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Calculate sin and cos using complex numbers

  1. Jun 11, 2005 #1
    How can I calculate cos 72° and sin 72° using complex numbers, and without the use of a calculator?

    I noticed that 5*72° = 360° so (cos 72° + i*sin 72°)^5 = 1. But, I don't quite know how to go from there.. :shy:
     
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  3. Jun 11, 2005 #2

    dextercioby

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    What are the 5 independent solutions to the equation

    [tex] z^{5}=1 [/tex]

    And which of the 5 solutions is useless...?

    Daniel.
     
  4. Jun 11, 2005 #3
    I haven't seen this type of problem before but here's what I'm thinking. Take [itex]cos\theta =p,\, sin\theta=q[/itex]. So expand out [itex](p+iq)^5=1[/itex]. So the imaginary part is 0 and the real part is 1. So collect the multiples of i and substitute [itex]q=\sqrt{1-p^2}[/tex]. You'll end up with something you can solve using the quadratic formula. Then it's just a matter of picking the root that makes the most sense. Remember you're looking for the primitive 5th root of 1. So consider the unit circle and think about which root is close to the real value you are looking for.

    Hope that's helpful
    Steven
     
  5. Jun 11, 2005 #4

    Hurkyl

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    Nifty; that sounds easier than doing some cleverness with a regular pentagon, which is how I figured out I could figure this out!
     
  6. Jun 12, 2005 #5
    Okay, I used your approach snoble, and solved it. As for the pentagon, thought about that too but got stuck there.

    Thanks all!
     
  7. Jun 12, 2005 #6

    Hurkyl

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    Similar triangles are your friend!
     
  8. Jun 12, 2005 #7
    euler's formula:

    [tex]e^{i x} = cos x + i sin x[/tex]
     
  9. Jun 12, 2005 #8

    dextercioby

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    Unless he started with it,he'll find it pretty useless.I'd say Hurkyl's geometrical method rocks...

    Daniel.
     
  10. Jun 12, 2005 #9
    Certainly the geometric method is the most straight forward but I didn't suggest it because the question asked about using complex numbers. The geometry of the trig functions can be seen while completely ignoring complex numbers. When it comes to student exercises clearly the process is more important than the result. Yes it is important to understand the geometry of complex numbers but that does not seem to be the point of this exercise.
     
  11. Jun 14, 2005 #10
    That formula arrives at a later chapter in the book, so strictly speaken I can't use it. :smile:

    And my geometry isn't what it used to be. 2 angles of 54° and one of 72°..don't really see how to calculate cos and sin of all this without a calculator and just the drawing, but that's ok. :wink:

    At the moment I'm studying algebra and had to review the subject of complex numbers to be able to understand complex matrices, so geometry isn't really a priority right now.
    I know that geometry is useful with algebra, but I don't think I'll need it on the exam..I hope.

    Thanks again.
     
  12. Jun 14, 2005 #11

    Hurkyl

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    The more I learn, the more I get the impression that geometry and algebra are inseparable. :biggrin:

    With the geometry problem, what you want to draw are the diagonals... not the lines joining the center to the vertices. You'll get a lot of similar triangles that way.
     
  13. Jun 20, 2005 #12

    lurflurf

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    First know you can do this either by drawing triangles or by expanding sin(5x) and that useing complex numbers (while fun) is not needed. Your observation is the same as the fact that z^5=1 has roots whose real and imaginary parts are cosines and sines of 72°,144°,216°,288°,360°. z^5=1 is hard to solve so let z=x+y i
    (x+y i)^5=1
    expand the left side. You will notice that the real and imaginary parts look similar with x and y interchanged. Equate real and imaginary parts.
    Re((x+y i)^5)=1
    Im((x+y i)^5=0
    Each part can be made into a one veriable equation using x^2+y^2=1.
    Solve for x or y using whichever equation is easier to solve (hint: one is much harder than the other.)
    You will get 5 values. (or two if you use 0<x,y<1 for values you want)
    Think about which is which one you want.
    Or you can check values in the original equation.
    use x^2+y^2=1 to get the other value.
     
  14. Jun 20, 2005 #13
    Well lurflurf, that's exactly what I did. From the 5 values, one is zero and 2 are negative (if I remember correctly). That leaves us with 2 positive values for both x and y. The way to get the right one is to substitute x or y (can't remember which one) into the other equation. If both sides are not equal, then that's the wrong one. :smile:
     
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