1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Turning Equations into Trigonometric Equations

  1. Jun 25, 2005 #1
    Hello all,

    I have been looking up the golden ratio and found most of what I needed on mathworld.

    The site states that [tex]\phi \ = \ \frac{1}{2}(1+\sqrt{5})[/tex].

    I can see how (despite the fact that I don't understand how the ratio:

    [tex]\phi \ = \ \frac{AC}{BC} \ = \ \frac{AB}{AC}[/tex] is formed but that is another matter).

    The problem I am having is turing this into either

    [tex]\phi \ = \ 2 cos (\frac{\pi}{5})[/tex] or [tex]\phi \ = \ \frac{1}{2} sec (\frac{2 \pi}{5})[/tex].

    Can anyone give me some hints please?

    Cheers. :smile:

    The Bob (2004 ©)
  2. jcsd
  3. Jun 25, 2005 #2
    As i understand the problem is to calculate [tex]\cos(\frac{\pi}{5})[/tex]
    It's possible using [tex]e^{i\phi}=\cos\phi+i\sin\phi[/tex].
    Take first phi=pi/5, and then solve for Cos(phi) from [tex]e^{5i\phi}=(\cos\phi+i\sin\phi)^5=-1[/tex].
    I made it by myself. It realy works :approve:
    Last edited: Jun 25, 2005
  4. Jun 25, 2005 #3
    It might well work and I appreciate it. However I want to 'derive' it. Someone must have done to get it in the first place and that is what I would like to understand and work out and be able to apply it to any future examples.

    I know that people work things out by knowing what they want, what they have and simply finding the middle steps but I want to pretend I know the form of what I want, what I have and have to find the middle steps. The difference is I do not know what it will come to (this is what I am pretending).

    The Bob (2004 ©)
  5. Jun 26, 2005 #4
    I found a method which will produce the answer without the slightest insight into why it works. The principal reference is http://www.andrews.edu/~calkins/math/webtexts/numb18.htm#SIN18 [Broken], where it shown by use of trick, Sin72 = cos 18, we are able to find the sin 18. What is done is that sin72=2cos36(sin36) = 2[cos(18)^2-sin(18)^2][2sin(18)cos(18).

    Now we continue to produce an answer in terms of sin(18), WHILE DIVIDING OUT Cos(18). This comes down to an equation for x=sin(18); [tex]8x^3-4x+1 =0.[/tex] Fortunately, x=1/2 is an answer, and so by synthetic division we are left with [tex]8x^2+4x-2=0.[/tex] This gives the result that we can use:

    [tex]sin(18)=\frac{-1+\sqrt(5)}{4}[/tex] And from this we produce the cos(18).

    We then use the formula cos(36) = cos(18)^2-sin(18)^2, to produce the result:

    [tex]cos(36) =\frac{1+\sqrt(5)}{4}[/tex] Since this happens to be (1/2)[tex]\phi[/tex], the golden ratio, we are done!
    Last edited by a moderator: May 2, 2017
  6. Jun 26, 2005 #5
    A second algebraic way to work is to consider [tex]\theta=e^\frac{2\pi}{5}[/tex]=cos 72 +isin72. Then since [tex]\theta[/tex] is the fifth root of unity, we have [tex](\theta-1)(\theta^4+\theta^3+\theta^2+\theta+1)=0[/tex] Since we are not concerned with 1 as a root, we look at the cyclotomic polynomial, which we divide by [tex]\theta^2 [/tex], giving:

    [tex]\theta^2+\theta^1+1+\theta^{-1}+\theta^{-2}=0[/tex]. Now since 2sin72 = [tex]\theta+\theta^{-1}=X, [/tex] we are able to substitue and get the equation: [tex]X^2+X-1=0[/tex], or [tex]X=\frac{-1+\sqrt5}{2}[/tex]

    Then we divide by 2, and use the half angle cosine formula, to arrive at

    cos(36)^2 =[tex]\frac{3+\sqrt5}{8} =(\frac{1+\sqrt5}{4})^2 [/tex]
    Last edited: Jun 26, 2005
  7. Jun 26, 2005 #6
    If you are looking for a case of the golden ratio, look no further than the Fibonacci numbers: 1,1,2,3,5,8,13, 21.....In this case we have: [tex]F_n=F_{n-1}+F_{n-2}[/tex]

    So we have the ratio [tex]\frac{F_{n}}{F_{n-1}}=1 +\frac{F_{n-2}}{F_{n-1}}[/tex]

    When the values get very large, the ratio settles down on a limit, and so we come closer and closer to the final ratio X. In the equation we always take the positive value for the ratio, which becomes:

    X=1+1/X. Or that [tex]X^2-X-1 =0[/tex] Thus [tex] X=\frac{1+\sqrt5}{2}[/tex]
    Last edited: Jun 27, 2005
  8. Jun 27, 2005 #7
    The Bob: I don't understand how the ratio is formed, but that is another matter.

    In a line of the form A_________B____C, we form the ratio of the larger part over the smaller, equals the ratio of the whole line divided by the larger part:

    [tex]\frac{AB}{BC}=\frac{AC}{AB}[/tex] Now if we let AB=1, and BC=X, we have the ratio: [tex]\frac{1}{X}=\frac{1+X}{1}[/tex]

    This gives [tex]X=\frac{-1+\sqrt5}{2}[/tex]. BUT, we are seeking the ratio, which is 1/X. Thus we invert the fraction and multiply top and bottom by

    [tex](1+\sqrt5)[/tex] to rationalize the denominator. This gives our desired ratio =[tex]\frac{1+\sqrt5}{2}[/tex]
    Last edited: Jun 27, 2005
  9. Jun 29, 2005 #8
    Thanks for this information Robert. I don't have time to read it now but I will do later and post any thoughts I get from it.


    The Bob (2004 ©)
  10. Jun 30, 2005 #9
    This goes over my head.

    I think this makes sense but, as you said, is cheating for what I am looking for.

    This one is also 'over-my-head' stuff.

    This also seems to confuse me but it might be my mind at this moment in time.

    Cheers for the information guys. :smile:

    The Bob (2004 ©)
    Last edited by a moderator: May 2, 2017
  11. Jul 1, 2005 #10
    Well if you want a popular text to look at, I recommend: The Golden Ratio , by Mario Livio, Broadway Books, 2002, which I found at Barnes and Noble. Here is a poem the book quotes:

    The golden mean is quite absurd
    It's not your ordinary surd
    If you invert it (This is fun),
    You'll get itself, reduced by one
    But if increased by unity
    This yields its square, take it from me.

    Paul S Bruckman, Fibonacci Quarterly 1977.
    Last edited: Jul 1, 2005
  12. Jul 2, 2005 #11
    I think what Bob is asking is how do you jump from variables and simple algebraic expressions to cos and sin functions. Usually, it is best understood geometrically.

    Take a simple right-angle triangle with the right-angle at the origin of a simple x/y graph (2 dimensions, cartesian coordinates). Let's make the long side x, and the tall side y. You have one other side, the diagonal, or hypoteneuse h.
    And there are three angles, each related to the proportions of the sides.

    All the functions Sin, Cos, Tan, etc. are formally defined by ratios, y/x, y/h, x/h, etc. So in a typical problem that involves a triangle, like a ladder leaning against a building or whatever, you can draw a sketch with a triangle and use knowledge of some sides, or angles to figure out the others. What happens on paper is that you start with a simple algebraic equation, and often end up with an equation that now has some Cos or Sin function in it, as a general solution. This simplifies a tedious problem to a simple equation where you just plug in an angle and you get the answer to some other thing like x or what have you.

    In the example above, I know one angle (90 degrees), two sides (y = height, x = distance from building), and maybe I want to know the angle etc. After solving a few of these problems using Pythagorean Theorem ( h^2 = x^2 + y^2 ) I might want to instead set up a simple formula where I just plug in the angle and one side, and get the other side, or the length of ladder I need. I have moved from specific cases, to a general formula using a Cos or Sin function.

    Similarly, what is going on with your example, is that someone has taken one way of making a Golden Ratio, and using the Pythagorean Theorem and some substitutions and algebraic steps has generated another formula based upon a trig function. This works (only) if you have a diagram that identifies the sides and angles properly.
  13. Jul 2, 2005 #12
    It was Euclid: http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV11.html, who figured out how to construct the regular pentagon.

    We can work with an isosceles triangle who's vertex angles is 1/2 that of its base angles. Thus we have X+2X+2X = 180, or that the vertex angle is 36. Now if we take one of the two 72 degree angles and bisect it, draw the line to the other side, and we now have created another isosceles triangle with angles of 72, 72, 36. A similar triangle. That is where these sines and cosines come in, and how we proceed to the golden ratio. http://mathforum.org/library/drmath/view/55113.html.

    There is a good diagram: http://mathworld.wolfram.com/GoldenGnomon.html A theorem tells us that the angles of the regular p-gon are 180x(n-2). Thus if n=5, the interior angels total 540, and divided by 5, each angles is108 degrees. From that, we can find the angles as shown in the diagram. The book mentioned before, The Golden Ratio goes into this.

    Actually the Ancients were fascinated by this golden ratio discovery, and it also bears on the fact that much later, Gauss, who constructed the 17 sided polygon, recognized that that we have a situation involving[ 2^(2^n)]+1, called a Fermat prime. 5 = 2^2+1, 17=2^4+1, so from the standpoint of Galois Theory, we are working with solvable groups involving square roots. (I guess this is getting quite a bit ahead in an explanation, but today we would have recognized that 5 =2^2 +1 is constructible involving square roots without having to do the construction, but to the Greeks it was a big thing.)
    Last edited: Jul 2, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Turning Equations into Trigonometric Equations
  1. Trigonometric equation (Replies: 2)