# Stuck on a trigonometry question

## Homework Statement

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

## 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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HallsofIvy
Homework Helper
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.

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

ehild

Last edited:
NascentOxygen
Staff Emeritus
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))
$2t = 2-2t^{2}$
$t^{2}+t-1=0 , t>0$
$t= \frac{-1+\sqrt{5}-1}{2\times1} = \frac{\sqrt{5}-1}{2}$

$Tan63°26' = 2$
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.

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...

I mean rounded off my bad sorry.

I mean rounded off my bad sorry.

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

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.

Yes, it was exactly..

NascentOxygen
Staff Emeritus