Stuck on a trigonometry question

The question in the OP is wrong. In summary, the post presents a conversation about finding the exact value of tan31°43', which is shown to be equal to \frac{\sqrt{5}-1}{2}. However, it is pointed out that the question is incorrect and the correct question should be finding the tangent of half of an angle with tangent 2.
  • #1
Mishi
7
0

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

ehild
 
Last edited:
  • #6
So the conclusion is that Mishi posted the wrong question?
 
  • #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]
 
  • #8
Mishi said:
[itex]Tan63°26' = 2 [/itex]

And how did you show that?? It isn't true as any calculator will show.
 
  • #9
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 ...
 

1. What is trigonometry?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is used to solve problems involving right triangles, as well as to model and analyze periodic phenomena.

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Trigonometry can be challenging for students because it involves complex concepts and formulas that require a strong foundation in algebra and geometry. It also requires a lot of practice and understanding of how to apply the concepts to solve problems.

3. How can I improve my understanding of trigonometry?

To improve your understanding of trigonometry, it is important to review and practice the basic trigonometric functions (sine, cosine, and tangent) and their properties. You should also work on solving a variety of problems to become more familiar with the different types of questions and how to approach them.

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