apologize is my problem looks jumbled or anything.

    Secondly this is not homework, so please don't think I'm trying to cheat!

    Anyways, I simply want to solve for x, and I am completely stumped:

    2*pi = x*sin( pi*(x-2)/2x )

    Sin is in radians, but I don't need an exact answer. If you can solve it leave all the pi's/fractions in it please. Thank you.
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  2. #2
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    Quote Originally Posted by Dormir View Post
    First off, I am new here and don't know how to use math symbols such as pi, so I apologize is my problem looks jumbled or anything.

    Secondly this is not homework, so please don't think I'm trying to cheat!

    Anyways, I simply want to solve for x, and I am completely stumped:

    2*pi = x*sin( pi*(x-2)/2x )

    Sin is in radians, but I don't need an exact answer. If you can solve it leave all the pi's/fractions in it please. Thank you.
    It does not look promising.

    2pi = x*sin[pi(x -2)/(2x)] ----------(i)
    2pi = x*sin[pi*x/(2x) -2pi/(2x)]
    2pi = x*sin[pi/2 - pi/x]
    2pi = x*cos(pi/x) -----------------(ii)

    In cases like this...in cases where there no "straightforward" solutions...I use iteration to get one value of x.
    I have my own iteration method but you can use the popular Newton's Method

    In my own method, I am getting x = 6.975 approx.
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  3. #3
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    Well I just finished high school trig so I don't exactly know Newton's method. I tried looking it up and couldn't find a good, simple explanation. But your answer does seem close to something I got, which means I messed up somewhere in getting that expression. I was actually hoping to get an irrational number...

    Thanks for the help though.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dormir View Post
    First off, I am new here and don't know how to use math symbols such as pi, so I apologize is my problem looks jumbled or anything.

    Secondly this is not homework, so please don't think I'm trying to cheat!

    Anyways, I simply want to solve for x, and I am completely stumped:

    2*pi = x*sin( pi*(x-2)/2x )

    Sin is in radians, but I don't need an exact answer. If you can solve it leave all the pi's/fractions in it please. Thank you.
    Quote Originally Posted by Dormir View Post
    Well I just finished high school trig so I don't exactly know Newton's method. I tried looking it up and couldn't find a good, simple explanation. But your answer does seem close to something I got, which means I messed up somewhere in getting that expression. I was actually hoping to get an irrational number...

    Thanks for the help though.
    I'm sure it will be rational...

    I noticed something when we get to this point:

    2\frac{\pi}{x}=\cos\left(\frac{\pi}{x}\right)

    If we let \varphi=\frac{\pi}{x}, the equation becomes

    2\varphi=\cos\varphi

    We will see that \varphi\in\mathbb{Q} (in english, \varphi is an element of the rational numbers).

    You may need to find some approximation technique that makes sense to you, or just use your calculator to find when this is true.

    My calculator tells me that this is the case when \varphi\approx.450183611295

    Thus, \frac{\pi}{x}=.450183611295\implies x=\frac{\pi}{.450183611295}\implies \color{red}\boxed{x\approx 6.97846961722}

    I hope this makes sense!

    --Chris
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  5. #5
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    That made very little sense to me. I'm not even sure how you got to your first step, though I was hoping for a non-calculator answer. I'll start from the beginning. I was trying to find a general formula for the ratio of a circle to a regular polygon that fit's exactly inside it as part of another problem. So it is (pi*r^2) / .5*aP where aP is apothem*perimeter. The interior angle (theta) of any x sided polygon, in radians, is pi(x-2)/2x. Making a triangle out of the apothem, the radius, and half of a side (s), and half of theta as our angle, we see that a = r sin (.5*theta). s = 2 r cos (.5*theta). P = sx, so P = 2rxcos(.5*theta). Put those into the first equation and the ratio = pi / ( x sin (.5*theta )*cos(.5* theta)). Using the identities we get 2*pi / sin (theta). When the ratio is 1, when the polygon and circle have equal areas, we can multiply both sides by the bottom and get 2pi = x*sin( pi(x-2)/2x). I hope I that made sense and I didn't make a mistake. I was able to use my calculator to find about your answer, but it didn't make sense to me, so I came here. If you could go step-by-step on your method I'd appreciate it.
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  6. #6
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    \pi is an Integer?!

    Better check the definition of Rational Numbers.
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  7. #7
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    Quote Originally Posted by Dormir View Post
    I'll start from the beginning. I was trying to find a general formula for the ratio of a circle to a regular polygon that fit's exactly inside it as part of another problem. So it is (pi*r^2) / .5*aP where aP is apothem*perimeter. The interior angle (theta) of any x sided polygon, in radians, is pi(x-2)/2x. Making a triangle out of the apothem, the radius, and half of a side (s), and half of theta as our angle, we see that a = r sin (.5*theta). s = 2 r cos (.5*theta). P = sx, so P = 2rxcos(.5*theta). Put those into the first equation and the ratio = pi / ( x sin (.5*theta )*cos(.5* theta)). Using the identities we get 2*pi / sin (theta). When the ratio is 1, when the polygon and circle have equal areas, we can multiply both sides by the bottom and get 2pi = x*sin( pi(x-2)/2x). I hope I that made sense and I didn't make a mistake.
    "The interior angle (theta) of any x sided polygon, in radians, is pi(x-2)/2x."
    That should be pi(x-2)/x only. Not over 2x.

    "When the ratio is 1, when the polygon and circle have equal areas,...."
    That kills your computations or assumptions or ideas.
    Because when "the area of the inscribed circle is equal to the area of the circumsbibing regular polygon", then there are no inscibed circle, no circumscribing polygon and no apothem.
    Then you cannot use 2pi = x*sin[pi(x-2)/x] because [(area of circle) /(area of polygon)] = 1 is not applicable here.
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  8. #8
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    I made some mistakes when writing my first post, so to clear some things up:

    I didn't mean to say the answer was rational or not, but possibly imaginary.

    For pi(x-2)/2x, the dividing by 2 is because only half of theta is used when making a triangle between the side, the apothem and the radius.

    I thought it may be incorrect from the beginning because a polygon wouldn't realistically fit completely inside a circle, but it was just an attempt to solve another problem I didn't post here. Thank you all for the replies.
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  9. #9
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    Quote Originally Posted by Dormir View Post
    I made some mistakes when writing my first post, so to clear some things up:

    I didn't mean to say the answer was rational or not, but possibly imaginary.

    For pi(x-2)/2x, the dividing by 2 is because only half of theta is used when making a triangle between the side, the apothem and the radius.

    I thought it may be incorrect from the beginning because a polygon wouldn't realistically fit completely inside a circle, but it was just an attempt to solve another problem I didn't post here. Thank you all for the replies.
    "For pi(x-2)/2x, the dividing by 2 is because only half of theta is used when making a triangle between the side, the apothem and the radius."

    That is right, but you wrote the interior angle is pi(x-2)/2x.

    Then you wrote,
    'Put those into the first equation and the ratio = pi / ( x sin (.5*theta )*cos(.5* theta))."
    That is pi / [x*(1/2)sin(2 *0.5*theta)]
    = (2pi) /[x*sin(theta)]
    So the angle is theta, the whole interior angle. Not (theta)/2.
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  10. #10
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    I said I wrote it wrong on my post, but I didn't while doingthe problem. No one has shown me how to work this out yet, does anyone have the time for a step by step solution?
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