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Homework Help: Finding the exact value of sec^2x

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The curves y=cosx and y=tanx intersect at point P whose x coordinate is [itex]\alpha[/itex]
    a) show the curves intersect at right angles at P.
    b) show that sec^2[itex]\alpha[/itex] = (1 + [itex]\sqrt{5}[/itex])/2

    2. Relevant equations
    I think they're all given up there ^

    3. The attempt at a solution
    I managed to do part a pretty easily, but I get stuck on part b.

    tan[itex]\alpha[/itex] = cos[itex]\alpha[/itex]

    cos^2[itex]\alpha[/itex] = sin[itex]\alpha[/itex]

    sin^2[itex]\alpha[/itex] + sin[itex]\alpha[/itex] - 1 = 0

    I'm not sure how to do go on from there.

  2. jcsd
  3. Oct 20, 2011 #2


    Staff: Mentor

    It looks like you're stuck on part a and haven't gotten to part b, yet.
    This equation is quadratic in sin(α). Let u = sin(α). What does the equation look like with this substitution?
  4. Oct 20, 2011 #3
    Alright I got it:

    sin[itex]\alpha[/itex] =[itex](-1 + \sqrt{5})/2[/itex]

    cos^2[itex]\alpha[/itex] = [itex](-1 + \sqrt{5})/2[/itex]

    Tan^2[itex]\alpha[/itex] = [itex](-1 + \sqrt{5})/2[/itex]

    1 + Tan^2[itex]\alpha[/itex] = 1 + [itex](-1 + \sqrt{5})/2[/itex]

    sec^2[itex]\alpha[/itex] = [itex](1 + \sqrt{5})/2[/itex]

  5. Oct 20, 2011 #4


    Staff: Mentor

    I don't think so.
    You should have two values.
    sin[itex]\alpha[/itex] =[itex](-1 \pm \sqrt{5})/2[/itex]
    How does this follow from the line above it?
  6. Oct 20, 2011 #5
    I didn't know how to do +-, so I left it as plus, answer is still the same though.

    sin[itex]\alpha[/itex] =[itex](-1 \pm \sqrt{5})/2[/itex]

    From cos^2[itex]\alpha[/itex] = sin[itex]\alpha[/itex]

    cos^2[itex]\alpha[/itex] = [itex](-1 \pm \sqrt{5})/2[/itex]

    As tanx = cosx when x = [itex]\alpha[/itex]

    tan^2[itex]\alpha[/itex] = [itex](-1 \pm \sqrt{5})/2[/itex]

    adding 1 to both sides

    1 + tan^2[itex]\alpha[/itex] = 1 + [itex](-1 \pm \sqrt{5})/2[/itex]

    sec^2[itex]\alpha[/itex] = [itex](1 \pm \sqrt{5})/2[/itex]

    but sec^2[itex]\alpha[/itex] is going to be positive, so its the positive solution

    sec^2[itex]\alpha[/itex] = [itex](1 + \sqrt{5})/2[/itex]
  7. Oct 20, 2011 #6


    Staff: Mentor

    OK, looks good. Your explanation helped me follow what you were doing without having to reproduce your work.

    Have you done part a? You have to show that the two curves intersect at a right angle.
  8. Oct 20, 2011 #7
    Yeah I did part a first. Just found the gradient of the 2 lines at [itex]\alpha[/itex] and they multiplied to give -1. I couldn't see how that would have helped me with part b though.
  9. Oct 23, 2011 #8


    User Avatar
    Homework Helper

    You could also have gotten there a little quicker by "flipping the equation over" to write

    [tex] sec^{2} \alpha = \frac {2}{-1 \pm \sqrt{5}} = \frac {2}{-1 \pm \sqrt{5}} \cdot \frac {-1 \mp \sqrt{5}}{-1 \mp \sqrt{5}} = \frac {2 ( -1 \mp \sqrt{5} )}{1 - 5} = \frac {1 \pm \sqrt{5} }{2} [/tex]

    and you discarded the negative result.

    Why am I bothering to remark on this at all? Because [itex] \frac {1 + \sqrt{5} }{2} [/itex] is a special number, the "golden ratio" [itex] \phi [/itex] (also labeled [itex] \tau [/itex] by some folks). Its reciprocal is [itex] \frac{1}{\phi} = \phi - 1 [/itex] * ; while it is not as "famous" as [itex]\pi[/itex] or e , it is nearly as ubiquitous and turns up in a lot of unexpected places (like here!). When the quadratic equation [itex]x^{2} \pm x - 1 = 0 [/itex] appears in some application, [itex] \phi [/itex] or its related numbers are not far away...

    * which is why your calculations behaved the way they did; and, BTW, the "rejected" value [itex] \frac {1 - \sqrt{5} }{2} [/itex] equals [itex] - \frac{1}{\phi} [/itex] !

    Well, it certainly wouldn't have been particularly obvious.... But when you evaluated the slopes of the tangent lines at [itex] x = \alpha[/itex], you had, for y = tan x ,

    [tex] m_{2} = \sec^{2} \alpha \Rightarrow \cos^{2} \alpha = \frac{1}{m_{2}} \Rightarrow \sin^{2} \alpha = 1 - \cos^{2} \alpha = 1 - \frac{1}{m_{2}} . [/tex]

    Since you also have, for the slope of y = cos x , [itex] m_{1} = -\sin \alpha [/itex] and the perpendicular tangent lines give us [itex] m_{1} = - \frac{1}{m_{2}} [/itex],
    you would have the equation

    [tex] \sin^{2} \alpha = 1 - \frac{1}{m_{2}} \Rightarrow m_{1}^{2} = 1 - \frac{1}{m_{2}} \Rightarrow \frac{1}{m_{2}^{2}} = 1 - \frac{1}{m_{2}} \Rightarrow 1 = m_{2}^{2} - m_{2} \Rightarrow m_{2}^{2} - m_{2} - 1 = 0 , [/tex]

    and so out pops [itex] m_{2} = \phi [/itex]. A wonderful number!
  10. Oct 24, 2011 #9
    Oddly enough I never thought of doing it at the time, though now that I see it, it seems to be the most obvious option.

    I never knew any of that. I have seen the number come up quite a few times, but I never thought it to be of any significance. I seem to recall seeing it a lot with complex numbers and finding the exact values of sinx etc. Is there any reason it isn't as well known? I am only in high school so I may find it in the future.

    Thanks for the input

    Thanks for that, I just didn't see that at all, didn't make the connection. Good to see how to do it though. Just had my exam today, so it's a bit too late to do anything about it :P
  11. Oct 24, 2011 #10


    User Avatar
    Homework Helper

    The "golden ratio" isn't quite as fundamental a number in our mathematical practice as [itex]\pi[/itex] or e are, so we don't run across it quite so often. But if you search on "Golden Ratio", you'll find that quite a lot has been written about it (just one example: the ratio of terms Fn+1 / Fn in the Fibonacci series tends to [itex]\phi[/itex] , as n goes to infinity). I'm kind of "sensitized" to such numbers, so I am always interested in the strange places where they crop up...

    There isn't any reason you should have suspected any of this going into the problem. It was when I saw That Number come up in this thread that I decided to investgate how it was related.
    Last edited: Oct 24, 2011
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