# FP1 HelpWatch

#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?

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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.
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9 years ago
#3
Update: it does work. You need to take out an "obvious"(!) factor from the resultant quintic. It's not too hard to spot.
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9 years ago
#4
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