FP1 Help Watch

azamally
Badges: 0
Rep:
?
#1
Report Thread starter 9 years ago
#1
Hey

I was given this problem in class while studying complex numbers but don't know where to start.

Prove that tan(pi/15) is a root of:
(t^4) - (6sqrt3)t^3 + 8(t^2) + (2sqrt3)t - 1 = 0

Any suggestions?

Sorry about the format. Thanks in advance.
0
quote
reply
generalebriety
Badges: 14
Rep:
?
#2
Report 9 years ago
#2
I haven't yet tried this, but my initial thought was the following. I don't know tan(pi/15), but I do know things like tan(pi/6), tan(pi/3), tan(pi/2). I also noticed that pi/3 was five times pi/15, i.e. tan(pi/3) = tan(5 * pi/15); moreover, initially I didn't have a clue where those sqrt(3) things were coming from, but now notice that tan(pi/3) = sqrt(3), and so tan(5 * pi/15) = sqrt(3).

So here's my idea, which I haven't checked, but I think will work: expand tan(5A) in terms of s = tan(A) (using complex numbers or whatever), and then put A = pi/15. Then just note that after rearrangement you show that s is a root of that equation you posted.
0
quote
reply
generalebriety
Badges: 14
Rep:
?
#3
Report 9 years ago
#3
Update: it does work. :p: You need to take out an "obvious"(!) factor from the resultant quintic. It's not too hard to spot.
0
quote
reply
DFranklin
Badges: 18
Rep:
?
#4
Report 9 years ago
#4
See also http://www.thestudentroom.co.uk/showthread.php?t=823453
0
quote
reply
X

Reply to thread

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18

Do you like exams?

Yes (205)
18.82%
No (661)
60.7%
Not really bothered about them (223)
20.48%

Watched Threads

View All