Results 1 to 3 of 3

Math Help - Expressing trigonometric functions as real radicals problem

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1

    Expressing trigonometric functions as real radicals problem

    This is another "challenge question" from my book on complex analysis. I didn't really get why the points outlined below the question text would work, but I tried to follow them anyway to see what would happen.

    I've attached the question text and my progress with this so far. Since z = 1 and z = -1 are obvious roots, I start by dividing by z^2 - 1. Then I think I've gotten to the point where I've solved the bi-quadratic equation for u, but I can't figure out how to get from there to a solution for z.

    Thanks for any help!
    Attached Thumbnails Attached Thumbnails Expressing trigonometric functions as real radicals problem-questiontext.jpg   Expressing trigonometric functions as real radicals problem-solutionattempt.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2009
    Posts
    195
    Thanks
    1

    Re: Expressing trigonometric functions as real radicals problem

    Nevermind, I figured it out and got the right answer in the end! Still, if anyone is familiar with the method involved and could explained where it is taken from, it would be very appreciated.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,406
    Thanks
    1294

    Re: Expressing trigonometric functions as real radicals problem

    Quote Originally Posted by gralla55 View Post
    This is another "challenge question" from my book on complex analysis. I didn't really get why the points outlined below the question text would work, but I tried to follow them anyway to see what would happen.

    I've attached the question text and my progress with this so far. Since z = 1 and z = -1 are obvious roots, I start by dividing by z^2 - 1. Then I think I've gotten to the point where I've solved the bi-quadratic equation for u, but I can't figure out how to get from there to a solution for z.

    Thanks for any help!
    Suppose we have a regular pentagon. It's well known that the angle sum of a polygon of \displaystyle n sides is \displaystyle (n - 2) \cdot 180^{\circ} , and so the angle sum of a pentagon is \displaystyle 540^{\circ}. Since it's a regular pentagon, all angles are equal, and so each angle is \displaystyle \frac{540^{\circ}}{5} = 108^{\circ}.



    If we join one vertex to all others as shown in the diagram, we can see that we split the pentagon into three triangles. The left-hand triangle is isosceles due to the pentagon being regular, and so the two remaining angles are equal. Since the angles add to \displaystyle 180^{\circ}, that means the remaining angles are each \displaystyle 36^{\circ}. The right-hand triangle is congruent to the left-hand triangle, and the angles in the middle triangle are found by evaluating what remains from the \displaystyle 108^{\circ} angles. The middle triangle is also clearly isosceles.



    If we focus on the middle triangle, if we call the bottom length \displaystyle l and the other sides \displaystyle a (which we can do because it is isosceles), then we can write \displaystyle a in terms of \displaystyle l using the sine rule.

    \displaystyle \begin{align*} \frac{a}{\sin{ \left( 72^{\circ} \right) } } &= \frac{l}{\sin{ \left( 36^{\circ} \right) } } \\ a &= \frac{ l\sin{ \left( 72^{\circ} \right) } }{ \sin{ \left( 36^{\circ} \right) } } \\ a &= \frac{ l \sin{ \left( 2 \cdot 36^{\circ} \right) } }{ \sin{ \left( 36^{\circ} \right) } } \\ a &= \frac{ 2l\sin{ \left( 36^{\circ} \right) }\cos{ \left( 36^{\circ} \right) } }{ \sin{ \left( 36^{\circ} \right) } } \\ a &= 2l\cos{ \left( 36^{\circ} \right) } \end{align*}



    If we bisect one of the \displaystyle 72^{\circ} angles, we split this triangle into two smaller triangles. Since we bisected a \displaystyle 72^{\circ} angle, that means the smallest angle of the smallest triangle is \displaystyle 36^{\circ}. This means the remaining angle is \displaystyle 72^{\circ}, giving another isosceles triangle. So the new segment we just created also is of length \displaystyle l. We will say that the final leg of this triangle is \displaystyle 1 unit in length. This triangle is also clearly similar to the original triangle.

    If we look at the other new triangle, since we bisected a \displaystyle 72^{\circ} angle, that means the angle that is contributed to this new triangle by the bisected angle is also \displaystyle 36^{\circ} and the final angle is \displaystyle 108^{\circ}, and thus is also an isosceles triangle. So that means the other length adjacent to the \displaystyle 108^{\circ} angle is also \displaystyle l.



    If we compare the two similar triangles, the length of \displaystyle 1 unit in the smaller triangle corresponds to a length of \displaystyle l units in the larger triangle, and so the scaling factor is \displaystyle l.

    We can also see that a length of \displaystyle l units in the smaller triangle corresponds to a length of \displaystyle a units in the larger triangle. So a = l \cdot l = l^2.

    But we can also see that \displaystyle a = l + 1, and so l^2 = l + 1. Solving this equation for \displaystyle l gives

    \displaystyle \begin{align*} l^2 &= l + 1 \\ l^2 - l - 1 &= 0 \\ l^2 - l + \left( -\frac{1}{2} \right)^2 - \left( -\frac{1}{2} \right) ^2 - 1 &= 0 \\ \left( l - \frac{1}{2} \right) ^2 - \frac{1}{4} - \frac{4}{4} &= 0 \\ \left( l - \frac{1}{2} \right) ^2 &= \frac{5}{4} \\ l - \frac{1}{2} &= \pm \frac{\sqrt{5}}{2} \\ l &= \frac{1 \pm \sqrt{5}}{2}  \end{align*}

    Since this is a length, it's obvious that only the positive solution is acceptable. Therefore \displaystyle l = \frac{1 + \sqrt{5}}{2}.

    We knew that \displaystyle a = l + 1, so \displaystyle a = \frac{3 + \sqrt{5}}{2}.

    We also found earlier that a = 2l\cos{\left( 36^{\circ} \right) }, and so

    \displaystyle \begin{align*} \frac{3 + \sqrt{5}}{2} &= 2 \left( \frac{1 + \sqrt{5}}{2} \right) \cos{ \left( 36^{\circ} \right) } \\ \frac{3 + \sqrt{5}}{2} &= \left( 1 + \sqrt{5} \right) \cos{ \left( 36^{\circ} \right) } \\ \frac{3 + \sqrt{5}}{2 \left( 1 + \sqrt{5} \right) } &= \cos{ \left( 36^{\circ} \right) } \\ \frac{ \left( 3 + \sqrt{5} \right) \left( 1 - \sqrt{5} \right) }{2 \left( 1 + \sqrt{5} \right) \left( 1 - \sqrt{5} \right) } &= \cos{ \left( 36^{\circ} \right) } \\ \frac{3 - 3\sqrt{5} + \sqrt{5} - 5}{2 \left( -4 \right) } &= \cos{ \left( 36^{\circ} \right) } \\ \frac{-2 - 2\sqrt{5}}{-8} &= \cos{ \left( 36^{\circ} \right) } \\ \frac{ -2 \left( 1 + \sqrt{5} \right) }{ -8 } &= \cos{ \left( 36^{\circ} \right) } \\ \cos{ \left( 36^{\circ} \right) } &= \frac{1 + \sqrt{5}}{4} \end{align*}

    And by Pythagoras, that means

    \displaystyle \begin{align*} \sin{ \left( 36^{\circ} \right) } &= \sqrt{ 1 - \left( \frac{1 + \sqrt{5}}{4} \right) ^2 } \\ &= \sqrt{ \frac{16 - 1 - 2\sqrt{5} - 5}{16} } \\ &= \sqrt{\frac{10 - 2\sqrt{5}}{16}} \\ &= \frac{\sqrt{10 -2\sqrt{5}}}{4} \end{align*}

    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Minimum problem with trigonometric functions.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 8th 2012, 10:32 AM
  2. Expressing functions [word problems]
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: September 18th 2011, 07:54 PM
  3. Trigonometric Functions of Real Numbers
    Posted in the Trigonometry Forum
    Replies: 9
    Last Post: January 31st 2010, 10:32 AM
  4. Radicals in trigonometric ratios
    Posted in the Geometry Forum
    Replies: 4
    Last Post: January 24th 2010, 02:09 PM
  5. Expressing functions as power series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 10th 2009, 06:08 PM

Search Tags


/mathhelpforum @mathhelpforum