Stuck on a trigonometry question

  • Thread starter Mishi
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  • #1
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Homework Statement


Show that the exact value of tan°43' is [itex]\frac{\sqrt{5}-1}{2}[/itex]


Homework Equations





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
 

Answers and Replies

  • #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
 
  • #3
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woops, sorry typo, tan31° 43' , i think that its possible, but i just forgot how to do it.
 
  • #4
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No, the question in your OP is incorrect. As Halls noticed.
 
  • #5
ehild
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[tex]\frac{\sqrt{5}-1}{2}=\tan(0.5\arctan(2))[/tex] - this is exact.

ehild
 
Last edited:
  • #6
NascentOxygen
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So the conclusion is that Mishi posted the wrong question?
 
  • #7
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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]
 
  • #8
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[itex]Tan63°26' = 2 [/itex]
And how did you show that?? It isn't true as any calculator will show.
 
  • #9
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Apparently it does tan63°26' = 1.99985903 which equates to 2.
 
  • #10
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Apparently it does tan63°26' = 1.99985903 which equates to 2.
.....

Since when is 1.99985903 equal to 2?? You have some weird definition of equality...
 
  • #11
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I mean rounded off my bad sorry.
 
  • #12
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I mean rounded off my bad sorry.
Your OP talked about the EXACT value, not rounded values.
 
  • #13
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Exact values for the final answer. I don't know, its my tutors question and answer, sorry if it was misleading.
 
  • #14
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Exact values for the final answer. I don't know, its my tutors question and answer, sorry if it was misleading.
Was the question in the OP the exact question your tutor asked?? In that case: find a better tutor.
 
  • #15
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Yes, it was exactly..
 
  • #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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