Recent content by Mishi

Yes, it was exactly..

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

I mean rounded off my bad sorry.

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

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} =...

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

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...