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Homework Help: Product=1 homework question

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    My calculator shows tan 6. tan 42. tan 66. tan 78 =1
    Ho is this possible?

    2. Relevant equations

    tan (90-x)=cot x

    3. The attempt at a solution
    but these angles are not supplementary. Ho is this possible?
     
  2. jcsd
  3. May 4, 2008 #2

    symbolipoint

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    What were you expecting?

    [tex] \[
    (.1051)(.9004)(2.246)(4.7046) = 0.99993...
    \]
    [/tex]

    That seems close enough to call "1".
     
  4. May 4, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi xphloem! Welcome to PF! :smile:

    How did you come across this? :smile:

    If we take inverses, we can rewrite this as tan 12 tan 24 tan 48 tan 96 = 1.

    So I suspect it has something to do with a regular pentagon, whose interior angles are 108º, and whose exterior angles are 72º.

    But that's as far as I've got! :rolleyes:
     
  5. May 5, 2008 #4
    It's actually 12 tan 24 tan 48 tan 96 = -1

    I've been able to prove cos 12 cos 24 cos 48 cos 96 = -1/16

    sin 12 sin 24 sin 48 sin 96 is equal to 1/16 but I can only prove this with a calculator :grumpy:

    sin(24) = 2sin(12)cos(12)

    sin(48) = 2sin(24)cos(24) = 4sin(12)cos(12)cos(24)

    sin(96) = 2sin(48)cos(48) = 8sin(12)cos(12)cos(24)cos(48)

    sin(192) = -sin(12) = 2sin(96)cos(96) = 16sin(12)cos(12)cos(24)cos(48)cos(96)

    divide by 16 sin(12) to get:

    cos 12 cos 24 cos 48 cos 96 = -1/16
     
  6. May 5, 2008 #5
    thanks all. I got this in a book called brain teasers. I have solved it myself. thanks for all the help!
     
  7. May 5, 2008 #6
    so how did you solve it? any hints?
     
  8. May 6, 2008 #7

    tiny-tim

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    ooh, kamerling, that's clever! :smile:

    Your splitting the problem into cos and sin products has given me the following idea:

    tanAtanBtanCtanD = -1
    iff cosAcosBcosCcosD + sinAsinBsinCsinD = 0​

    But (4cosAcosBcosCcosD + sinAsinBsinCsinD)

    = [cos(A+B) + cos(A-B)][cos(C+D) + cos(C-D)] + [cos(A+B) - cos(A-B)][cos(C+D) - cos(C-D)]

    = 2[cos(A+B)cos(C+D) + cos(A-B)cos(C-D)]

    = cos(A+B+C+D) + cos(-A-B+C+D) + cos(A-B-C+D) + cos(-A+B-C+D)

    (putting E = A + B + C + D)

    = cosE + cos(E - 2(A+B)) + cos(E - 2(B+C)) + cos(E - 2(C+A)).

    So, in particular, tanAtanBtanCtanD = -1 if A+B+C+D = 180º and:
    cos2(B+C) + cos2(C+A) + cos2(A+B) = -1;​

    or if A+B+C+D = 90º and:
    sin2(B+C) + sin2(C+A) + sin2(A+B)) = 0.​

    (For example, 12º 24º 48º and 96º, because cos72º + cos120º + cos144º = cos72º - cos60º - cos36º = -1)​

    can anyone find a simpler proof
    or some geometric explanation for this? :smile:
     
  9. May 6, 2008 #8
    I got this from the following equations:
    2 sin A sin B= cos (A-B)-cos(A+B)
    2 cos A cos B=cos (A-B)+cos(A+B)

    Divide the (i) by (ii)
    to obtain the result :)
     
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