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Stuck on a trigonometry question

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the exact value of tan°43' is [itex]\frac{\sqrt{5}-1}{2}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]tan2x = \frac{2tanx}{1-tan^{2}x}[/itex]
    [itex]\frac{2tan31° 43}{1-tan^{2}31° 43'}[/itex]



    From here, i got stuck or I'm doing it wrongly, i forgot how to do it.

    Btw, I'm new here, please tell me if i posted it in the wrong place xD
     
  2. jcsd
  3. Jan 20, 2012 #2

    HallsofIvy

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    I don't know what you mean by "°43' ". Later you use 31° 43'. Is that what you meant? In any case, 2*(31° 43')= 63° 23' which does not have a simple tangent so I don't think you will have accomplished anything. However, I think that your real prioblem is that what you are trying to prove simply isn't true!

    The tangent of 31° 43' is 0.6180145062306733743600526613162 while [itex](\sqrt{5}- 1)/2p[/itex] is 0.61803398874989484820458683436564. They different after the first four decimal places
     
  4. Jan 20, 2012 #3
    woops, sorry typo, tan31° 43' , i think that its possible, but i just forgot how to do it.
     
  5. Jan 20, 2012 #4

    micromass

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    No, the question in your OP is incorrect. As Halls noticed.
     
  6. Jan 20, 2012 #5

    ehild

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    [tex]\frac{\sqrt{5}-1}{2}=\tan(0.5\arctan(2))[/tex] - this is exact.

    ehild
     
    Last edited: Jan 20, 2012
  7. Jan 20, 2012 #6

    NascentOxygen

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    So the conclusion is that Mishi posted the wrong question?
     
  8. Jan 20, 2012 #7
    No, my question is correct:
    Show that the exact value of tan31°43' is [itex]\frac{\sqrt{5}-1}{2}[/itex]
    I got the solution to it.

    [itex]Tan63°26' = 2 [/itex]
    [itex]tanα[/itex] = [itex]\frac{2tanα}{1-tan^{2}α}= 2[/itex]
    where (where α=31°43' (acute angle))
    [itex]2t = 2-2t^{2}[/itex]
    [itex]t^{2}+t-1=0 , t>0 [/itex]
    [itex]t= \frac{-1+\sqrt{5}-1}{2\times1} = \frac{\sqrt{5}-1}{2}[/itex]
     
  9. Jan 20, 2012 #8

    micromass

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    And how did you show that?? It isn't true as any calculator will show.
     
  10. Jan 20, 2012 #9
    Apparently it does tan63°26' = 1.99985903 which equates to 2.
     
  11. Jan 20, 2012 #10

    micromass

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    .....

    Since when is 1.99985903 equal to 2?? You have some weird definition of equality...
     
  12. Jan 20, 2012 #11
    I mean rounded off my bad sorry.
     
  13. Jan 20, 2012 #12

    micromass

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    Your OP talked about the EXACT value, not rounded values.
     
  14. Jan 20, 2012 #13
    Exact values for the final answer. I don't know, its my tutors question and answer, sorry if it was misleading.
     
  15. Jan 20, 2012 #14

    micromass

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    Was the question in the OP the exact question your tutor asked?? In that case: find a better tutor.
     
  16. Jan 20, 2012 #15
    Yes, it was exactly..
     
  17. Jan 20, 2012 #16

    NascentOxygen

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    So the correct question should have been: Given an angle having tangent = 2, show that the tangent of half of this angle is given by the expression ....
     
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