Is Tan 43° Exactly Equal to (sqrt(5) - 1)/2?

AI Thread Summary
The discussion centers on whether tan 43° is exactly equal to (sqrt(5) - 1)/2. Participants analyze the mathematical approach, noting that the tangent of 31° 43' is approximately 0.618, while (sqrt(5) - 1)/2 is approximately 0.61803, indicating they are not equal. There is confusion regarding the correct interpretation of angles and the calculations involved, with some participants suggesting the original question may be flawed. Ultimately, the consensus is that the question posed was misleading, and a better formulation would clarify the relationship between the tangent of an angle and its half. The conversation highlights the importance of precision in mathematical expressions and problem statements.
Mishi
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Homework Statement


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


Homework Equations





The Attempt at a Solution


tan2x = \frac{2tanx}{1-tan^{2}x}
\frac{2tan31° 43}{1-tan^{2}31° 43'}



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
 
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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 (\sqrt{5}- 1)/2p is 0.61803398874989484820458683436564. They different after the first four decimal places
 
woops, sorry typo, tan31° 43' , i think that its possible, but i just forgot how to do it.
 
No, the question in your OP is incorrect. As Halls noticed.
 
\frac{\sqrt{5}-1}{2}=\tan(0.5\arctan(2)) - this is exact.

ehild
 
Last edited:
So the conclusion is that Mishi posted the wrong question?
 
No, my question is correct:
Show that the exact value of tan31°43' is \frac{\sqrt{5}-1}{2}
I got the solution to it.

Tan63°26' = 2
tanα = \frac{2tanα}{1-tan^{2}α}= 2
where (where α=31°43' (acute angle))
2t = 2-2t^{2}
t^{2}+t-1=0 , t>0
t= \frac{-1+\sqrt{5}-1}{2\times1} = \frac{\sqrt{5}-1}{2}
 
Mishi said:
Tan63°26' = 2

And how did you show that?? It isn't true as any calculator will show.
 
Apparently it does tan63°26' = 1.99985903 which equates to 2.
 
  • #10
Mishi said:
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
I mean rounded off my bad sorry.
 
  • #12
Mishi said:
I mean rounded off my bad sorry.

Your OP talked about the EXACT value, not rounded values.
 
  • #13
Exact values for the final answer. I don't know, its my tutors question and answer, sorry if it was misleading.
 
  • #14
Mishi said:
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
Yes, it was exactly..
 
  • #16
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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