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Prove tthat tan 50 * tan 60 * tan 70 = tan 80

  1. Feb 11, 2005 #1
    hiiiiiiiiiii
    i am here with a very tough question
    prove tthat
    tan 50 * tan 60 * tan 70 = tan 80
    cheers
    abc
     
  2. jcsd
  3. Feb 11, 2005 #2
    Are you sure I get
    50*Tan[60]*Tan[70] = 19.538
    Tan[80]=9.00365
    If in radians and the below if in degrees:
    50*Tan[60]*Tan[70] = 237.939
    Tan[80]=5.67128

    A play on words maybe I am not catching it anyone else?

    Edit: oops, Thanks gerben.
     
    Last edited: Feb 11, 2005
  4. Feb 11, 2005 #3
    tan 50 * tan 60 * tan 70 = tan 80 (in degrees)
    Not
    50 * tan 60 * tan 70 = tan 80
     
  5. Feb 11, 2005 #4
    Last edited by a moderator: Feb 11, 2005
  6. Feb 11, 2005 #5

    AKG

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    Homework Helper

    Icebreaker, you simply show that the quantities are equal to within whatever error Google's calculator computes things. I think the proof should involve the use of the following identity (with x = 10) :

    [tex]\tan (nx) = \frac{\tan [(n - 1)x] + \tan (x)}{1 - \tan [(n - 1)x]\tan (x)}[/tex]
     
  7. Feb 12, 2005 #6
    We know tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))

    So tan(60+x)=(tan(60)+tan(x))/(1-tan(60)tan(x)) -- all numbers in degrees

    with a corresponding formula for tan(60-x). Multiplying the two, using tan(60)^2=3 and putting t=tan(x)

    tan(60-x)tan(60+x)=(3+t^2)/(1-3t^2)

    Now putting x=10 degrees, and so t=tan(10)
    we have tan(30)=(3t-t^3)/(1-3t^2) and tan(60)=1/tan(30) so

    tan(50)tan(60)tan(70)=[(3+t^2)/(1-3^t^2)]/[(3t-t^3)/(1-3t^2)]=1/t
    =1/tan(10)=tan(80)
     
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