Stuck on a trigonometry question

1. Jan 20, 2012

Mishi

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

2. Relevant equations

3. 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

2. Jan 20, 2012

HallsofIvy

Staff Emeritus
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

3. Jan 20, 2012

Mishi

woops, sorry typo, tan31° 43' , i think that its possible, but i just forgot how to do it.

4. Jan 20, 2012

micromass

Staff Emeritus
No, the question in your OP is incorrect. As Halls noticed.

5. Jan 20, 2012

ehild

$$\frac{\sqrt{5}-1}{2}=\tan(0.5\arctan(2))$$ - this is exact.

ehild

Last edited: Jan 20, 2012
6. Jan 20, 2012

Staff: Mentor

So the conclusion is that Mishi posted the wrong question?

7. Jan 20, 2012

Mishi

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))
$2t = 2-2t^{2}$
$t^{2}+t-1=0 , t>0$
$t= \frac{-1+\sqrt{5}-1}{2\times1} = \frac{\sqrt{5}-1}{2}$

8. Jan 20, 2012

micromass

Staff Emeritus
And how did you show that?? It isn't true as any calculator will show.

9. Jan 20, 2012

Mishi

Apparently it does tan63°26' = 1.99985903 which equates to 2.

10. Jan 20, 2012

micromass

Staff Emeritus
.....

Since when is 1.99985903 equal to 2?? You have some weird definition of equality...

11. Jan 20, 2012

Mishi

I mean rounded off my bad sorry.

12. Jan 20, 2012

micromass

Staff Emeritus

13. Jan 20, 2012

Mishi

Exact values for the final answer. I don't know, its my tutors question and answer, sorry if it was misleading.

14. Jan 20, 2012

micromass

Staff Emeritus
Was the question in the OP the exact question your tutor asked?? In that case: find a better tutor.

15. Jan 20, 2012

Mishi

Yes, it was exactly..

16. Jan 20, 2012

Staff: Mentor

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 ....